1 Introduction
1.1 Classical results of Lucas and Kummer
It is a well-known result of Lucas [Reference LucasLuc78] that, for all nonnegative integers
$m,n$
and all primes
$p$
, we have
$$\begin{eqnarray}\binom{m}{n}\equiv \mathop{\prod }_{i=0}^{k}\binom{m_{i}}{n_{i}}\hspace{0.6em}{\rm mod}\hspace{0.2em}p,\end{eqnarray}$$
where
$m=m_{0}+m_{1}p+\cdots +m_{k}p^{k}$
and
$n=n_{0}+n_{1}p+\cdots +n_{k}p^{k}$
are the base-
$p$
expansions of
$m$
and
$n$
.
In particular, a prime
$p$
divides the binomial
$\binom{m}{n}$
if and only if there is
$0\leqslant i\leqslant k$
such that
$m_{i}<n_{i}$
. Precisely, Kummer proved in [Reference KummerKum52] that, for all nonnegative integers
$m\geqslant n$
, the
$p$
-adic valuationFootnote
1
of the binomial
$\binom{m}{n}$
is the number of carries which occur when
$n$
is added to
$m-n$
in base
$p$
. As a consequence, we have
$$\begin{eqnarray}\binom{m}{n}\in p^{\unicode[STIX]{x1D6FC}}\mathbb{Z},\quad \text{where }\unicode[STIX]{x1D6FC}=\#\left\{0\leqslant i\leqslant k:\binom{m_{i}}{n_{i}}=0\right\}.\end{eqnarray}$$
In this article, we show that many sequences
$(A(n))_{n\geqslant 0}$
of Apéry-like numbers satisfy congruences similar to (1.1), that is, for all nonnegative integers
$n$
and all primes
$p$
, we have
$$\begin{eqnarray}A(n)\equiv \mathop{\prod }_{i=0}^{k}A(n_{i})\hspace{0.6em}{\rm mod}\hspace{0.2em}p,\end{eqnarray}$$
where
$n=n_{0}+n_{1}p+\cdots +n_{k}p^{k}$
is the base-
$p$
expansion of
$n$
. Furthermore, we prove that an analogue of (1.2) holds for those numbers, that is,
$$\begin{eqnarray}A(n)\in p^{\unicode[STIX]{x1D6FC}}\mathbb{Z},\quad \text{where }\unicode[STIX]{x1D6FC}=\#\{0\leqslant i\leqslant k:A(n_{i})\equiv 0\;\text{mod}\;p\},\end{eqnarray}$$
which proves Beukers’ conjectures on the
$p$
-adic valuation of Apéry numbers.
1.2 Beukers’ conjectures on Apéry numbers
For all nonnegative integers
$n$
, we set
$$\begin{eqnarray}A_{1}(n):=\mathop{\sum }_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}\quad \text{and}\quad A_{2}(n):=\mathop{\sum }_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}.\end{eqnarray}$$
These sequences were used in
$1979$
by Apéry in his proofs of the irrationality of
$\unicode[STIX]{x1D701}(3)$
and
$\unicode[STIX]{x1D701}(2)$
(see [Reference ApéryApé79]). In the
$1980$
s, several congruences satisfied by these sequences were proved (see, for example, [Reference BeukersBeu85, Reference BeukersBeu87, Reference Chowla, Cowles and CowlesCCC80, Reference GesselGes82, Reference MimuraMim83]). In particular, Gessel proved in [Reference GesselGes82] that
$A_{1}$
satisfies the
$p$
-Lucas property for all prime numbers
$p$
, that is, for any prime
$p$
, all
$v$
in
$\{0,\ldots ,p-1\}$
and all nonnegative integers
$n$
, we have
$$\begin{eqnarray}A_{1}(v+np)\equiv A_{1}(v)A_{1}(n)\hspace{0.6em}{\rm mod}\hspace{0.2em}p.\end{eqnarray}$$
Thereby, if
$n=n_{0}+n_{1}p+\cdots +n_{N}p^{N}$
is the base-
$p$
expansion of
$n$
, then we obtain
$$\begin{eqnarray}A_{1}(n)\equiv A_{1}(n_{0})\cdots A_{1}(n_{N})\hspace{0.6em}{\rm mod}\hspace{0.2em}p.\end{eqnarray}$$
In particular,
$p$
divides
$A_{1}(n)$
if and only if there exists
$k$
in
$\{0,\ldots ,N\}$
such that
$p$
divides
$A_{1}(n_{k})$
. Beukers stated in [Reference BeukersBeu86] two conjectures, when
$p=5$
or
$11$
, which generalize this property.Footnote
2
Before stating these conjectures, we observe that the set of all
$v$
in
$\{0,\ldots ,4\}$
(respectively
$v$
in
$\{0,\ldots ,10\}$
) satisfying
$A_{1}(v)\equiv 0\;\text{mod}\;5$
(respectively
$A_{1}(v)\equiv 0\;\text{mod}\;11$
) is
$\{1,3\}$
(respectively
$\{5\}$
).
Conjecture A (Beukers [Reference BeukersBeu86]).
Let
$n$
be a nonnegative integer whose base-
$5$
expansion is
$n=n_{0}+n_{1}5+\cdots +n_{N}5^{N}$
. Let
$\unicode[STIX]{x1D6FC}$
be the number of
$k$
in
$\{0,\ldots ,N\}$
such that
$n_{k}=1$
or
$3$
. Then
$5^{\unicode[STIX]{x1D6FC}}$
divides
$A_{1}(n)$
.
Conjecture B (Beukers [Reference BeukersBeu86]).
Let
$n$
be a nonnegative integer whose base-
$11$
expansion is
$n=n_{0}+n_{1}11+\cdots +n_{N}11^{N}$
. Let
$\unicode[STIX]{x1D6FC}$
be the number of
$k$
in
$\{0,\ldots ,N\}$
such that
$n_{k}=5$
. Then
$11^{\unicode[STIX]{x1D6FC}}$
divides
$A_{1}(n)$
.
Similarly, Sequence
$A_{2}$
satisfies the
$p$
-Lucas property for all primes
$p$
. Furthermore, Beukers and Stienstra proved in [Reference Beukers and StienstraBS85] that, if
$p\equiv 3\;\text{mod}\;4$
, then
$A_{2}((p-1)/2)\equiv 0\;\text{mod}\;p$
, and Beukers stated in [Reference BeukersBeu86] the following conjecture.
Conjecture C. Let
$p$
be a prime number satisfying
$p\equiv 3\;\text{mod}\;4$
. Let
$n$
be a nonnegative integer whose base-
$p$
expansion is
$n=n_{0}+n_{1}p+\cdots +n_{N}p^{N}$
. Let
$\unicode[STIX]{x1D6FC}$
be the number of
$k$
in
$\{0,\ldots ,N\}$
such that
$n_{k}=(p-1)/2$
. Then
$p^{\unicode[STIX]{x1D6FC}}$
divides
$A_{2}(n)$
.
Conjectures A–C have been extended to generalized Apéry numbers and any prime
$p$
by Deutsch and Sagan in [Reference Deutsch and SaganDS06, Conjecture 5.13] but this extension is false for at least one generalization of Apéry numbers. Indeed, a counterexample is given by
$$\begin{eqnarray}A(n)=\mathop{\sum }_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{3},\end{eqnarray}$$
since
$A(1)=9\equiv 0\;\text{mod}\;3$
but
$A(4)=A(1+3)=1152501$
is not divisible by
$3^{2}$
.
The main aim of this article is to prove Theorem 1, stated in § 1.4, which confirms and generalizes Conjectures A–C. First, we introduce some notations which we use throughout this article.
1.3 Notations
In order to study arithmetic properties of sums of products of binomial coefficients, such as Apéry numbers, we first study families, indexed by
$\mathbb{N}^{d}$
, of ratios of factorials of linear forms with integer coefficients. For example, we will obtain congruences for
$A_{1}(n)$
by studying the factorial ratios
$$\begin{eqnarray}\frac{(2n_{1}+n_{2})!^{2}}{n_{1}!^{4}n_{2}!^{2}}\quad (n_{1},n_{2}\in \mathbb{N}),\end{eqnarray}$$
as we have the useful formula
$$\begin{eqnarray}A_{1}(n)=\mathop{\sum }_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}=\mathop{\sum }_{n_{1}+n_{2}=n}\frac{(2n_{1}+n_{2})!^{2}}{n_{1}!^{4}n_{2}!^{2}}.\end{eqnarray}$$
Let
$d$
be a positive integer. Given tuples of vectors in
$\mathbb{N}^{d}$
,
$e=(\mathbf{e}_{1},\ldots ,\mathbf{e}_{u})$
and
$f=(\mathbf{f}_{1},\ldots ,\mathbf{f}_{v})$
, we shall prove congruences for the factorial ratios
$$\begin{eqnarray}{\mathcal{Q}}_{e,f}(\mathbf{n}):=\frac{\mathop{\prod }_{i=1}^{u}(\mathbf{e}_{i}\cdot \mathbf{n})!}{\mathop{\prod }_{i=1}^{v}(\mathbf{f}_{i}\cdot \mathbf{n})!}\quad (\mathbf{n}\in \mathbb{N}^{d})\end{eqnarray}$$
to deduce arithmetic properties of the numbersFootnote 3
$$\begin{eqnarray}\mathfrak{S}_{e,f}(n):=\mathop{\sum }_{\mathbf{n}\in \mathbb{N}^{d},|\mathbf{n}|=n}{\mathcal{Q}}_{e,f}(\mathbf{n})\quad (n\in \mathbb{N}).\end{eqnarray}$$
Here
$\cdot$
denotes the standard scalar product on
$\mathbb{R}^{d}$
and
$|\mathbf{n}|=n_{1}+\cdots +n_{d}$
if
$\mathbf{n}=(n_{1},\ldots ,n_{d})$
. For example, we obtain that
$\mathfrak{S}_{e,f}(n)=A_{1}(n)$
with the tuples
$$\begin{eqnarray}e=((2,1),(2,1))\quad \text{and}\quad f=((1,0),(1,0),(1,0),(1,0),(0,1),(0,1)).\end{eqnarray}$$
Because of the summation in (1.4), it is usually difficult to study arithmetic properties of
$\mathfrak{S}_{e,f}(n)$
; however, we will show that, in many interesting cases, we can transfer the
$p$
-Lucas property from
${\mathcal{Q}}_{e,f}(\mathbf{n})$
to
$\mathfrak{S}_{e,f}(n)$
. To that purpose, we define the
$p$
-Lucas property for families of
$p$
-adic integers indexed by
$\mathbb{N}^{d}$
. For all primes
$p$
, we write
$\mathbb{Z}_{p}$
for the ring of
$p$
-adic integers.
If
$A=(A(\mathbf{n}))_{\mathbf{n}\in \mathbb{N}^{d}}$
is a
$\mathbb{Z}_{p}$
-valued family, then we say that
$A$
satisfies the
$p$
-Lucas property if, for all vectors
$\mathbf{v}$
in
$\{0,\ldots ,p-1\}^{d}$
and
$\mathbf{n}$
in
$\mathbb{N}^{d}$
, we have
$$\begin{eqnarray}A(\mathbf{v}+\mathbf{n}p)\equiv A(\mathbf{v})A(\mathbf{n})\hspace{0.6em}{\rm mod}\hspace{0.2em}p\mathbb{Z}_{p}.\end{eqnarray}$$
If
$\mathbf{n}$
is nonzero, then we say that
$\mathbf{n}=\mathbf{n}_{0}+\mathbf{n}_{1}p+\cdots +\mathbf{n}_{N}p^{N}$
is the base-
$p$
expansion of
$\mathbf{n}$
if, for all
$i$
in
$\{0,\ldots ,N\}$
, we have
$\mathbf{n}_{i}\in \{0,\ldots ,p-1\}^{d}$
and
$\mathbf{n}_{N}\neq \mathbf{0}$
, where
$\mathbf{0}:=(0,\ldots ,0)$
. Hence, if
$A$
satisfies the
$p$
-Lucas property, then we have
$$\begin{eqnarray}A(\mathbf{n})\equiv A(\mathbf{n}_{0})\cdots A(\mathbf{n}_{N})\hspace{0.6em}{\rm mod}\hspace{0.2em}p\mathbb{Z}_{p}.\end{eqnarray}$$
We write
${\mathcal{Z}}_{p}(A)$
for the set of all vectors
$\mathbf{v}$
in
$\{0,\ldots ,p-1\}^{d}$
such that
$A(\mathbf{v})$
belongs to
$p\mathbb{Z}_{p}$
. Hence,
$A(\mathbf{n})$
is in
$p\mathbb{Z}_{p}$
if and only if at least one
$\mathbf{n}_{i}$
,
$0\leqslant i\leqslant N$
, belongs to
${\mathcal{Z}}_{p}(A)$
. To state our generalization of Conjectures A–C, we define the following counting function. For every nonzero vector
$\mathbf{n}$
in
$\mathbb{N}^{d}$
whose base-
$p$
expansion is
$\mathbf{n}=\mathbf{n}_{0}+\mathbf{n}_{1}p+\cdots +\mathbf{n}_{N}p^{N}$
, we write
$\unicode[STIX]{x1D6FC}_{p}(A,\mathbf{n})$
for the number of
$i$
in
$\{0,\ldots ,N\}$
such that
$\mathbf{n}_{i}\in {\mathcal{Z}}_{p}(A)$
, and we set
$\unicode[STIX]{x1D6FC}_{p}(A,\mathbf{0})=0$
. Thereby, to prove Conjectures A–C, it is enough to show that
$A_{i}(n)\in p^{\unicode[STIX]{x1D6FC}_{p}(A_{i},n)}\mathbb{Z}$
with
$i=1,p=5$
or
$11$
and
$i=2$
,
$p\equiv 3\;\text{mod}\;4$
.
Our generalization of Beukers’ conjectures will apply to sequences
$\mathfrak{S}_{e,f}$
restricted to the following two conditions.
The first condition (the
$r$
-admissibility) ensures that we can transfer the
$p$
-Lucas property from
${\mathcal{Q}}_{e,f}(\mathbf{n})$
to
$\mathfrak{S}_{e,f}(n)$
. If
$\mathbf{m}=(m_{1},\ldots ,m_{d})$
and
$\mathbf{n}=(n_{1},\ldots ,n_{d})$
belong to
$\mathbb{R}^{d}$
, then we write
$\mathbf{m}\geqslant \mathbf{n}$
if, for all
$i$
in
$\{1,\ldots ,d\}$
, we have
$m_{i}\geqslant n_{i}$
. Furthermore, we set
$\mathbf{1}:=(1,\ldots ,1)\in \mathbb{N}^{d}$
and we write
$\mathbf{1}_{k}$
for the vector in
$\mathbb{N}^{d}$
, all of whose coordinates equal zero except the
$k$
th, which is
$1$
. Let
${\mathcal{S}}:=\{1\leqslant i\leqslant u:\mathbf{e}_{i}\geqslant \mathbf{1}\}$
. For every positive integer
$r$
, we say that
$e$
is
$r$
-admissible if
$$\begin{eqnarray}\#{\mathcal{S}}+\min _{1\leqslant k\leqslant d}\#\{1\leqslant i\leqslant u:i\notin {\mathcal{S}}\text{ and }\mathbf{e}_{i}\geqslant d\mathbf{1}_{k}\}\geqslant r.\end{eqnarray}$$
We will use this definition with
$r=1$
or
$2$
. In the case of the Apéry numbers
$A_{1}(n)$
, we study the family
${\mathcal{Q}}_{e,f}$
with the tuple
$e=((2,1),(2,1))$
, so that
$\#{\mathcal{S}}=2$
and
$e$
is
$2$
-admissible. As another example, we will also prove a result similar to Beukers’ conjectures for the sequence
$$\begin{eqnarray}A_{6}(n):=\mathop{\sum }_{k=0}^{n}\binom{n}{k}\binom{2k}{k}\binom{2(n-k)}{n-k}.\end{eqnarray}$$
We can write
$$\begin{eqnarray}A_{6}(n)=\mathop{\sum }_{n_{1}+n_{2}=n}\frac{(n_{1}+n_{2})!(2n_{1})!(2n_{2})!}{n_{1}!^{3}n_{2}!^{3}},\end{eqnarray}$$
so that
$A_{6}(n)=\mathfrak{S}_{e,f}(n)$
with
$e=((1,1),(2,0),(0,2))$
. In this case, we have
$d=2$
,
$\#{\mathcal{S}}=1$
but
$e$
is also
$2$
-admissible because for
$k=1$
or
$2$
we have
$\#\{2\leqslant i\leqslant 3:\mathbf{e}_{i}\geqslant 2\mathbf{1}_{k}\}=1$
.
The second condition is of differential type. To apply our main result, we need the generating series of
$(\mathfrak{S}_{e,f}(n))_{n\geqslant 0}$
to be annihilated by a differential operator of a special form that we describe below. We set
$\unicode[STIX]{x1D703}:=z(d/dz)$
and we say that a differential operator
${\mathcal{L}}$
in
$\mathbb{Z}_{p}[z,\unicode[STIX]{x1D703}]$
is of type I if there is a nonnegative integer
$q$
such that:
-
–
${\mathcal{L}}=P_{0}(\unicode[STIX]{x1D703})+zP_{1}(\unicode[STIX]{x1D703})+\cdots +z^{q}P_{q}(\unicode[STIX]{x1D703})$
with
$P_{k}(X)\in \mathbb{Z}_{p}[X]$
for
$0\leqslant k\leqslant q$
; -
–
$P_{0}(\mathbb{Z}_{p}^{\times })\subset \mathbb{Z}_{p}^{\times }$
; -
– for all
$k$
in
$\{2,\ldots ,q\}$
, we have
$P_{k}(X)\in \prod _{i=1}^{k-1}(X+i)^{2}\mathbb{Z}_{p}[X]$
.
We say that a differential operator
${\mathcal{L}}$
in
$\mathbb{Z}_{p}[z,\unicode[STIX]{x1D703}]$
is of type II if:
-
–
${\mathcal{L}}=P_{0}(\unicode[STIX]{x1D703})+zP_{1}(\unicode[STIX]{x1D703})+z^{2}P_{2}(\unicode[STIX]{x1D703})$
with
$P_{k}(X)\in \mathbb{Z}_{p}[X]$
for
$0\leqslant k\leqslant 2$
; -
–
$P_{0}(\mathbb{Z}_{p}^{\times })\subset \mathbb{Z}_{p}^{\times }$
; -
–
$P_{2}(X)\in (X+1)\mathbb{Z}_{p}[X]$
.
For example, the generating series of
$(A_{1}(n))_{n\geqslant 0}$
is annihilated by the differential operator
$$\begin{eqnarray}{\mathcal{L}}_{1}=\unicode[STIX]{x1D703}^{3}-z(34\unicode[STIX]{x1D703}^{3}+51\unicode[STIX]{x1D703}^{2}+27\unicode[STIX]{x1D703}+5)+z^{2}(\unicode[STIX]{x1D703}+1)^{3},\end{eqnarray}$$
which is of type I for every prime
$p$
. We will also prove a result similar to Beukers’ conjectures for the numbers
$$\begin{eqnarray}A_{5}(n)=\mathop{\sum }_{k=0}^{n}\binom{n}{k}^{4}.\end{eqnarray}$$
The generating series of
$A_{5}$
is annihilated by the differential operator
$$\begin{eqnarray}{\mathcal{L}}_{5}=\unicode[STIX]{x1D703}^{3}-z2(2\unicode[STIX]{x1D703}+1)(3\unicode[STIX]{x1D703}^{2}+3\unicode[STIX]{x1D703}+1)-z^{2}4(\unicode[STIX]{x1D703}+1)(4\unicode[STIX]{x1D703}+5)(4\unicode[STIX]{x1D703}+3),\end{eqnarray}$$
which is of type II for every prime
$p$
.
Our main result confirms Conjectures A–C, and also provides surprising similar properties for some deformations of Apéry-like numbers. For example, while proving that, for every prime
$p$
and all nonnegative integers
$n$
, we have
$$\begin{eqnarray}A_{1}(n)=\mathop{\sum }_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}\in p^{\unicode[STIX]{x1D6FC}_{p}(A_{1},n)}\mathbb{Z},\end{eqnarray}$$
we will also show that, for every nonnegative integer
$a$
, we have
$$\begin{eqnarray}\mathop{\sum }_{k=0}^{n}k^{a}\binom{n}{k}^{2}\binom{n+k}{k}^{2}\in p^{\unicode[STIX]{x1D6FC}_{p}(A_{1},n)-1}\mathbb{Z}.\end{eqnarray}$$
More generally, we will obtain congruences for deformations
$\mathfrak{S}_{e,f}^{g}$
of the sequences
$\mathfrak{S}_{e,f}$
defined as follows. For any prime
$p$
, we write
$\mathfrak{F}_{p}^{d}$
for the set of all functions
$g:\mathbb{N}^{d}\rightarrow \mathbb{Z}_{p}$
such that, for all nonnegative integers
$K$
, there exists a sequence
$(P_{K,k})_{k\geqslant 0}$
of polynomial functions with coefficients in
$\mathbb{Z}_{p}$
which converges pointwise to
$g$
on
$\{0,\ldots ,K\}^{d}$
. For all tuples
$e$
and
$f$
of vectors in
$\mathbb{N}^{d}$
, all
$g\in \mathfrak{F}_{p}^{d}$
and all nonnegative integers
$m$
, we set
$$\begin{eqnarray}\mathfrak{S}_{e,f}^{g}(m):=\mathop{\sum }_{\mathbf{n}\in \mathbb{N}^{d},|\mathbf{n}|=m}{\mathcal{Q}}_{e,f}(\mathbf{n})g(\mathbf{n}).\end{eqnarray}$$
1.4 Main results
In the rest of the article, if
$e=(\mathbf{e}_{1},\ldots ,\mathbf{e}_{u})$
is a tuple of vectors in
$\mathbb{N}^{d}$
, then we set
$|e|:=\mathbf{e}_{1}+\cdots +\mathbf{e}_{u}$
. The main result of this article is the following.
Theorem 1. Let
$e$
and
$f=(\mathbf{1}_{k_{1}},\ldots ,\mathbf{1}_{k_{v}})$
be two disjoint tuples of vectors in
$\mathbb{N}^{d}$
such that
$|e|=|f|$
, for all
$i$
in
$\{1,\ldots ,v\}$
,
$k_{i}$
is in
$\{1,\ldots ,d\}$
, and
$e$
is
$2$
-admissible. Let
$p$
be a fixed prime. Assume that the generating series of
$\mathfrak{S}_{e,f}$
is annihilated by a differential operator
${\mathcal{L}}\in \mathbb{Z}_{p}[z,\unicode[STIX]{x1D703}]$
such that at least one of the following conditions holds:
-
–
${\mathcal{L}}$
is of type I; -
–
${\mathcal{L}}$
is of type II and
$p-1\in {\mathcal{Z}}_{p}(\mathfrak{S}_{e,f})$
.
Then, for all nonnegative integers
$n$
and all functions
$g$
in
$\mathfrak{F}_{p}^{d}$
, we have
$$\begin{eqnarray}\mathfrak{S}_{e,f}(n)\in p^{\unicode[STIX]{x1D6FC}_{p}(\mathfrak{S}_{e,f},n)}\mathbb{Z}\quad \text{and}\quad \mathfrak{S}_{e,f}^{g}(n)\in p^{\unicode[STIX]{x1D6FC}_{p}(\mathfrak{S}_{e,f},n)-1}\mathbb{Z}_{p}.\end{eqnarray}$$
In § 1.6, we show that Theorem 1 applies to many classical sequences. In particular, Theorem 1 implies Conjectures A–C. Indeed, we have
$A_{1}=\mathfrak{S}_{e_{1},f_{1}}$
and
$A_{2}=\mathfrak{S}_{e_{2},f_{2}}$
with
$d=2$
,
$$\begin{eqnarray}e_{1}=((2,1),(2,1))\quad \text{and}\quad f_{1}=((1,0),(1,0),(1,0),(1,0),(0,1),(0,1))\end{eqnarray}$$
and
$$\begin{eqnarray}e_{2}=((2,1),(1,1))\quad \text{and}\quad f_{2}=((1,0),(1,0),(1,0),(0,1),(0,1)).\end{eqnarray}$$
Furthermore, it is well known that
$f_{A_{1}}$
, respectively
$f_{A_{2}}$
, is annihilated by the differential operator
${\mathcal{L}}_{1}$
, respectively
${\mathcal{L}}_{2}$
, defined by
$$\begin{eqnarray}{\mathcal{L}}_{1}=\unicode[STIX]{x1D703}^{3}-z(34\unicode[STIX]{x1D703}^{3}+51\unicode[STIX]{x1D703}^{2}+27\unicode[STIX]{x1D703}+5)+z^{2}(\unicode[STIX]{x1D703}+1)^{3}\end{eqnarray}$$
and
$$\begin{eqnarray}{\mathcal{L}}_{2}=\unicode[STIX]{x1D703}^{2}-z(11\unicode[STIX]{x1D703}^{2}+11\unicode[STIX]{x1D703}+3)-z^{2}(\unicode[STIX]{x1D703}+1)^{2}.\end{eqnarray}$$
Since
${\mathcal{L}}_{1}$
and
${\mathcal{L}}_{2}$
are of type I for all primes
$p$
, the conditions of Theorem 1 are satisfied by
$A_{1}$
and
$A_{2}$
, and Conjectures A–C hold. In addition, for all primes
$p$
and all nonnegative integers
$n$
and
$a$
, we obtain that
$$\begin{eqnarray}\mathop{\sum }_{k=0}^{n}k^{a}\binom{n}{k}^{2}\binom{n+k}{k}^{2}\in p^{\unicode[STIX]{x1D6FC}_{p}(A_{1},n)-1}\mathbb{Z}\quad \text{and}\quad \mathop{\sum }_{k=0}^{n}k^{a}\binom{n}{k}^{2}\binom{n+k}{k}\in p^{\unicode[STIX]{x1D6FC}_{p}(A_{2},n)-1}\mathbb{Z}.\end{eqnarray}$$
We provide a similar result which applies to the constant terms of powers of certain Laurent polynomials. Consider a Laurent polynomial
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}(\mathbf{x})=\mathop{\sum }_{i=1}^{k}\unicode[STIX]{x1D6FC}_{i}\mathbf{x}^{\mathbf{a}_{i}}\in \mathbb{Z}_{p}[x_{1}^{\pm },\ldots ,x_{d}^{\pm }],\end{eqnarray}$$
where
$\mathbf{a}_{i}\in \mathbb{Z}^{d}$
and
$\unicode[STIX]{x1D6FC}_{i}\neq 0$
for
$i$
in
$\{1,\ldots ,k\}$
. Recall that the Newton polyhedron of
$\unicode[STIX]{x1D6EC}$
is the convex hull of
$\{\mathbf{a}_{1},\ldots ,\mathbf{a}_{k}\}$
in
$\mathbb{R}^{d}$
. Hence, we have the following result.
Theorem 2. Let
$p$
be a fixed prime. Let
$\unicode[STIX]{x1D6EC}(\mathbf{x})\in \mathbb{Z}_{p}[x_{1}^{\pm },\ldots ,x_{d}^{\pm }]$
be a Laurent polynomial and consider the sequence of the constant terms of powers of
$\unicode[STIX]{x1D6EC}$
defined, for all nonnegative integers
$n$
, by
$$\begin{eqnarray}A(n):=[\unicode[STIX]{x1D6EC}(\mathbf{x})^{n}]_{\boldsymbol{ 0}}.\end{eqnarray}$$
Assume that the Newton polyhedron of
$\unicode[STIX]{x1D6EC}$
contains the origin as its only interior integral point, and that
$f_{A}$
is annihilated by a differential operator
${\mathcal{L}}$
in
$\mathbb{Z}_{p}[z,\unicode[STIX]{x1D703}]$
such that at least one of the following conditions holds:
-
–
${\mathcal{L}}$
is of type I; -
–
${\mathcal{L}}$
is of type II and
$p-1\in {\mathcal{Z}}_{p}(A)$
.
Then, for all nonnegative integers
$n$
, we have
$$\begin{eqnarray}A(n)\in p^{\unicode[STIX]{x1D6FC}_{p}(A,n)}\mathbb{Z}_{p}.\end{eqnarray}$$
For example, Theorem 2 applies to Apéry numbers
$A_{1}$
thanks to the following formula of Lairez (personal communication, 2013):
$$\begin{eqnarray}A_{1}(n)=\biggl[\biggl(\frac{(1+z)(yz+z+1)(1+x)(xy+x+y)}{xyz}\biggr)^{n}\biggr]_{(0,0,0)}.\end{eqnarray}$$
By a result of Samol and van Straten [Reference Samol and van StratenSvS15], if
$\unicode[STIX]{x1D6EC}(\mathbf{x})\in \mathbb{Z}_{p}[x_{1}^{\pm },\ldots ,x_{d}^{\pm }]$
contains the origin as its only interior integral point, then
$([\unicode[STIX]{x1D6EC}(\mathbf{x})^{n}]_{\mathbf{0}})_{n\geqslant 0}$
satisfies the
$p$
-Lucas property, which is essential for the proof of Theorem 2. Likewise, the proof of Theorem 1 rests on the fact that
$\mathfrak{S}_{e,f}$
satisfies the
$p$
-Lucas property when
$|e|=|f|$
,
$e$
is
$2$
-admissible and
$f=(\mathbf{1}_{k_{1}},\ldots ,\mathbf{1}_{k_{v}})$
. Since those results deal with multisums of factorial ratios, it seems natural to study similar arithmetic properties for simpler numbers such as families of factorial ratios. To that purpose, we prove Theorem 3 below, which gives an effective criterion for
${\mathcal{Q}}_{e,f}$
to satisfy the
$p$
-Lucas property for almost all primes
$p$
.Footnote
4
Furthermore, Theorem 3 shows that if
$A:={\mathcal{Q}}_{e,f}$
satisfies the
$p$
-Lucas property for almost all primes
$p$
, then, for all nonnegative integers
$n$
and all primes
$p$
, we have
$A(n)\in p^{\unicode[STIX]{x1D6FC}_{p}(A,n)}\mathbb{Z}$
.
To state this result, we introduce some additional notations. For all tuples
$e$
and
$f$
of vectors in
$\mathbb{N}^{d}$
, we write
$\unicode[STIX]{x1D6E5}_{e,f}$
for Landau’s function defined, for all
$\mathbf{x}$
in
$\mathbb{R}^{d}$
, by
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x}):=\mathop{\sum }_{i=1}^{u}\lfloor \mathbf{e}_{i}\cdot \mathbf{x}\rfloor -\mathop{\sum }_{i=1}^{v}\lfloor \mathbf{f}_{i}\cdot \mathbf{x}\rfloor \in \mathbb{Z},\end{eqnarray}$$
where
$\lfloor \cdot \rfloor$
denotes the floor function. Therefore, according to Landau’s criterion [Reference LandauLan00] and a result of the author [Reference DelaygueDel13], we have the following dichotomy.
-
– If, for all
$\mathbf{x}$
in
$[0,1]^{d}$
, we have
$\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x})\geqslant 0$
, then
${\mathcal{Q}}_{e,f}$
is a family of integers. -
– If there exists
$\mathbf{x}$
in
$[0,1]^{d}$
such that
$\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x})\leqslant -1$
, then there are only finitely many primes
$p$
such that
${\mathcal{Q}}_{e,f}$
is a family of
$p$
-adic integers.
In the rest of the article, we write
${\mathcal{D}}_{e,f}$
for the semi-algebraic set of all
$\mathbf{x}$
in
$[\!0,1\!)^{d}$
such that there exists a component
$\mathbf{d}$
of
$e$
or
$f$
satisfying
$\mathbf{d}\cdot \mathbf{x}\geqslant 1$
. Observe that
$\unicode[STIX]{x1D6E5}_{e,f}$
vanishes on the nonempty set
$[\!0,1\!)^{d}\setminus {\mathcal{D}}_{e,f}$
.
Theorem 3. Let
$e$
and
$f$
be disjoint tuples of vectors in
$\mathbb{N}^{d}$
such that
${\mathcal{Q}}_{e,f}$
is a family of integers. Then we have the following dichotomy.
-
(i) If
$|e|=|f|$
and if, for all
$\mathbf{x}$
in
${\mathcal{D}}_{e,f}$
, we have
$\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x})\geqslant 1$
, then, for all primes
$p$
,
${\mathcal{Q}}_{e,f}$
satisfies the
$p$
-Lucas property. -
(ii) If
$|e|\neq |f|$
or if there exists
$\mathbf{x}$
in
${\mathcal{D}}_{e,f}$
such that
$\unicode[STIX]{x1D6E5}_{e,f}(x)=0$
, then there are only finitely many primes
$p$
such that
${\mathcal{Q}}_{e,f}$
satisfies the
$p$
-Lucas property.
Furthermore, if
${\mathcal{Q}}_{e,f}$
satisfies the
$p$
-Lucas property for all primes
$p$
, then, for all
$\mathbf{n}$
in
$\mathbb{N}^{d}$
and every prime
$p$
, we have
$$\begin{eqnarray}{\mathcal{Q}}_{e,f}(\mathbf{n})\in p^{\unicode[STIX]{x1D6FC}_{p}({\mathcal{Q}}_{e,f},\mathbf{n})}\mathbb{Z}.\end{eqnarray}$$
Remark.
Theorem 3 implies that
${\mathcal{Q}}_{e,f}$
satisfies the
$p$
-Lucas property for all primes
$p$
if and only if all Taylor coefficients at the origin of the associated mirror maps
$z_{e,f,k}$
,
$1\leqslant k\leqslant d$
, are integers (see [Reference DelaygueDel13, Theorems 1 and 3]). Indeed, if
$\unicode[STIX]{x1D6E5}_{e,f}$
is nonnegative on
$[0,1]^{d}$
and if
$|e|\neq |f|$
, then there exists
$k$
in
$\{1,\ldots ,d\}$
such that the
$k$
th component of
$|e|$
is greater than the
$k$
th component of
$|f|$
.
Coster proved in [Reference CosterCos88] results similar to Theorems 1–3 for the coefficients of certain algebraic power series. Namely, given a prime
$p\geqslant 3$
, integers
$a_{1},\ldots ,a_{p-1}$
and a sequence
$A$
such that
$$\begin{eqnarray}f_{A}(z)=(1+a_{1}z+\cdots +a_{p-1}z^{p-1})^{1/(1-p)},\end{eqnarray}$$
Coster proved that, for all nonnegative integers
$n$
, we have
$$\begin{eqnarray}v_{p}(A(n))\geqslant \biggl\lfloor\frac{\unicode[STIX]{x1D6FC}_{p}(A,n)+1}{2}\biggr\rfloor.\end{eqnarray}$$
1.5 Auxiliary results
The proof of Theorem 1 rests on three important results. The first one is stated rather formally but we believe that it may be useful to study results similar to Beukers’ conjectures for other sequences. Throughout this article, if
$(A(n))_{n\geqslant 0}$
is a sequence taking its values in
$\mathbb{Z}$
or
$\mathbb{Z}_{p}$
, then, for all negative integers
$n$
, we set
$A(n):=0$
.
Proposition 1. Let
$p$
be a fixed prime and
$A$
a
$\mathbb{Z}_{p}$
-valued sequence satisfying the
$p$
-Lucas property with
$A(0)$
in
$\mathbb{Z}_{p}^{\times }$
. Let
$\mathfrak{A}$
be the
$\mathbb{Z}_{p}$
-module spanned by
$A$
. Assume that:
-
(a) there exists a set
$\mathfrak{B}$
of
$\mathbb{Z}_{p}$
-valued sequences with
$\mathfrak{A}\subset \mathfrak{B}$
such that, for all
$B$
in
$\mathfrak{B}$
, all
$v$
in
$\{0,\ldots ,p-1\}$
and all positive integers
$n$
, there exist
$A^{\prime }$
in
$\mathfrak{A}$
and a sequence
$(B_{k})_{k\geqslant 0}$
,
$B_{k}$
in
$\mathfrak{B}$
, such that
$$\begin{eqnarray}B(v+np)=A^{\prime }(n)+\mathop{\sum }_{k=0}^{\infty }p^{k+1}B_{k}(n-k);\end{eqnarray}$$
-
(b)
$f_{A}(z)$
is annihilated by a differential operator
${\mathcal{L}}$
in
$\mathbb{Z}_{p}[z,\unicode[STIX]{x1D703}]$
such that at least one of the following conditions holds:-
∗
${\mathcal{L}}$
is of type I; -
∗
${\mathcal{L}}$
is of type II and
$p-1\in {\mathcal{Z}}_{p}(A)$
.
-
Then, for all
$B$
in
$\mathfrak{B}$
and all nonnegative integers
$n$
, we have
$$\begin{eqnarray}A(n)\in p^{\unicode[STIX]{x1D6FC}_{p}(A,n)}\mathbb{Z}_{p}\quad \text{and}\quad B(n)\in p^{\unicode[STIX]{x1D6FC}_{p}(A,n)-1}\mathbb{Z}_{p}.\end{eqnarray}$$
We will apply Proposition 1 with
$A=\mathfrak{S}_{e,f}$
for some tuples
$e$
and
$f$
satisfying the conditions of Theorem 1 for a fixed prime
$p$
. Then we will choose the set
$\mathfrak{B}$
to be the set of the deformations
$\mathfrak{S}_{e,f}^{g}$
for
$g$
in
$\mathfrak{F}_{p}^{d}$
. Taking
$g$
to be a constant in
$\mathbb{Z}_{p}$
shows that the set
$\mathfrak{B}$
contains the
$\mathbb{Z}_{p}$
-module
$\mathfrak{A}$
spanned by
$A$
. The main difficulty in this article is to show, by
$p$
-adic techniques, that Assertion (a) in Proposition 1 holds with these choices. In particular, we shall prove and use several times the following result.
Proposition 2. Let
$p$
be a fixed prime. We write
$\unicode[STIX]{x1D6E4}_{p}$
for the
$p$
-adic Gamma function. Then there exists a function
$g$
in
$\mathfrak{F}_{p}^{2}$
such that, for all nonnegative integers
$n$
and
$m$
, we have
$$\begin{eqnarray}\frac{\unicode[STIX]{x1D6E4}_{p}((m+n)p)}{\unicode[STIX]{x1D6E4}_{p}(mp)\unicode[STIX]{x1D6E4}_{p}(np)}=1+g(m,n)p.\end{eqnarray}$$
Our proof of Theorem 2 does not use Proposition 1 but rests on the beautiful result of Mellit and Vlasenko [Reference Mellit and VlasenkoMV16, Lemma 1], which gives useful congruences modulo powers of
$p$
for some constant terms of powers of Laurent polynomials. In this case, the
$p$
-adic difficulties are hidden in the result of Mellit and Vlasenko.
Finally, we give a general result to prove the
$p$
-Lucas property for many sums of products of binomial coefficients. We recall that a tuple
$e=(\mathbf{e}_{1},\ldots ,\mathbf{e}_{u})$
of vectors in
$\mathbb{N}^{d}$
is
$1$
-admissible if either
$\mathbf{e}_{i}\geqslant \mathbf{1}$
for some
$i$
or if, for every
$k$
in
$\{1,\ldots ,d\}$
, we have
$\mathbf{e}_{i}\geqslant d\mathbf{1}_{k}$
for some
$i$
.
Proposition 3. Let
$e$
and
$f$
be disjoint tuples of vectors in
$\mathbb{N}^{d}$
such that
$|e|=|f|$
and, for all
$\mathbf{x}$
in
${\mathcal{D}}_{e,f}$
,
$\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x})\geqslant 1$
. Assume that
$e$
is
$1$
-admissible. Then
$\mathfrak{S}_{e,f}$
is integer-valued and satisfies the
$p$
-Lucas property for all primes
$p$
.
1.6 Application of Theorem 1
By applying Theorem 1, we obtain results similar to Conjectures A–C for numbers satisfying Apéry-like recurrence relations, which we list below. Characters in brackets in the last column of the following table form the sequence number in the On-line Encyclopedia of Integer Sequences [OEIS13].
All differential operators listed in the above table are of type I for all primes
$p$
, except the one associated with
$A_{5}(n):=\sum _{k=0}^{n}\binom{n}{k}^{4}$
, which reads
$$\begin{eqnarray}{\mathcal{L}}_{5}=\unicode[STIX]{x1D703}^{3}-z2(2\unicode[STIX]{x1D703}+1)(3\unicode[STIX]{x1D703}^{2}+3\unicode[STIX]{x1D703}+1)-z^{2}4(\unicode[STIX]{x1D703}+1)(4\unicode[STIX]{x1D703}+5)(4\unicode[STIX]{x1D703}+3).\end{eqnarray}$$
Hence,
${\mathcal{L}}_{5}$
is of type II for all primes
$p$
. By a result of Calkin [Reference CalkinCal98, Proposition 3], for all primes
$p$
, we have
$A_{5}(p-1)\equiv 0\;\text{mod}\;p$
, i.e.
$p-1$
is in
${\mathcal{Z}}_{p}(A_{5})$
. Thus, we can apply Theorem 1 to
$A_{5}$
.
Observe that the generating function of the central binomial coefficients is annihilated by the differential operator
${\mathcal{L}}=\unicode[STIX]{x1D703}-z(4\unicode[STIX]{x1D703}+2)$
, which is of type I for all primes
$p$
.
We set
$A_{6}(n):=\sum _{k=0}^{n}\binom{n}{k}\binom{2k}{k}\binom{2(n-k)}{n-k}$
. In 1885, Catalan gave in [Reference CatalanCat85] a recurrence relation for the Catalan–Larcombe–French sequence
$2^{n}A_{6}(n)$
from which we deduce a recurrence relation for
$A_{6}(n)$
(see also Case (d) in [Reference Almkvist and ZudilinAZ06]). According to this relation,
$A_{6}(n)$
is also Sequence E in Zagier’s list [Reference ZagierZag09], that is,
$$\begin{eqnarray}A_{6}(n)=\mathop{\sum }_{k=0}^{\lfloor n/2\rfloor }4^{n-2k}\binom{n}{2k}\binom{2k}{k}^{2}.\end{eqnarray}$$
Furthermore, according to [Reference Richmond and ShallitRS09], Domb numbers
$A_{8}(n)=\sum _{k=0}^{n}\binom{n}{k}^{2}\binom{2k}{k}\binom{2(n-k)}{n-k}$
are also the numbers of abelian squares of length
$2n$
over an alphabet with four letters.
Now we consider the numbers
$C_{i}(n)$
of abelian squares of length
$2n$
over an alphabet with
$i$
letters, which, for all positive integers
$i\geqslant 2$
, satisfy (see [Reference Richmond and ShallitRS09])
$$\begin{eqnarray}C_{i}(n)=\underset{k_{1},\ldots ,k_{i}\in \mathbb{N}}{\mathop{\sum }_{k_{1}+\cdots +k_{i}=n}}\biggl(\frac{n!}{k_{1}!\cdots k_{i}!}\biggr)^{2}.\end{eqnarray}$$
According to [Reference Borwein, Nuyens, Straub and WanBNSW11],
$C_{i}(n)$
is also the
$(2n)$
th moment of the distance to the origin after
$i$
steps travelled by a walk in the plane with unit steps in random directions.
To apply Theorem 1 to
$C_{i}$
, it suffices to show that its generating series
$f_{C_{i}}$
is annihilated by a differential operator of type I for all primes
$p$
. Indeed, by [Reference Borwein, Nuyens, Straub and WanBNSW11, Proposition 1 and Theorem 2], for all
$j\geqslant 2$
,
$C_{j}(n)$
satisfies the recurrence relation of order
$\lceil j/2\rceil$
with polynomial coefficients of degree
$j-1$
:
$$\begin{eqnarray}n^{j-1}C_{j}(n)+\mathop{\sum }_{i\geqslant 1}\biggl(n^{j-1}\mathop{\sum }_{\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{i}}\mathop{\prod }_{k=1}^{i}(-\unicode[STIX]{x1D6FC}_{k})(j+1-\unicode[STIX]{x1D6FC}_{k})\biggl(\frac{n-k}{n-k+1}\biggr)^{\unicode[STIX]{x1D6FC}_{k}-1}\biggr)C_{j}(n-i)=0,\end{eqnarray}$$
where the sum is over all sequences of positive integers
$\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{i}$
satisfying
$\unicode[STIX]{x1D6FC}_{k}\leqslant j$
and
$\unicode[STIX]{x1D6FC}_{k+1}\leqslant \unicode[STIX]{x1D6FC}_{k}-2$
. We consider
$i\geqslant 2$
and
$i$
positive integers
$\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{i}\leqslant j$
satisfying
$\unicode[STIX]{x1D6FC}_{k+1}\leqslant \unicode[STIX]{x1D6FC}_{k}-2$
. We have
$$\begin{eqnarray}n^{j-1}\mathop{\prod }_{k=1}^{i}\biggl(\frac{n-k}{n-k+1}\biggr)^{\unicode[STIX]{x1D6FC}_{k}-1}=\frac{n^{j-1}}{n^{\unicode[STIX]{x1D6FC}_{1}-1}}\biggl(\mathop{\prod }_{k=1}^{i-1}(n-k)^{\unicode[STIX]{x1D6FC}_{k}-\unicode[STIX]{x1D6FC}_{k+1}}\biggr)(n-i)^{\unicode[STIX]{x1D6FC}_{i}-1}\end{eqnarray}$$
with
$j-\unicode[STIX]{x1D6FC}_{1}\geqslant 0$
,
$\unicode[STIX]{x1D6FC}_{k}-\unicode[STIX]{x1D6FC}_{k+1}\geqslant 2$
and
$\unicode[STIX]{x1D6FC}_{i}-1\geqslant 0$
. Then
$f_{C_{j}}(z)$
is annihilated by a differential operator
${\mathcal{L}}=P_{0}(\unicode[STIX]{x1D703})+zP_{1}(\unicode[STIX]{x1D703})+\cdots +z^{q}P_{q}(\unicode[STIX]{x1D703})$
with
$P_{0}(\unicode[STIX]{x1D703})=\unicode[STIX]{x1D703}^{j-1}$
and, for all
$i\geqslant 2$
,
$$\begin{eqnarray}P_{i}(\unicode[STIX]{x1D703})\in \mathop{\prod }_{k=1}^{i-1}(\unicode[STIX]{x1D703}+i-k)^{2}\mathbb{Z}[\unicode[STIX]{x1D703}]\subset \mathop{\prod }_{k=1}^{i-1}(\unicode[STIX]{x1D703}+k)^{2}\mathbb{Z}[\unicode[STIX]{x1D703}],\end{eqnarray}$$
so that
${\mathcal{L}}$
is of type I for all primes
$p$
, as expected.
1.7 Structure of the article
In § 2, we use several results of [Reference DelaygueDel13] to prove Theorem 3. Section 3 is devoted to the proofs of Theorem 2 and Proposition 1. In particular, we prove Lemma 1, which points out the role played by differential operators in our proofs. In § 4, we prove Theorem 1 by applying Proposition 1 to
$\mathfrak{S}_{e,f}$
. It is the most technical part of this article.
2 Proof of Theorem 3
First, we prove that if
$|e|=|f|$
, then, for all primes
$p$
, all
$\mathbf{a}$
in
$\{0,\ldots ,p-1\}^{d}$
and all
$\mathbf{n}$
in
$\mathbb{N}^{d}$
, we have
$$\begin{eqnarray}\frac{{\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{n}p)}{{\mathcal{Q}}_{e,f}(\mathbf{a}){\mathcal{Q}}_{e,f}(\mathbf{n})}\in \frac{\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{j=1}^{\lfloor \mathbf{e}_{i}\cdot \mathbf{a}/p\rfloor }(1+(\mathbf{e}_{i}\cdot \mathbf{n})/j)}{\mathop{\prod }_{i=1}^{v}\mathop{\prod }_{j=1}^{\lfloor \mathbf{f}_{i}\cdot \mathbf{a}/p\rfloor }(1+(\mathbf{f}_{i}\cdot \mathbf{n})/j)}(1+p\mathbb{Z}_{p}).\end{eqnarray}$$
Indeed, we have
$$\begin{eqnarray}\frac{{\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{n}p)}{{\mathcal{Q}}_{e,f}(\mathbf{a}){\mathcal{Q}}_{e,f}(\mathbf{n})}=\frac{{\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{n}p)}{{\mathcal{Q}}_{e,f}(\mathbf{a}){\mathcal{Q}}_{e,f}(\mathbf{n}p)}\cdot \frac{{\mathcal{Q}}_{e,f}(\mathbf{n}p)}{{\mathcal{Q}}_{e,f}(\mathbf{n})}.\end{eqnarray}$$
Since
$|e|=|f|$
, we can apply [Reference DelaygueDel13, Lemma 7]Footnote
5
with
$\mathbf{c}=\mathbf{0}$
,
$\mathbf{m}=\mathbf{n}$
and
$s=0$
, which yields
$$\begin{eqnarray}\frac{{\mathcal{Q}}_{e,f}(\mathbf{n}p)}{{\mathcal{Q}}_{e,f}(\mathbf{n})}\in 1+p\mathbb{Z}_{p}.\end{eqnarray}$$
Furthermore, we have
$$\begin{eqnarray}\displaystyle \frac{{\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{n}p)}{{\mathcal{Q}}_{e,f}(\mathbf{a}){\mathcal{Q}}_{e,f}(\mathbf{n}p)} & = & \displaystyle \frac{1}{{\mathcal{Q}}_{e,f}(\mathbf{a})}\frac{\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{j=1}^{\mathbf{e}_{i}\cdot \mathbf{a}}(j+\mathbf{e}_{i}\cdot \mathbf{n}p)}{\mathop{\prod }_{i=1}^{v}\mathop{\prod }_{j=1}^{\mathbf{f}_{i}\cdot \mathbf{a}}(j+\mathbf{f}_{i}\cdot \mathbf{n}p)}\nonumber\\ \displaystyle & = & \displaystyle \frac{\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{j=1}^{\mathbf{e}_{i}\cdot \mathbf{a}}(1+(\mathbf{e}_{i}\cdot \mathbf{n}p)/j)}{\mathop{\prod }_{i=1}^{v}\mathop{\prod }_{j=1}^{\mathbf{f}_{i}\cdot \mathbf{a}}(1+(\mathbf{f}_{i}\cdot \mathbf{n}p)/j)}\nonumber\\ \displaystyle & \in & \displaystyle \frac{\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{j=1}^{\lfloor \mathbf{e}_{i}\cdot \mathbf{a}/p\rfloor }(1+(\mathbf{e}_{i}\cdot \mathbf{n})/j)}{\mathop{\prod }_{i=1}^{v}\mathop{\prod }_{j=1}^{\lfloor \mathbf{f}_{i}\cdot \mathbf{a}/p\rfloor }(1+(\mathbf{f}_{i}\cdot \mathbf{n})/j)}(1+p\mathbb{Z}_{p})\nonumber\end{eqnarray}$$
because, if
$p$
does not divide
$j$
, then
$1+(\mathbf{e}_{i}\cdot \mathbf{n}p)/j$
belongs to
$1+p\mathbb{Z}_{p}$
. This finishes the proof of (2.1).
Now we prove Assertion (i) in Theorem 3. Let
$p$
be a fixed prime number. It is well known that, for all nonnegative integers
$n$
, we have
$$\begin{eqnarray}v_{p}(n!)=\mathop{\sum }_{\ell =1}^{\infty }\biggl\lfloor\frac{n}{p^{\ell }}\biggr\rfloor.\end{eqnarray}$$
We remind the reader that the Landau function
$\unicode[STIX]{x1D6E5}_{e,f}$
is defined by
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x})=\mathop{\sum }_{i=1}^{u}\lfloor \mathbf{e}_{i}\cdot \mathbf{x}\rfloor -\mathop{\sum }_{j=1}^{v}\lfloor \mathbf{f}_{j}\cdot \mathbf{x}\rfloor \quad (\mathbf{x}\in \mathbb{R}^{d}).\end{eqnarray}$$
Thus, for all vectors
$\mathbf{n}$
in
$\mathbb{N}^{d}$
, we have
$$\begin{eqnarray}v_{p}({\mathcal{Q}}_{e,f}(\mathbf{n}))=\mathop{\sum }_{\ell =1}^{\infty }\unicode[STIX]{x1D6E5}_{e,f}\biggl(\frac{\mathbf{n}}{p^{\ell }}\biggr).\end{eqnarray}$$
Fix
$\mathbf{n}$
in
$\mathbb{N}^{d}$
and
$\mathbf{a}$
in
$\{0,\ldots ,p-1\}^{d}$
. Let
$\{\cdot \}$
denote the fractional part function. For any vector of real numbers
$\mathbf{x}=(x_{1},\ldots ,x_{d})$
, we set
$\{\mathbf{x}\}:=(\{x_{1}\},\ldots ,\{x_{d}\})$
. Since
$|e|=|f|$
, we have
$$\begin{eqnarray}v_{p}({\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{n}p))=\mathop{\sum }_{\ell =1}^{\infty }\unicode[STIX]{x1D6E5}_{e,f}\biggl(\biggl\{\frac{\mathbf{a}+\mathbf{n}p}{p^{\ell }}\biggr\}\biggr)\geqslant \unicode[STIX]{x1D6E5}_{e,f}\biggl(\frac{\mathbf{a}}{p}\biggr)\end{eqnarray}$$
because
$\unicode[STIX]{x1D6E5}_{e,f}$
is nonnegative on
$[0,1]^{d}$
. By assumption, if
$\mathbf{x}$
belongs to
${\mathcal{D}}_{e,f}$
, then
$\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x})\geqslant 1$
. On the one hand, if
$\mathbf{a}/p$
is in
${\mathcal{D}}_{e,f}$
, then both
${\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{n}p)$
and
${\mathcal{Q}}_{e,f}(\mathbf{a}){\mathcal{Q}}_{e,f}(\mathbf{n})$
are congruent to
$0$
modulo
$p$
. On the other hand, if
$\mathbf{a}/p$
is not in
${\mathcal{D}}_{e,f}$
, then, by definition, for all
$\mathbf{d}$
in
$e$
or
$f$
, we have
$\lfloor \mathbf{d}\cdot \mathbf{a}/p\rfloor =0$
, so that (2.1) yields
$$\begin{eqnarray}{\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{n}p)\equiv {\mathcal{Q}}_{e,f}(\mathbf{a}){\mathcal{Q}}_{e,f}(\mathbf{n})\hspace{0.6em}{\rm mod}\hspace{0.2em}p\mathbb{Z}_{p},\end{eqnarray}$$
as expected. This proves Assertion (i) in Theorem 3.
Now we prove Assertion (ii) in Theorem 3. If
$\mathbf{m}=(m_{1},\ldots ,m_{d})$
is a vector in
$\mathbb{R}^{d}$
and
$k\in \{1,\ldots ,d\}$
, then we set
$\mathbf{m}^{(k)}:=m_{k}$
. If
$|e|\neq |f|$
, then, since
$\unicode[STIX]{x1D6E5}_{e,f}$
is nonnegative on
$[0,1]^{d}$
, there exists
$k$
in
$\{1,\ldots ,d\}$
such that
$|e|^{(k)}-|f|^{(k)}=\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{1}_{k})\geqslant 1$
. Thereby, for almost all primes
$p$
, we have
$$\begin{eqnarray}v_{p}({\mathcal{Q}}_{e,f}(\mathbf{1}_{k}+\mathbf{1}_{k}p))=\mathop{\sum }_{\ell =1}^{\infty }\unicode[STIX]{x1D6E5}_{e,f}\biggl(\frac{\mathbf{1}_{k}+\mathbf{1}_{k}p}{p^{\ell }}\biggr)\geqslant \unicode[STIX]{x1D6E5}_{e,f}\biggl(\frac{\mathbf{1}_{k}}{p}+\mathbf{1}_{k}\biggr)\geqslant 1,\end{eqnarray}$$
but
$v_{p}({\mathcal{Q}}_{e,f}(\mathbf{1}_{k}))=0$
, so that
${\mathcal{Q}}_{e,f}$
does not satisfy the
$p$
-Lucas property.
Throughout the rest of this proof, we assume that
$|e|=|f|$
. According to [Reference DelaygueDel13, § 7.3.2], there exist
$k$
in
$\{1,\ldots ,d\}$
and a rational fraction
$R(X)$
in
$\mathbb{Q}(X)$
,
$R(X)\neq 1$
, such that, for all large enough prime numbers
$p$
, we can choose
$\mathbf{a}_{p}$
in
$\{0,\ldots ,p-1\}^{d}$
satisfying
${\mathcal{Q}}_{e,f}(\mathbf{a}_{p})\in \mathbb{Z}_{p}^{\times }$
and such that, for all nonnegative integers
$n$
, we have (see [Reference DelaygueDel13, (7.10)])
$$\begin{eqnarray}{\mathcal{Q}}_{e,f}(\mathbf{a}_{p}+\mathbf{1}_{k}np)\in R(n){\mathcal{Q}}_{e,f}(\mathbf{a}_{p}){\mathcal{Q}}_{e,f}(\mathbf{1}_{k}n)(1+p\mathbb{Z}_{p}).\end{eqnarray}$$
We fix a nonnegative integer
$n$
satisfying
$R(n)\neq 1$
. For almost all primes
$p$
, the numbers
$R(n)$
,
${\mathcal{Q}}_{e,f}(\mathbf{1}_{k}n)$
and
${\mathcal{Q}}_{e,f}(\mathbf{a}_{p})$
are invertible in
$\mathbb{Z}_{p}$
, and
$R(n)\not \equiv 1\;\text{mod}\;p\mathbb{Z}_{p}$
. Thus, we obtain
$$\begin{eqnarray}{\mathcal{Q}}_{e,f}(\mathbf{a}_{p}+\mathbf{1}_{k}np)\not \equiv {\mathcal{Q}}_{e,f}(\mathbf{a}_{p}){\mathcal{Q}}_{e,f}(\mathbf{1}_{k}n)\hspace{0.6em}{\rm mod}\hspace{0.2em}p\mathbb{Z}_{p},\end{eqnarray}$$
which finishes the proof of Assertion (ii) in Theorem 3.
Now we assume that
$|e|=|f|$
and that, for all
$\mathbf{x}$
in
${\mathcal{D}}_{e,f}$
, we have
$\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x})\geqslant 1$
. Hence, for every prime
$p$
, we have
$$\begin{eqnarray}{\mathcal{Z}}_{p}({\mathcal{Q}}_{e,f})=\{\mathbf{v}\in \{0,\ldots ,p-1\}^{d}:\mathbf{v}/p\in {\mathcal{D}}_{e,f}\}.\end{eqnarray}$$
Furthermore, if
$\mathbf{v}/p$
belongs to
${\mathcal{D}}_{e,f}$
, then, for all positive integers
$N$
and all vectors
$\mathbf{a}_{0},\ldots ,\mathbf{a}_{N-1}$
in
$\{0,\ldots ,p-1\}^{d}$
, we have
$$\begin{eqnarray}\biggl\{\frac{\mathbf{a}_{0}+\mathbf{a}_{1}p+\cdots +\mathbf{a}_{N-1}p^{N-1}+\mathbf{v}p^{N}}{p^{N+1}}\biggr\}=\frac{\mathbf{a}_{0}+\mathbf{a}_{1}p+\cdots +\mathbf{a}_{N-1}p^{N-1}+\mathbf{v}p^{N}}{p^{N+1}}\geqslant \frac{\mathbf{v}}{p},\end{eqnarray}$$
so that
$$\begin{eqnarray}\biggl\{\frac{\mathbf{a}_{0}+\mathbf{a}_{1}p+\cdots +\mathbf{a}_{N-1}p^{N-1}+\mathbf{v}p^{N}}{p^{N+1}}\biggr\}\in {\mathcal{D}}_{e,f}.\end{eqnarray}$$
Hence, for every
$\mathbf{n}$
in
$\mathbb{N}^{d}$
,
$\mathbf{n}=\sum _{k=0}^{\infty }\mathbf{n}_{k}p^{k}$
with
$\mathbf{n}_{k}\in \{0,\ldots ,p-1\}^{d}$
, we have
$$\begin{eqnarray}v_{p}({\mathcal{Q}}_{e,f}(\mathbf{n}))=\mathop{\sum }_{\ell =1}^{\infty }\unicode[STIX]{x1D6E5}_{e,f}\biggl(\biggl\{\frac{\mathop{\sum }_{k=0}^{\ell -1}\mathbf{n}_{k}p^{k}}{p^{\ell }}\biggr\}\biggr)\geqslant \unicode[STIX]{x1D6FC}_{p}({\mathcal{Q}}_{e,f},\mathbf{n})\end{eqnarray}$$
and Theorem 3 is proved.
3 Proofs of Theorem 2 and Proposition 1
3.1 Induction via Apéry-like recurrence relations
In this section, we fix a prime
$p$
. We remind the reader that if
$A$
is a
$\mathbb{Z}_{p}$
-valued sequence, then
${\mathcal{Z}}_{p}(A)$
is the set of the digits
$v\in \{0,\ldots ,p-1\}$
such that
$A(v)\in p\mathbb{Z}_{p}$
. If
$n$
is a nonnegative integer whose base-
$p$
expansion is
$n=n_{0}+n_{1}p+\cdots +n_{N}p^{N}$
, then
$\unicode[STIX]{x1D6FC}_{p}(A,n)$
is the number of
$i$
in
$\{0,\ldots ,N\}$
such that
$n_{i}$
belongs to
${\mathcal{Z}}_{p}(A)$
.
If
$A$
is a
$\mathbb{Z}_{p}$
-valued sequence, then, for all nonnegative integers
$r$
, we write
${\mathcal{U}}_{A}(r)$
for the assertion ‘For all
$n,i\in \mathbb{N}$
,
$i\leqslant r$
, if
$\unicode[STIX]{x1D6FC}_{p}(A,n)\geqslant i$
, then
$A(n)\in p^{i}\mathbb{Z}_{p}$
.’. As a first step, we shall prove the following result.
Lemma 1. Let
$A$
be a
$\mathbb{Z}_{p}$
-valued sequence satisfying the
$p$
-Lucas property with
$A(0)$
in
$\mathbb{Z}_{p}^{\times }$
. Assume that the generating series of
$A$
is annihilated by a differential operator
${\mathcal{L}}\in \mathbb{Z}_{p}[z,\unicode[STIX]{x1D703}]$
such that at least one of the following conditions holds:
-
–
${\mathcal{L}}$
is of type I; -
–
${\mathcal{L}}$
is of type II and
$p-1\in {\mathcal{Z}}_{p}(A)$
.
Let
$r$
be a nonnegative integer such that
${\mathcal{U}}_{A}(r)$
holds. Then, for all
$n_{0}$
in
${\mathcal{Z}}_{p}(A)$
and all nonnegative integers
$m$
satisfying
$\unicode[STIX]{x1D6FC}_{p}(A,m)\geqslant r$
, we have
$$\begin{eqnarray}A(n_{0}+mp)\in p^{r+1}\mathbb{Z}_{p}.\end{eqnarray}$$
Proof. Since
$A$
satisfies the
$p$
-Lucas property, we can assume that
$r$
is nonzero. The generating series of
$A$
is annihilated by a differential operator
${\mathcal{L}}=P_{0}(\unicode[STIX]{x1D703})+zP_{1}(\unicode[STIX]{x1D703})+\cdots +z^{q}P_{q}(\unicode[STIX]{x1D703})$
with
$P_{k}(X)$
in
$\mathbb{Z}_{p}[X]$
and
$P_{0}(\mathbb{Z}_{p}^{\times })\subset \mathbb{Z}_{p}^{\times }$
. Thus, for every nonnegative integer
$n$
, we have
$$\begin{eqnarray}\mathop{\sum }_{k=0}^{q}P_{k}(n-k)A(n-k)=0.\end{eqnarray}$$
We fix a nonnegative integer
$m$
satisfying
$\unicode[STIX]{x1D6FC}_{p}(A,m)\geqslant r$
. In particular, since
$r$
is nonzero and
$A(0)$
is invertible in
$\mathbb{Z}_{p}$
, we have
$m\geqslant 1$
. Furthermore, for all
$v$
in
$\{0,\ldots ,p-1\}$
, we also have
$\unicode[STIX]{x1D6FC}_{p}(A,v+mp)\geqslant r$
. According to
${\mathcal{U}}_{A}(r)$
, we obtain that, for all
$v$
in
$\{0,\ldots ,p-1\}$
,
$A(v+mp)$
belongs to
$p^{r}\mathbb{Z}_{p}$
, so that
$A(v+mp)=:\unicode[STIX]{x1D6FD}(v,m)p^{r}$
with
$\unicode[STIX]{x1D6FD}(v,m)\in \mathbb{Z}_{p}$
.
By (3.1), for all
$v$
in
$\{q,\ldots ,p-1\}$
, we have
$$\begin{eqnarray}\displaystyle 0=\mathop{\sum }_{k=0}^{q}P_{k}(v-k+mp)A(v-k+mp) & = & \displaystyle p^{r}\mathop{\sum }_{k=0}^{q}P_{k}(v-k+mp)\unicode[STIX]{x1D6FD}(v-k,m)\nonumber\\ \displaystyle & \equiv & \displaystyle p^{r}\mathop{\sum }_{k=0}^{q}P_{k}(v-k)\unicode[STIX]{x1D6FD}(v-k,m)\quad \text{mod}\;p^{r+1}\mathbb{Z}_{p}\nonumber\end{eqnarray}$$
because, for all polynomials
$P$
in
$\mathbb{Z}_{p}[X]$
and all integers
$a$
and
$c$
, we have
$P(a+cp)\equiv P(a)\;\text{mod}\;p\mathbb{Z}_{p}$
. Thus, for all
$v$
in
$\{q,\ldots ,p-1\}$
, we obtain
$$\begin{eqnarray}\mathop{\sum }_{k=0}^{q}P_{k}(v-k)\unicode[STIX]{x1D6FD}(v-k,m)\equiv 0\quad \text{mod}\;p\mathbb{Z}_{p}.\end{eqnarray}$$
We claim that if
$v$
is in
$\{1,\ldots ,q-1\}$
, then, for all
$k$
in
$\{v+1,\ldots ,q\}$
, we have
$$\begin{eqnarray}P_{k}(v+mp-k)A(v+mp-k)\in p^{r+1}\mathbb{Z}_{p}.\end{eqnarray}$$
Indeed, on the one hand, if
${\mathcal{L}}$
is of type II, then we have
$q=2$
and
$P_{2}(X)$
belongs to
$(X+1)\mathbb{Z}_{p}[X]$
, which yields
$$\begin{eqnarray}P_{2}(-1+mp)A(-1+mp)\in pA(p-1+(m-1)p)\mathbb{Z}_{p}.\end{eqnarray}$$
Since
$0$
is not in
${\mathcal{Z}}_{p}(A)$
, we have
$\unicode[STIX]{x1D6FC}_{p}(A,m-1)\geqslant r-1$
, which, together with
$p-1\in {\mathcal{Z}}_{p}(A)$
, leads to
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{p}(A,p-1+(m-1)p)\geqslant r.\end{eqnarray}$$
According to
${\mathcal{U}}_{A}(r)$
, we obtain that
$pA(p-1+(m-1)p)$
is in
$p^{r+1}\mathbb{Z}_{p}$
, as expected. On the other hand, if
${\mathcal{L}}$
is of type I, then, for all
$v$
in
$\{1,\ldots ,q-1\}$
and all
$k$
in
$\{v+1,\ldots ,q\}$
, we have
$$\begin{eqnarray}P_{k}(X)\in \mathop{\prod }_{i=1}^{k-1}(X+i)^{2}\mathbb{Z}_{p}[X],\end{eqnarray}$$
so that
$$\begin{eqnarray}v_{p}(P_{k}(v+mp-k))\geqslant v_{p}\biggl(\mathop{\prod }_{i=1}^{k-1}(v+mp-k+i)^{2}\biggr).\end{eqnarray}$$
Writing
$k-v=a+bp$
with
$a$
in
$\{0,\ldots ,p-1\}$
and
$b$
in
$\mathbb{N}$
, we obtain
$k-1\geqslant a+bp$
, so that
$$\begin{eqnarray}v_{p}\biggl(\mathop{\prod }_{i=1}^{k-1}(mp+i-a-bp)\biggr)\geqslant \left\{\begin{array}{@{}ll@{}}b\quad & \text{if }a=0,\\ b+1\quad & \text{if }a\geqslant 1,\end{array}\right.\end{eqnarray}$$
which yields
$$\begin{eqnarray}v_{p}(P_{k}(v+mp-k))\geqslant \left\{\begin{array}{@{}ll@{}}2b\quad & \text{if }a=0,\\ 2b+2\quad & \text{if }a\geqslant 1.\end{array}\right.\end{eqnarray}$$
Thus, to prove (3.3), it is enough to show that
$$\begin{eqnarray}A(v+mp-k)\in \left\{\begin{array}{@{}ll@{}}p^{r+1-2b}\mathbb{Z}_{p}\quad & \text{if }a=0,\\ p^{r-1-2b}\mathbb{Z}_{p}\quad & \text{if }a\geqslant 1.\end{array}\right.\end{eqnarray}$$
By definition of
$a$
and
$b$
, we have
$v+mp-k=-a+(m-b)p$
with
$a$
in
$\{0,\ldots ,p-1\}$
. If
$-a+(m-b)p$
is negative, then
$A(v+mp-k)=0$
and (3.4) holds. By assumption, we have
$\unicode[STIX]{x1D6FC}_{p}(A,m)\geqslant r$
and
$0\notin {\mathcal{Z}}_{p}(A)$
. Hence, if
$m-b$
is nonnegative, then we have
$\unicode[STIX]{x1D6FC}_{p}(A,m-b)\geqslant r-b$
. Thus, we have either
$a=0$
and
$\unicode[STIX]{x1D6FC}_{p}(A,v+mp-k)\geqslant r-b$
, or
$a,m-b\geqslant 1$
and
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{p}(A,v+mp-k)=\unicode[STIX]{x1D6FC}_{p}(A,p-a+(m-b-1)p)\geqslant r-b-1.\end{eqnarray}$$
Hence, Assertion
${\mathcal{U}}_{A}(r)$
yields
$$\begin{eqnarray}A(v+mp-k)\in \left\{\begin{array}{@{}ll@{}}p^{r-b}\mathbb{Z}_{p}\quad & \text{if }a=0,\\ p^{r-1-b}\mathbb{Z}_{p}\quad & \text{if }a\geqslant 1.\end{array}\right.\end{eqnarray}$$
If
$a=0$
, then
$b\geqslant 1$
and
$-b\geqslant 1-2b$
, so that (3.4) holds and (3.3) is proved.
By (3.3), for all nonnegative integers
$v$
satisfying
$1\leqslant v\leqslant \min (q-1,p-1)$
, we have
$$\begin{eqnarray}\displaystyle 0 & = & \displaystyle \mathop{\sum }_{k=0}^{q}P_{k}(v-k+mp)A(v-k+mp)\nonumber\\ \displaystyle & \equiv & \displaystyle \mathop{\sum }_{k=0}^{v}P_{k}(v-k+mp)A(v-k+mp)\quad \text{mod}\;p^{r+1}\mathbb{Z}_{p}\nonumber\\ \displaystyle & \equiv & \displaystyle p^{r}\mathop{\sum }_{k=0}^{v}P_{k}(v-k+mp)\unicode[STIX]{x1D6FD}(v-k,m)\quad \text{mod}\;p^{r+1}\mathbb{Z}_{p}\nonumber\\ \displaystyle & \equiv & \displaystyle p^{r}\mathop{\sum }_{k=0}^{v}P_{k}(v-k)\unicode[STIX]{x1D6FD}(v-k,m)\quad \text{mod}\;p^{r+1}\mathbb{Z}_{p}.\nonumber\end{eqnarray}$$
Thus, for all nonnegative integers
$v$
satisfying
$1\leqslant v\leqslant \min (q-1,p-1)$
, we have
$$\begin{eqnarray}\mathop{\sum }_{k=0}^{v}P_{k}(v-k)\unicode[STIX]{x1D6FD}(v-k,m)\equiv 0\quad \text{mod}\;p\mathbb{Z}_{p}.\end{eqnarray}$$
Both sequences
$(\unicode[STIX]{x1D6FD}(v,m))_{0\leqslant v\leqslant p-1}$
and
$(A(v))_{0\leqslant v\leqslant p-1}$
satisfy (3.2) and (3.5). Furthermore, for all
$v$
in
$\{1,\ldots ,p-1\}$
,
$P_{0}(v)$
and
$A(0)$
are invertible in
$\mathbb{Z}_{p}$
. Hence, there exists
$\unicode[STIX]{x1D6FE}(m)$
in
$\{0,\ldots ,p-1\}$
such that, for all
$v$
in
$\{0,\ldots ,p-1\}$
, we have
$\unicode[STIX]{x1D6FD}(v,m)\equiv A(v)\unicode[STIX]{x1D6FE}(m)\;\text{mod}\;p\mathbb{Z}_{p}$
, so that
$$\begin{eqnarray}A(v+mp)\equiv A(v)\unicode[STIX]{x1D6FE}(m)p^{r}\quad \text{mod}\;p^{r+1}\mathbb{Z}_{p}.\end{eqnarray}$$
Since
$n_{0}$
is in
${\mathcal{Z}}_{p}(A)$
, we have
$A(n_{0})\in p\mathbb{Z}_{p}$
, so that
$A(n_{0}+mp)$
belongs to
$p^{r+1}\mathbb{Z}_{p}$
and Lemma 1 is proved.◻
3.2 Proof of Theorem 2
Let
$p$
be a fixed prime number. For every positive integer
$n$
, we set
$\ell (n):=\lfloor \log _{p}(n)\rfloor +1$
, the length of the expansion of
$n$
to the base
$p$
, and
$\ell (0):=1$
. For all nonnegative integers
$n_{1},\ldots ,n_{r}$
, we set
$$\begin{eqnarray}n_{1}\ast \cdots \ast n_{r}:=n_{1}+n_{2}p^{\ell (n_{1})}+\cdots +n_{r}p^{\ell (n_{1})+\cdots +\ell (n_{r-1})},\end{eqnarray}$$
so that the expansion of
$n_{1}\ast \cdots \ast n_{r}$
to the base
$p$
is the concatenation of the respective expansions of
$n_{1},\ldots ,n_{r}$
. Then, by a result of Mellit and Vlasenko [Reference Mellit and VlasenkoMV16, Lemma 1], there exists a
$\mathbb{Z}_{p}$
-valued sequence
$(c_{n})_{n\geqslant 0}$
such that, for all positive integers
$n$
, we have
$$\begin{eqnarray}A(n)=\underset{1\leqslant r\leqslant \ell (n),n_{r}>0}{\mathop{\sum }_{n_{1}\ast \cdots \ast n_{r}=n}}c_{n_{1}}\cdots c_{n_{r}}\quad \text{and}\quad c_{n}\equiv 0\quad \text{mod}\;p^{\ell (n)-1}\mathbb{Z}_{p}.\end{eqnarray}$$
For every nonnegative integer
$r$
, we write
${\mathcal{U}}(r)$
for the assertion: ‘For all
$n,i\in \mathbb{N}$
,
$i\leqslant r$
, if
$\unicode[STIX]{x1D6FC}_{p}(A,n)\geqslant i$
, then
$A(n),c_{n}\in p^{i}\mathbb{Z}_{p}$
.’. To prove Theorem 2, it suffices to show that, for all nonnegative integers
$r$
, Assertion
${\mathcal{U}}(r)$
holds.
First, we prove
${\mathcal{U}}(1)$
. By [Reference Mellit and VlasenkoMV16, Theorem 1],
$A$
satisfies the
$p$
-Lucas property. In addition, if
$v$
is in
${\mathcal{Z}}_{p}(A)$
, then
$v$
is nonzero because
$A(0)=1$
and, by (3.6), we have
$c_{v}=A(v)\in p\mathbb{Z}_{p}$
. Now, if a nonnegative integer
$n$
satisfies
$\ell (n)=2$
and
$\unicode[STIX]{x1D6FC}_{p}(A,n)\geqslant 1$
, then (3.6) yields
$A(n)\equiv c_{n}\;\text{mod}\;p\mathbb{Z}_{p}$
, so that
$c_{n}$
is in
$p\mathbb{Z}_{p}$
. Hence, by induction on
$\ell (n)$
, we obtain that, for all nonnegative integers
$n$
satisfying
$\unicode[STIX]{x1D6FC}_{p}(A,n)\geqslant 1$
,
$c_{n}$
belongs to
$p\mathbb{Z}_{p}$
, so that
${\mathcal{U}}(1)$
holds.
Let
$r$
be a positive integer such that
${\mathcal{U}}(r)$
holds. We shall prove that
${\mathcal{U}}(r+1)$
is true. For all positive integers
$M$
, we write
${\mathcal{U}}_{M}(r+1)$
for the assertion:
$$\begin{eqnarray}\text{`For all }n,i\in \mathbb{N},n\leqslant M,i\leqslant r+1,\text{ if }\unicode[STIX]{x1D6FC}_{p}(A,n)\geqslant i,\text{ then }A(n),c_{n}\in p^{i}\mathbb{Z}_{p}.\text{'}\end{eqnarray}$$
Hence,
${\mathcal{U}}_{M}(r+1)$
is true if
$\ell (M)\leqslant r$
. Let
$M$
be a positive integer such that
${\mathcal{U}}_{M}(r+1)$
holds. We shall prove
${\mathcal{U}}_{M+1}(r+1)$
. By Assertions
${\mathcal{U}}(r)$
and
${\mathcal{U}}_{M}(r+1)$
, it suffices to prove that if
$\unicode[STIX]{x1D6FC}_{p}(A,M+1)$
is greater than
$r$
, then
$A(M+1)$
and
$c_{M+1}$
belong to
$p^{r+1}\mathbb{Z}_{p}$
. In the rest of the proof, we assume that
$\unicode[STIX]{x1D6FC}_{p}(A,M+1)$
is greater than
$r$
.
If
$u$
and
$n_{1},\ldots ,n_{u}$
are nonnegative integers satisfying
$2\leqslant u\leqslant \ell (M+1)$
and
$n_{1}\ast \cdots \ast ~n_{u}=M+1$
with
$n_{u}>0$
, then, for all
$i$
in
$\{1,\ldots ,u\}$
, we have
$n_{i}\leqslant M$
and
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}_{p}(A,n_{1})+\cdots +\unicode[STIX]{x1D6FC}_{p}(A,n_{u})=\unicode[STIX]{x1D6FC}_{p}(A,M+1)\geqslant r+1.\end{eqnarray}$$
Then there exist a positive integer
$k$
and integers
$1\leqslant a_{1}<\cdots <a_{k}\leqslant u$
and
$1\leqslant i_{1},\ldots ,i_{k}\leqslant r+1$
such that
$\unicode[STIX]{x1D6FC}_{p}(A,n_{a_{j}})\geqslant i_{j}$
and
$i_{1}+\cdots +i_{k}\geqslant r+1$
. Thereby, Assertion
${\mathcal{U}}_{M}(r+1)$
yields
$c_{n_{1}}\cdots c_{n_{u}}\in p^{r+1}\mathbb{Z}_{p}$
, so that
$$\begin{eqnarray}\underset{2\leqslant u\leqslant \ell (M+1),n_{u}>0}{\mathop{\sum }_{n_{1}\ast \cdots \ast n_{u}=M+1}}c_{n_{1}}\cdots c_{n_{u}}\in p^{r+1}\mathbb{Z}_{p}.\end{eqnarray}$$
By (3.6), we obtain
$$\begin{eqnarray}A(M+1)\equiv c_{M+1}\quad \text{mod}\;p^{r+1}\mathbb{Z}_{p}\quad \text{and}\quad c_{M+1}\equiv 0\quad \text{mod}\;p^{\ell (M+1)-1}\mathbb{Z}_{p}.\end{eqnarray}$$
Hence, it suffices to consider the case
$\ell (M+1)=r+1$
. In particular, we have
$M+1=v+mp$
, where
$v$
is in
${\mathcal{Z}}_{p}(A)$
and
$m$
is a nonnegative integer satisfying
$\unicode[STIX]{x1D6FC}_{p}(A,m)=r$
. Since
${\mathcal{U}}(r)$
holds, Lemma 1 yields
$A(M+1)\in p^{r+1}\mathbb{Z}_{p}$
. Thus, we also have
$c_{M+1}\in p^{r+1}\mathbb{Z}_{p}$
and Assertion
${\mathcal{U}}_{M+1}(r+1)$
holds. This finishes the proof of
${\mathcal{U}}(r+1)$
and so that of Theorem 2.◻
3.3 Proof of Proposition 1
Let
$p$
be a prime and
$A$
a
$\mathbb{Z}_{p}$
-valued sequence satisfying the hypotheses of Proposition 1. For every nonnegative integer
$n$
, we write
$\unicode[STIX]{x1D6FC}(n)$
, respectively
${\mathcal{Z}}$
, as a shorthand for
$\unicode[STIX]{x1D6FC}_{p}(A,n)$
, respectively for
${\mathcal{Z}}_{p}(A)$
. For every nonnegative integer
$r$
, we define Assertions
$$\begin{eqnarray}{\mathcal{U}}(r):\text{`For all }n,i\in \mathbb{N},i\leqslant r,\text{if }\unicode[STIX]{x1D6FC}(n)\geqslant i,\text{then }A(n)\in p^{i}\mathbb{Z}_{p}.\text{'}\end{eqnarray}$$
and
$$\begin{eqnarray}{\mathcal{V}}(r):\text{`For all }n,i\in \mathbb{N},i\leqslant r,\text{and all }B\in \mathfrak{B},\text{if }\unicode[STIX]{x1D6FC}(n)\geqslant i,\text{then }B(n)\in p^{i-1}\mathbb{Z}_{p}.\text{'}\end{eqnarray}$$
To prove Proposition 1, we have to show that, for all nonnegative integers
$r$
, Assertions
${\mathcal{U}}(r)$
and
${\mathcal{V}}(r)$
are true. We shall prove those assertions by induction on
$r$
.
Observe that Assertions
${\mathcal{U}}(0)$
,
${\mathcal{V}}(0)$
and
${\mathcal{V}}(1)$
are trivial. Furthermore, since
$A$
satisfies the
$p$
-Lucas property, Assertion
${\mathcal{U}}(1)$
holds. Let
$r_{0}$
be a fixed positive integer,
$r_{0}\geqslant 2$
, such that Assertions
${\mathcal{U}}(r_{0}-1)$
and
${\mathcal{V}}(r_{0}-1)$
are true. First, we prove Assertion
${\mathcal{V}}(r_{0})$
.
Let
$B$
in
$\mathfrak{B}$
and
$m$
in
$\mathbb{N}$
be such that
$\unicode[STIX]{x1D6FC}(m)\geqslant r_{0}$
. We write
$m=v+np$
with
$v$
in
$\{0,\ldots ,p-1\}$
. Since
$r_{0}\geqslant 2$
and
$0$
does not belong to
${\mathcal{Z}}$
, we have
$n\geqslant 1$
and, by Assertion (a) in Proposition 1, there exist
$A^{\prime }$
in
$\mathfrak{A}$
and a sequence
$(B_{k})_{k\geqslant 0}$
, with
$B_{k}$
in
$\mathfrak{B}$
, such that
$$\begin{eqnarray}B(v+np)=A^{\prime }(n)+\mathop{\sum }_{k=0}^{\infty }p^{k+1}B_{k}(n-k).\end{eqnarray}$$
In addition, we have
$\unicode[STIX]{x1D6FC}(n)\geqslant r_{0}-1$
and, since
$0$
is not in
${\mathcal{Z}}$
, we have
$\unicode[STIX]{x1D6FC}(n-1)\geqslant r_{0}-2$
. By induction, for all nonnegative integers
$k$
satisfying
$k\leqslant n$
, we have
$\unicode[STIX]{x1D6FC}(n-k)\geqslant r_{0}-1-k$
. Thus, by (3.7) in combination with
${\mathcal{U}}(r_{0}-1)$
and
${\mathcal{V}}(r_{0}-1)$
, we obtain
$$\begin{eqnarray}A^{\prime }(n)\in p^{r_{0}-1}\mathbb{Z}\quad \text{and}\quad p^{k+1}B_{k}(n-k)\in p^{k+1+r_{0}-2-k}\mathbb{Z}_{p}\subset p^{r_{0}-1}\mathbb{Z}_{p},\end{eqnarray}$$
so that
$B(v+np)$
belongs to
$p^{r_{0}-1}\mathbb{Z}_{p}$
and
${\mathcal{V}}(r_{0})$
is true.
Now we prove Assertion
${\mathcal{U}}(r_{0})$
. We write
${\mathcal{U}}_{N}(r_{0})$
for the assertion:
$$\begin{eqnarray}\text{`For all }n,i\in \mathbb{N},n\leqslant N,i\leqslant r_{0},\text{ if }\unicode[STIX]{x1D6FC}(n)\geqslant i,\text{ then }A(n)\in p^{i}\mathbb{Z}_{p}.\text{'}\end{eqnarray}$$
We shall prove
${\mathcal{U}}_{N}(r_{0})$
by induction on
$N$
. Assertion
${\mathcal{U}}_{1}(r_{0})$
holds. Let
$N$
be a positive integer such that
${\mathcal{U}}_{N}(r_{0})$
is true. Let
$n:=n_{0}+mp\leqslant N+1$
with
$n_{0}$
in
$\{0,\ldots ,p-1\}$
and
$m$
in
$\mathbb{N}$
. We can assume that
$\unicode[STIX]{x1D6FC}(n)\geqslant r_{0}$
.
If
$n_{0}$
is in
${\mathcal{Z}}$
, then we have
$\unicode[STIX]{x1D6FC}(m)\geqslant r_{0}-1$
and, by Lemma 1, we obtain that
$A(n)$
belongs to
$p^{r_{0}}\mathbb{Z}_{p}$
, as expected. If
$n_{0}$
is not in
${\mathcal{Z}}$
, then we have
$\unicode[STIX]{x1D6FC}(m)\geqslant r_{0}$
. By Assertion (a) in Proposition 1, there exist
$A^{\prime }$
in
$\mathfrak{A}$
and a sequence
$(B_{k})_{k\geqslant 0}$
with
$B_{k}$
in
$\mathfrak{B}$
such that
$$\begin{eqnarray}A(n)=A^{\prime }(m)+\mathop{\sum }_{k=0}^{\infty }p^{k+1}B_{k}(m-k).\end{eqnarray}$$
We have
$m\leqslant N$
,
$\unicode[STIX]{x1D6FC}(m)\geqslant r_{0}$
and
$\unicode[STIX]{x1D6FC}(m-k)\geqslant r_{0}-k$
; hence, by Assertions
${\mathcal{U}}_{N}(r_{0})$
and
${\mathcal{V}}(r_{0})$
, we obtain that
$A(n)$
belongs to
$p^{r_{0}}\mathbb{Z}_{p}$
. This finishes the induction on
$N$
and proves
${\mathcal{U}}(r_{0})$
. Therefore, by induction on
$r_{0}$
, Proposition 1 is proved.◻
4 Proof of Theorem 1
To prove Theorem 1, we shall apply Proposition 1 to
$\mathfrak{S}_{e,f}$
. As a first step, we prove that this sequence satisfies the
$p$
-Lucas property.
Proof of Proposition 3.
For all
$\mathbf{x}$
in
$[0,1]^{d}$
, we have
$\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x})=\unicode[STIX]{x1D6E5}_{e,f}(\{\mathbf{x}\})\geqslant 0$
, so that, by Landau’s criterion,
${\mathcal{Q}}_{e,f}$
is integer-valued. Let
$p$
be a fixed prime,
$v$
in
$\{0,\ldots ,p-1\}$
and
$n$
a nonnegative integer. We have
$$\begin{eqnarray}\mathfrak{S}_{e,f}(v+np)=\underset{k_{i}\in \mathbb{N}}{\mathop{\sum }_{k_{1}+\cdots +k_{d}=v+np}}{\mathcal{Q}}_{e,f}(k_{1},\ldots ,k_{d}).\end{eqnarray}$$
Write
$k_{i}=a_{i}+m_{i}p$
with
$a_{i}$
in
$\{0,\ldots ,p-1\}$
and
$m_{i}$
in
$\mathbb{N}$
. If
$a_{1}+\cdots +a_{d}\neq v$
, then we have
$a_{1}+\cdots +a_{d}\geqslant p$
and there exists
$i$
in
$\{1,\ldots ,d\}$
such that
$a_{i}\geqslant p/d$
. Write
$\mathbf{a}=(a_{1},\ldots ,a_{d})$
, so that
$\mathbf{1}\cdot \mathbf{a}/p\geqslant 1$
and
$d\mathbf{1}_{i}\cdot \mathbf{a}/p\geqslant 1$
. Since
$e=(\mathbf{e}_{1},\ldots ,\mathbf{e}_{u})$
is
$1$
-admissible, there exists a
$j$
in
$\{1,\ldots ,u\}$
such that either
$\mathbf{e}_{j}\geqslant \mathbf{1}$
or
$\mathbf{e}_{j}\geqslant d\mathbf{1}_{i}$
. Hence,
$\mathbf{e}_{j}\cdot \mathbf{a}/p\geqslant 1$
and
$\mathbf{a}/p$
belongs to
${\mathcal{D}}_{e,f}$
, so that
$\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{a}/p)\geqslant 1$
and
${\mathcal{Q}}_{e,f}(k_{1},\ldots ,k_{d})$
is in
$p\mathbb{Z}_{p}$
. In addition, by Theorem 3,
${\mathcal{Q}}_{e,f}$
satisfies the
$p$
-Lucas property for all primes
$p$
. Hence, we obtain
$$\begin{eqnarray}\displaystyle \mathfrak{S}_{e,f}(v+np) & \equiv & \displaystyle \underset{0\leqslant a_{i}\leqslant p-1}{\mathop{\sum }_{a_{1}+\cdots +a_{d}=v}}\;\underset{m_{i}\in \mathbb{N}}{\mathop{\sum }_{m_{1}+\cdots +m_{d}=n}}{\mathcal{Q}}_{e,f}(a_{1}+m_{1}p,\ldots ,a_{d}+m_{d}p)\quad \text{mod}\;p\mathbb{Z}_{p}\nonumber\\ \displaystyle & \equiv & \displaystyle \underset{0\leqslant a_{i}\leqslant p-1}{\mathop{\sum }_{a_{1}+\cdots +a_{d}=v}}\;\underset{m_{i}\in \mathbb{N}}{\mathop{\sum }_{m_{1}+\cdots +m_{d}=n}}{\mathcal{Q}}_{e,f}(a_{1},\ldots ,a_{d}){\mathcal{Q}}_{e,f}(m_{1},\ldots ,m_{d})\quad \text{mod}\;p\mathbb{Z}_{p}\nonumber\\ \displaystyle & \equiv & \displaystyle \mathfrak{S}_{e,f}(v)\mathfrak{S}_{e,f}(n)\quad \text{mod}\;p\mathbb{Z}_{p}.\nonumber\end{eqnarray}$$
This finishes the proof of Proposition 3. ◻
If
$e$
is
$2$
-admissible, then
$e$
is also
$1$
-admissible. Furthermore, if
$f=(\mathbf{1}_{k_{1}},\ldots ,\mathbf{1}_{k_{v}})$
, then, for all
$\mathbf{x}$
in
${\mathcal{D}}_{e,f}$
, we have
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{x})=\mathop{\sum }_{i=1}^{u}\lfloor \mathbf{e}_{i}\cdot \mathbf{x}\rfloor \geqslant 1.\end{eqnarray}$$
Hence, if
$e$
and
$f$
satisfy the conditions of Theorem 1, then Proposition 3 implies that, for all primes
$p$
,
$\mathfrak{S}_{e,f}$
has the
$p$
-Lucas property and
$\mathfrak{S}_{e,f}(0)=1$
is invertible in
$\mathbb{Z}_{p}$
. Thereby, to prove Theorem 1, it remains to prove that
$\mathfrak{S}_{e,f}$
satisfies Condition (a) in Proposition 1 with the set
$$\begin{eqnarray}\mathfrak{B}=\{\mathfrak{S}_{e,f}^{g}:g\in \mathfrak{F}_{p}^{d}\}.\end{eqnarray}$$
First, we prove that some special functions belong to
$\mathfrak{F}_{p}^{1}$
.
4.1 Special functions in
$\mathfrak{F}_{p}^{1}$
For all primes
$p$
, we write
$|\cdot |_{p}$
for the ultrametric norm on
$\mathbb{Q}_{p}$
(the field of
$p$
-adic numbers) defined by
$|a|_{p}:=p^{-v_{p}(a)}$
. Note that
$(\mathbb{Z}_{p},|\cdot |_{p})$
is a compact space. Furthermore, if
$(c_{n})_{n\geqslant 0}$
is a
$\mathbb{Z}_{p}$
-valued sequence, then
$\sum _{n=0}^{\infty }c_{n}$
is convergent in
$(\mathbb{Z}_{p},|\cdot |_{p})$
if and only if
$|c_{n}|_{p}$
tends to
$0$
as
$n$
tends to infinity. In addition, if
$\sum _{n=0}^{\infty }c_{n}$
converges, then
$(c_{n})_{n\in \mathbb{N}}$
is a summable family in
$(\mathbb{Z}_{p},|\cdot |_{p})$
.
In the rest of the article, for all primes
$p$
and all positive integers
$k$
, we set
$\unicode[STIX]{x1D6F9}_{p,k,0}(0)=1$
,
$\unicode[STIX]{x1D6F9}_{p,k,i}(0)=0$
for
$i\geqslant 1$
and, for all nonnegative integers
$i$
and
$m$
,
$m\geqslant 1$
, we set
$$\begin{eqnarray}\unicode[STIX]{x1D6F9}_{p,k,i}(m):=(-1)^{i}\unicode[STIX]{x1D70E}_{m,i}\biggl(\frac{1}{k},\frac{1}{k+p},\ldots ,\frac{1}{k+(m-1)p}\biggr),\end{eqnarray}$$
where
$\unicode[STIX]{x1D70E}_{m,i}$
is the
$i$
th elementary symmetric polynomial of
$m$
variables. Let us remind the reader that, for all nonnegative integers
$m$
and
$i$
satisfying
$i>m\geqslant 1$
, we have
$\unicode[STIX]{x1D70E}_{m,i}=0$
.
The aim of this section is to prove that, for all primes
$p$
, all
$k$
in
$\{1,\ldots ,p-1\}$
and all nonnegative integers
$i$
, we have
$$\begin{eqnarray}i!\unicode[STIX]{x1D6F9}_{p,k,i}\in \mathfrak{F}_{p}^{1},\end{eqnarray}$$
that is, for every nonnegative integer
$M$
, there exists a sequence of polynomial functions with coefficients in
$\mathbb{Z}_{p}$
which converges pointwise to
$i!\unicode[STIX]{x1D6F9}_{p,k,i}$
on
$\{0,\ldots ,M\}$
.
Proof of (4.1).
Throughout this proof, we fix a prime number
$p$
and an integer
$k$
in
$\{1,\ldots ,p-1\}$
. Furthermore, for all nonnegative integers
$i$
, we use
$\unicode[STIX]{x1D6F9}_{i}$
as a shorthand for
$\unicode[STIX]{x1D6F9}_{p,k,i}$
and
$\mathbb{N}_{{\geqslant}i}$
as a shorthand for the set of integers larger than or equal to
$i$
. We shall prove (4.1) by induction on
$i$
. To that end, for all nonnegative integers
$i$
, we write
${\mathcal{A}}_{i}$
for the following assertion:
‘There exists a sequence
$(T_{i,r})_{r\geqslant 0}$
of polynomial functions with coefficients in
$\mathbb{Z}_{p}$
which converges uniformly to
$i!\unicode[STIX]{x1D6F9}_{i}$
on
$\mathbb{N}.$
’
First, observe that, for all nonnegative integers
$m$
, we have
$\unicode[STIX]{x1D6F9}_{0}(m)=1$
, so that Assertion
${\mathcal{A}}_{0}$
is true. Let
$i$
be a fixed positive integer such that Assertions
${\mathcal{A}}_{0},\ldots ,{\mathcal{A}}_{i-1}$
are true. According to the Newton–Girard formulas, for all integers
$m\geqslant i$
, we have
$$\begin{eqnarray}i(-1)^{i}\unicode[STIX]{x1D70E}_{m,i}(X_{1},\ldots ,X_{m})=-\mathop{\sum }_{t=1}^{i}(-1)^{i-t}\unicode[STIX]{x1D70E}_{m,i-t}(X_{1},\ldots ,X_{m})\unicode[STIX]{x1D6EC}_{t}(X_{1},\ldots ,X_{m}),\end{eqnarray}$$
where
$\unicode[STIX]{x1D6EC}_{t}(X_{1},\ldots ,X_{m}):=X_{1}^{t}+\cdots +X_{m}^{t}$
. Thereby, for all integers
$m\geqslant i$
, we have
$$\begin{eqnarray}i\unicode[STIX]{x1D6F9}_{i}(m)=-\mathop{\sum }_{t=1}^{i}\unicode[STIX]{x1D6F9}_{i-t}(m)\unicode[STIX]{x1D6EC}_{t}\biggl(\frac{1}{k},\ldots ,\frac{1}{k+(m-1)p}\biggr).\end{eqnarray}$$
For all nonnegative integers
$j$
and
$t$
, we have
$$\begin{eqnarray}\frac{1}{(k+jp)^{t}}=\frac{1}{k^{t}}\frac{1}{(1+(j/k)p)^{t}}=\frac{1}{k^{t}}+\mathop{\sum }_{s=1}^{\infty }\frac{(-1)^{s}}{k^{t}}\binom{t-1+s}{s}\biggl(\frac{j}{k}\biggr)^{s}p^{s},\end{eqnarray}$$
where the right-hand side of (4.3) is a convergent series in
$(\mathbb{Z}_{p},|\cdot |_{p})$
because
$k$
is invertible in
$\mathbb{Z}_{p}$
. Therefore, we obtain that
$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EC}_{t}\biggl(\frac{1}{k},\ldots ,\frac{1}{k+(m-1)p}\biggr) & = & \displaystyle \frac{m}{k^{t}}+\mathop{\sum }_{j=0}^{m-1}\mathop{\sum }_{s=1}^{\infty }\frac{(-1)^{s}}{k^{t}}\binom{t-1+s}{s}\biggl(\frac{j}{k}\biggr)^{s}p^{s}\nonumber\\ \displaystyle & = & \displaystyle \frac{m}{k^{t}}+\mathop{\sum }_{s=1}^{\infty }\frac{(-1)^{s}}{k^{t+s}}\binom{t-1+s}{s}p^{s}\biggl(\mathop{\sum }_{j=0}^{m-1}j^{s}\biggr).\end{eqnarray}$$
According to Faulhaber’s formula (see [Reference Conway and GuyCG96]), for all positive integers
$s$
, we have
$$\begin{eqnarray}p^{s}\mathop{\sum }_{j=0}^{m-1}j^{s}=\mathop{\sum }_{c=1}^{s+1}(-1)^{s+1-c}\binom{s+1}{c}p^{s}\frac{B_{s+1-c}}{s+1}(m-1)^{c},\end{eqnarray}$$
where
$B_{k}$
is the
$k$
th first Bernoulli number. For all positive integers
$s$
and
$t$
, we set
$R_{0,t}(X):=X/k^{t}$
and
$$\begin{eqnarray}R_{s,t}(X):=\frac{1}{k^{t+s}}\binom{t-1+s}{s}\mathop{\sum }_{c=1}^{s+1}(-1)^{1-c}\binom{s+1}{c}p^{s}\frac{B_{s+1-c}}{s+1}(X-1)^{c},\end{eqnarray}$$
so that
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{t}\biggl(\frac{1}{k},\ldots ,\frac{1}{k+(m-1)p}\biggr)=\mathop{\sum }_{s=0}^{\infty }R_{s,t}(m).\end{eqnarray}$$
In the rest of this article, for all polynomials
$P(X)=\sum _{n=0}^{N}a_{n}X^{n}$
in
$\mathbb{Z}_{p}[X]$
, we set
$$\begin{eqnarray}\Vert P\Vert _{p}:=\max \{|a_{n}|_{p}:0\leqslant n\leqslant N\}.\end{eqnarray}$$
We claim that, for all nonnegative integers
$s$
and
$t$
,
$t\geqslant 1$
, we have
$$\begin{eqnarray}R_{s,t}(X)\in \mathbb{Z}_{p}[X],\quad \Vert R_{s,t}\Vert _{p}\underset{s\rightarrow \infty }{\longrightarrow }0\quad \text{and}\quad R_{s,t}(0)=0.\end{eqnarray}$$
Indeed, on the one hand, if
$p=2$
and
$s=1$
, then we have
$$\begin{eqnarray}R_{1,t}(X)=\frac{-t}{k^{t+1}}(X-1+(X-1)^{2})\in X\mathbb{Z}_{2}[X].\end{eqnarray}$$
On the other hand, if
$p\geqslant 3$
or
$s\geqslant 2$
, then we have
$p^{s}>s+1$
, so that
$v_{p}(s+1)\leqslant s-1$
. Furthermore, according to the von Staudt–Clausen theorem, we have
$v_{p}(B_{s+1-c})\geqslant -1$
. Thus, the coefficients of
$R_{s,t}(X)$
belong to
$\mathbb{Z}_{p}$
. To be more precise, we have
$v_{p}(s+1)\leqslant \log _{p}(s+1)$
, so that
$\Vert R_{s,t}\Vert _{p}\underset{s\rightarrow \infty }{\longrightarrow }0$
, as expected. In addition, we have
$$\begin{eqnarray}\displaystyle R_{s,t}(0) & = & \displaystyle -\frac{p^{s}}{(s+1)k^{t+s}}\binom{t-1+s}{s}\mathop{\sum }_{c=1}^{s+1}\binom{s+1}{c}B_{s+1-c}\nonumber\\ \displaystyle & = & \displaystyle -\frac{p^{s}}{(s+1)k^{t+s}}\binom{t-1+s}{s}\mathop{\sum }_{d=0}^{s}\binom{s+1}{d}B_{d}=0,\nonumber\end{eqnarray}$$
where we used the well-known relation satisfied by the Bernoulli numbers
$$\begin{eqnarray}\mathop{\sum }_{d=0}^{s}\binom{s+1}{d}B_{d}=0\quad (s\geqslant 1).\end{eqnarray}$$
According to
${\mathcal{A}}_{0},\ldots ,{\mathcal{A}}_{i-1}$
, for all
$j$
in
$\{0,\ldots ,i-1\}$
, there exists a sequence
$(T_{j,r})_{r\geqslant 0}$
of polynomial functions with coefficients in
$\mathbb{Z}_{p}$
which converges uniformly to
$j!\unicode[STIX]{x1D6F9}_{j}$
on
$\mathbb{N}$
. According to (4.2) and (4.5), for all nonnegative integers
$N$
, there exists
$S_{N}$
in
$\mathbb{N}$
such that, for all
$r\geqslant S_{N}$
and all
$m\geqslant i$
, we have
$$\begin{eqnarray}i!\unicode[STIX]{x1D6F9}_{i}(m)\equiv -\mathop{\sum }_{t=1}^{i}\frac{(i-1)!}{(i-t)!}T_{i-t,r}(m)\mathop{\sum }_{s=0}^{r}R_{s,t}(m)\quad \text{mod}\;p^{N}\mathbb{Z}_{p}.\end{eqnarray}$$
Thus, the sequence
$(T_{i,r})_{r\geqslant 0}$
of polynomial functions with coefficients in
$\mathbb{Z}_{p}$
, defined by
$$\begin{eqnarray}T_{i,r}(x):=-\mathop{\sum }_{t=1}^{i}\frac{(i-1)!}{(i-t)!}T_{i-t,r}(x)\mathop{\sum }_{s=0}^{r}R_{s,t}(x)\quad (x,r\in \mathbb{N})\end{eqnarray}$$
converges uniformly to
$i!\unicode[STIX]{x1D6F9}_{i}$
on
$\mathbb{N}_{{\geqslant}i}$
. To prove
${\mathcal{A}}_{i}$
, it suffices to show that, for all
$m$
in
$\{0,\ldots ,i-1\}$
, we have
$$\begin{eqnarray}T_{i,r}(m)\underset{r\rightarrow \infty }{\longrightarrow }0.\end{eqnarray}$$
Observe that (4.6) and (4.5) lead to
$T_{i,r}(0)=0$
. In particular, if
$i=1$
, then (4.7) holds. Now we assume that
$i\geqslant 2$
. For all
$m\geqslant 2$
, we have
$$\begin{eqnarray}\displaystyle \mathop{\sum }_{j=0}^{m}\unicode[STIX]{x1D6F9}_{j}(m)X^{j} & = & \displaystyle \mathop{\prod }_{w=0}^{m-1}\biggl(1-\frac{X}{k+wp}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \biggl(1-\frac{X}{k+(m-1)p}\biggr)\mathop{\prod }_{w=0}^{m-2}\biggl(1-\frac{X}{k+wp}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \biggl(1-\frac{X}{k+(m-1)p}\biggr)\mathop{\sum }_{j=0}^{m-1}\unicode[STIX]{x1D6F9}_{j}(m-1)X^{j}.\nonumber\end{eqnarray}$$
Thereby, for all
$j$
in
$\{1,\ldots ,m-1\}$
, we obtain that
$$\begin{eqnarray}\unicode[STIX]{x1D6F9}_{j}(m)=\unicode[STIX]{x1D6F9}_{j}(m-1)-\frac{\unicode[STIX]{x1D6F9}_{j-1}(m-1)}{k+(m-1)p}\end{eqnarray}$$
with
$$\begin{eqnarray}\frac{1}{k+(m-1)p}=\mathop{\sum }_{s=0}^{\infty }\frac{(-1)^{s}}{k^{s+1}}p^{s}(m-1)^{s}.\end{eqnarray}$$
Thus, there exists a sequence
$(U_{r})_{r\geqslant 0}$
of polynomials with coefficients in
$\mathbb{Z}_{p}$
such that, for all positive integers
$N$
, there exists a nonnegative integer
$S_{N}$
such that, for all
$r\geqslant S_{N}$
and all
$m\geqslant i+1$
, we have
$$\begin{eqnarray}T_{i,r}(m)\equiv T_{i,r}(m-1)-T_{i-1,r}(m-1)U_{r}(m-1)\quad \text{mod}\;p^{N}\mathbb{Z}_{p}.\end{eqnarray}$$
But, if
$V_{1}(X)$
and
$V_{2}(X)$
are polynomials with coefficients in
$\mathbb{Z}_{p}$
and if there exists a nonnegative integer
$a$
such that, for all
$m\geqslant a$
, we have
$V_{1}(m)\equiv V_{2}(m)\;\text{mod}\;p^{N}\mathbb{Z}_{p}$
, then, for all integers
$n$
, we have
$V_{1}(n)\equiv V_{2}(n)\;\text{mod}\;p^{N}\mathbb{Z}_{p}$
. Indeed, let
$n$
be an integer; there exists a nonnegative integer
$v$
such that
$n+vp^{N}\geqslant a$
. Thus, we obtain that
$$\begin{eqnarray}V_{1}(n)\equiv V_{1}(n+vp^{N})\equiv V_{2}(n+vp^{N})\equiv V_{2}(n)\quad \text{mod}\;p^{N}\mathbb{Z}_{p}.\end{eqnarray}$$
In particular, (4.8) also holds for all positive integers
$m$
.
Furthermore, according to
${\mathcal{A}}_{i-1}$
, for all
$m$
in
$\{0,\ldots ,i-2\}$
,
$T_{i-1,r}(m)$
tends to zero as
$r$
tends to infinity. Thus, for all positive integers
$N$
, there exists a nonnegative integer
$S_{N}$
such that, for all
$r\geqslant S_{N}$
and all
$m$
in
$\{1,\ldots ,i-1\}$
, we have
$$\begin{eqnarray}T_{i,r}(m)\equiv T_{i,r}(m-1)\quad \text{mod}\;p^{N}\mathbb{Z}_{p}.\end{eqnarray}$$
Since
$T_{i,r}(0)=0$
, we obtain that
$T_{i,r}(m)\equiv 0\;\text{mod}\;p^{N}\mathbb{Z}_{p}$
for all
$m$
in
$\{0,\ldots ,i-1\}$
and
$r\geqslant S_{N}$
, so that (4.7) holds. This finishes the induction on
$i$
and proves (4.1).◻
4.2 On the
$p$
-adic Gamma function
For every prime
$p$
, we write
$\unicode[STIX]{x1D6E4}_{p}$
for the
$p$
-adic Gamma function, so that, for all nonnegative integers
$n$
, we have
$$\begin{eqnarray}\unicode[STIX]{x1D6E4}_{p}(n)=(-1)^{n}\underset{p\nmid \unicode[STIX]{x1D706}}{\mathop{\prod }_{\unicode[STIX]{x1D706}=1}^{n-1}}\unicode[STIX]{x1D706}.\end{eqnarray}$$
The aim of this section is to prove Proposition 2.
Proof of Proposition 2.
Let
$p$
be a fixed prime number. For all nonnegative integers
$n$
and
$m$
, we have
$$\begin{eqnarray}\displaystyle \frac{\unicode[STIX]{x1D6E4}_{p}((m+n)p)}{\unicode[STIX]{x1D6E4}_{p}(mp)\unicode[STIX]{x1D6E4}_{p}(np)} & = & \displaystyle \biggl(\underset{p\nmid \unicode[STIX]{x1D706}}{\mathop{\prod }_{\unicode[STIX]{x1D706}=np}^{(m+n)p}}\unicode[STIX]{x1D706}\biggr)\bigg/\biggl(\underset{p\nmid \unicode[STIX]{x1D706}}{\mathop{\prod }_{\unicode[STIX]{x1D706}=1}^{mp}}\unicode[STIX]{x1D706}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \biggl(\underset{p\nmid \unicode[STIX]{x1D706}}{\mathop{\prod }_{\unicode[STIX]{x1D706}=1}^{mp}}(np+\unicode[STIX]{x1D706})\biggr)\bigg/\biggl(\underset{p\nmid \unicode[STIX]{x1D706}}{\mathop{\prod }_{\unicode[STIX]{x1D706}=1}^{mp}}\unicode[STIX]{x1D706}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \underset{p\nmid \unicode[STIX]{x1D706}}{\mathop{\prod }_{\unicode[STIX]{x1D706}=1}^{mp}}\biggl(1+\frac{np}{\unicode[STIX]{x1D706}}\biggr).\end{eqnarray}$$
Let
$X,T_{1},\ldots ,T_{m}$
be
$m+1$
variables. Then we have
$$\begin{eqnarray}\mathop{\prod }_{j=1}^{m}(X-T_{j})=X^{m}+\mathop{\sum }_{i=1}^{\infty }(-1)^{i}\unicode[STIX]{x1D70E}_{m,i}(T_{1},\ldots ,T_{m})X^{m-i}.\end{eqnarray}$$
Therefore, we obtain that
$$\begin{eqnarray}\displaystyle \underset{p\nmid \unicode[STIX]{x1D706}}{\mathop{\prod }_{\unicode[STIX]{x1D706}=1}^{mp}}\biggl(1+\frac{np}{\unicode[STIX]{x1D706}}\biggr) & = & \displaystyle \mathop{\prod }_{k=1}^{p-1}\mathop{\prod }_{\unicode[STIX]{x1D714}=0}^{m-1}\biggl(1+\frac{np}{k+\unicode[STIX]{x1D714}p}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\prod }_{k=1}^{p-1}\biggl(1+\mathop{\sum }_{i=1}^{\infty }(-1)^{i}\unicode[STIX]{x1D70E}_{m,i}\biggl(\frac{-np}{k},\ldots ,\frac{-np}{k+(m-1)p}\biggr)\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\prod }_{k=1}^{p-1}\biggl(1+\mathop{\sum }_{i=1}^{\infty }(-1)^{i}n^{i}p^{i}\unicode[STIX]{x1D6F9}_{p,k,i}(m)\biggr).\end{eqnarray}$$
Let
$k$
in
$\{1,\ldots ,p-1\}$
be fixed. By (4.1), for all positive integers
$i$
, there exists a sequence
$(P_{i,\ell })_{\ell \geqslant 0}$
of polynomial functions with coefficients in
$\mathbb{Z}_{p}$
which converges pointwise to
$i!\unicode[STIX]{x1D6F9}_{p,k,i}$
. We fix a nonnegative integer
$K$
. For all positive integers
$N$
, we set
$$\begin{eqnarray}f_{N}(x,y):=1+\mathop{\sum }_{i=1}^{K+1}(-1)^{i}x^{i}\frac{p^{i}}{i!}P_{i,N}(y).\end{eqnarray}$$
If
$n$
and
$m$
belong to
$\{0,\ldots ,K\}$
, then we have
$$\begin{eqnarray}1+\mathop{\sum }_{i=1}^{\infty }(-1)^{i}n^{i}p^{i}\unicode[STIX]{x1D6F9}_{p,k,i}(m)-f_{N}(n,m)=\mathop{\sum }_{i=1}^{K+1}(-1)^{i}n^{i}\frac{p^{i}}{i!}(i!\unicode[STIX]{x1D6F9}_{p,k,i}(m)-P_{i,N}(m))\underset{N\rightarrow \infty }{\longrightarrow }0.\end{eqnarray}$$
Furthermore, we have
$f_{N}(x,y)\in 1+p\mathbb{Z}_{p}[x,y]$
. Indeed, if
$i=i_{0}+i_{1}p+\cdots +i_{a}p^{a}$
with
$i_{j}$
in
$\{0,\ldots ,p-1\}$
, then we set
$\mathfrak{s}_{p}(i):=i_{0}+\cdots +i_{a}$
, so that, for all positive integers
$i$
, we have
$$\begin{eqnarray}i-v_{p}(i!)=i-\frac{i-\mathfrak{s}_{p}(i)}{p-1}=\frac{i(p-2)+\mathfrak{s}_{p}(i)}{p-1}>0.\end{eqnarray}$$
Hence, by (4.10), we obtain that there exists a function
$g$
in
$\mathfrak{F}_{p}^{2}$
such that, for all nonnegative integers
$n$
and
$m$
, we have
$$\begin{eqnarray}\underset{p\nmid \unicode[STIX]{x1D706}}{\mathop{\prod }_{\unicode[STIX]{x1D706}=1}^{mp}}\biggl(1+\frac{np}{\unicode[STIX]{x1D706}}\biggr)=1+g(n,m)p,\end{eqnarray}$$
which, together with (4.9), finishes the proof of Proposition 2. ◻
4.3 Last step in the proof of Theorem 1
Let
$e$
and
$f=(\mathbf{1}_{k_{1}},\ldots ,\mathbf{1}_{k_{v}})$
be two disjoint tuples of vectors in
$\mathbb{N}^{d}$
such that
$|e|=|f|$
, for all
$i$
in
$\{1,\ldots ,v\}$
,
$k_{i}$
is in
$\{1,\ldots ,d\}$
, and
$e$
is
$2$
-admissible. Let
$p$
be a fixed prime and
$\mathfrak{A}$
the
$\mathbb{Z}_{p}$
-module spanned by
$\mathfrak{S}_{e,f}$
. We set
$\mathfrak{B}:=\{\mathfrak{S}_{e,f}^{g}:g\in \mathfrak{F}_{p}^{d}\}$
, which is obviously constituted of
$\mathbb{Z}_{p}$
-valued sequences and contains
$\mathfrak{A}$
. To finish the proof of Theorem 1, we shall prove that
$\mathfrak{S}_{e,f}$
and
$\mathfrak{B}$
satisfy Condition (a) in Proposition 1. Hence, we have to show that, for all
$B$
in
$\mathfrak{B}$
, all
$v$
in
$\{0,\ldots ,p-1\}$
and all positive integers
$n$
, there exist
$A^{\prime }$
in
$\mathfrak{A}$
and a sequence
$(B_{k})_{k\geqslant 0}$
,
$B_{k}$
in
$\mathfrak{B}$
, such that
$$\begin{eqnarray}B(v+np)=A^{\prime }(n)+\mathop{\sum }_{k=0}^{\infty }p^{k+1}B_{k}(n-k).\end{eqnarray}$$
Let
$g$
be a fixed function in
$\mathfrak{F}_{p}^{d}$
that is a function
$g:\mathbb{N}^{d}\rightarrow \mathbb{Z}_{p}$
such that, for all nonnegative integers
$K$
, there exists a sequence of polynomial functions with coefficients in
$\mathbb{Z}_{p}$
which converges pointwise to
$g$
on
$\{0,\ldots ,K\}^{d}$
. In the rest of the proof, we write
$\mathbb{Z}_{p}+p\mathfrak{F}_{p}^{d}$
for the set of functions of the form
$\unicode[STIX]{x1D6FC}+ph$
, where
$\unicode[STIX]{x1D6FC}$
is a constant in
$\mathbb{Z}_{p}$
and
$h$
belongs to
$\mathfrak{F}_{p}^{d}$
. Observe that
$\mathbb{Z}_{p}+p\mathfrak{F}_{p}^{d}$
is a ring. We consider the sequence
$B:=\mathfrak{S}_{e,f}^{g}$
. Let
$\mathbf{a}$
be in
$\{0,\ldots ,p-1\}^{d}$
and
$\mathbf{m}$
in
$\mathbb{N}^{d}$
. First, we shall prove that, for every
$\mathbf{a}$
in
$\{0,\ldots ,p-1\}^{d}$
, there exists a function
$\unicode[STIX]{x1D70F}_{\mathbf{a}}$
in
$\mathbb{Z}_{p}+p\mathfrak{F}_{p}^{d}$
such that, for all
$v$
in
$\{0,\ldots ,p-1\}$
and
$n$
in
$\mathbb{N}$
, we have
$$\begin{eqnarray}\mathfrak{S}_{e,f}^{g}(v+np)=\mathop{\sum }_{\mathbf{0}\leqslant \mathbf{a}\leqslant \mathbf{1}(p-1)}\underset{|\mathbf{a}+\mathbf{m}p|=v+np}{\mathop{\sum }_{\mathbf{m}\geqslant \mathbf{0}}}{\mathcal{Q}}_{e,f}(\mathbf{m})\unicode[STIX]{x1D70F}_{\mathbf{a}}(\mathbf{m}).\end{eqnarray}$$
To that end, we express
${\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{m}p)$
as a product of
${\mathcal{Q}}_{e,f}(\mathbf{m})$
and elements of
$\mathbb{Z}_{p}+p\mathfrak{F}_{p}^{d}$
. We have
$$\begin{eqnarray}{\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{m}p)=\frac{\mathop{\prod }_{i=1}^{u}(\mathbf{e}_{i}\cdot \mathbf{m}p)!\mathop{\prod }_{k=1}^{\mathbf{e}_{i}\cdot \mathbf{a}}(\mathbf{e}_{i}\cdot \mathbf{m}p+k)}{\mathop{\prod }_{i=1}^{v}(\mathbf{f}_{i}\cdot \mathbf{m}p)!\mathop{\prod }_{k=1}^{\mathbf{f}_{i}\cdot \mathbf{a}}(\mathbf{f}_{i}\cdot \mathbf{m}p+k)}.\end{eqnarray}$$
For every nonnegative integer
$n$
, we have
$$\begin{eqnarray}\frac{(np)!}{n!}=p^{n}(-1)^{np}\unicode[STIX]{x1D6E4}_{p}(np),\end{eqnarray}$$
so that we have
$$\begin{eqnarray}\frac{\mathop{\prod }_{i=1}^{u}(\mathbf{e}_{i}\cdot \mathbf{m}p)!}{\mathop{\prod }_{i=1}^{v}(\mathbf{f}_{i}\cdot \mathbf{m}p)!}=p^{(|e|-|f|)\cdot \mathbf{m}}{\mathcal{Q}}_{e,f}(\mathbf{m})\frac{\mathop{\prod }_{i=1}^{u}(-1)^{\mathbf{e}_{i}\cdot \mathbf{m}p}\unicode[STIX]{x1D6E4}_{p}(\mathbf{e}_{i}\cdot \mathbf{m}p)}{\mathop{\prod }_{i=1}^{v}(-1)^{\mathbf{f}_{i}\cdot \mathbf{m}p}\unicode[STIX]{x1D6E4}_{p}(\mathbf{f}_{i}\cdot \mathbf{m}p)}.\end{eqnarray}$$
Furthermore, we have
$$\begin{eqnarray}\frac{\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{k=1}^{\mathbf{e}_{i}\cdot \mathbf{a}}(\mathbf{e}_{i}\cdot \mathbf{m}p+k)}{\mathop{\prod }_{i=1}^{v}\mathop{\prod }_{k=1}^{\mathbf{f}_{i}\cdot \mathbf{a}}(\mathbf{f}_{i}\cdot \mathbf{m}p+k)}=\frac{\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{k=1,p\nmid k}^{\mathbf{e}_{i}\cdot \mathbf{a}}(\mathbf{e}_{i}\cdot \mathbf{m}p+k)}{\mathop{\prod }_{i=1}^{v}\mathop{\prod }_{k=1,p\nmid k}^{\mathbf{f}_{i}\cdot \mathbf{a}}(\mathbf{f}_{i}\cdot \mathbf{m}p+k)}\cdot p^{\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{a}/p)}\frac{\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{k=1}^{\lfloor \mathbf{e}_{i}\cdot \mathbf{a}/p\rfloor }(\mathbf{e}_{i}\cdot \mathbf{m}+k)}{\mathop{\prod }_{i=1}^{v}\mathop{\prod }_{k=1}^{\lfloor \mathbf{f}_{i}\cdot \mathbf{a}/p\rfloor }(\mathbf{f}_{i}\cdot \mathbf{m}+k)}.\end{eqnarray}$$
Since
$|e|=|f|$
, we obtain that
$$\begin{eqnarray}\frac{\mathop{\prod }_{i=1}^{u}(-1)^{\mathbf{e}_{i}\cdot \mathbf{m}p}\unicode[STIX]{x1D6E4}_{p}(\mathbf{e}_{i}\cdot \mathbf{m}p)}{\mathop{\prod }_{i=1}^{v}(-1)^{\mathbf{f}_{i}\cdot \mathbf{m}p}\unicode[STIX]{x1D6E4}_{p}(\mathbf{f}_{i}\cdot \mathbf{m}p)}=\frac{\mathop{\prod }_{i=1}^{u}\unicode[STIX]{x1D6E4}_{p}(\mathbf{e}_{i}\cdot \mathbf{m}p)}{\mathop{\prod }_{i=1}^{v}\unicode[STIX]{x1D6E4}_{p}(\mathbf{f}_{i}\cdot \mathbf{m}p)}.\end{eqnarray}$$
Let
$\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{d}$
be nonnegative integers with
$\unicode[STIX]{x1D6FC}_{i_{0}}\geqslant 1$
for some
$i_{0}$
in
$\{1,\ldots ,d\}$
. By Proposition 2, there exists a function
$h$
in
$\mathfrak{F}_{p}^{d}$
such that, for all nonnegative integers
$m_{1},\ldots ,m_{d}$
, we have
$$\begin{eqnarray}\frac{\unicode[STIX]{x1D6E4}_{p}((\unicode[STIX]{x1D6FC}_{1}m_{1}+\cdots +\unicode[STIX]{x1D6FC}_{d}m_{d})p)}{\unicode[STIX]{x1D6E4}_{p}((\unicode[STIX]{x1D6FC}_{1}m_{1}+\cdots +(\unicode[STIX]{x1D6FC}_{i_{0}}-1)m_{i_{0}}+\cdots +\unicode[STIX]{x1D6FC}_{d}m_{d})p)\unicode[STIX]{x1D6E4}_{p}(m_{i_{0}}p)}=1+h(m_{1},\ldots ,m_{d})p.\end{eqnarray}$$
Hence, there exists a function
$h^{\prime }$
in
$\mathfrak{F}_{p}^{d}$
such that, for all nonnegative integers
$m_{1},\ldots ,m_{d}$
, we have
$$\begin{eqnarray}\frac{\unicode[STIX]{x1D6E4}_{p}((\unicode[STIX]{x1D6FC}_{1}m_{1}+\cdots +\unicode[STIX]{x1D6FC}_{d}m_{d})p)}{\unicode[STIX]{x1D6E4}_{p}(m_{1}p)^{\unicode[STIX]{x1D6FC}_{1}}\cdots \unicode[STIX]{x1D6E4}_{p}(m_{d}p)^{\unicode[STIX]{x1D6FC}_{d}}}=1+h^{\prime }(m_{1},\ldots ,m_{d})p.\end{eqnarray}$$
Since
$f$
is only constituted by vectors
$\mathbf{1}_{k}$
, there exists
$g^{\prime }$
in
$\mathfrak{F}_{p}^{d}$
such that, for all
$\mathbf{m}$
in
$\mathbb{N}^{d}$
, we have
$$\begin{eqnarray}\frac{\mathop{\prod }_{i=1}^{u}\unicode[STIX]{x1D6E4}_{p}(\mathbf{e}_{i}\cdot \mathbf{m}p)}{\mathop{\prod }_{i=1}^{v}\unicode[STIX]{x1D6E4}_{p}(\mathbf{f}_{i}\cdot \mathbf{m}p)}=1+g^{\prime }(\mathbf{m})p.\end{eqnarray}$$
Furthermore, if
$k$
is an integer coprime to
$p$
and
$\mathbf{d}$
a vector in
$\mathbb{N}^{d}$
, then, for every
$\mathbf{m}$
in
$\mathbb{N}^{d}$
, we have
$$\begin{eqnarray}\frac{1}{\mathbf{d}\cdot \mathbf{m}p+k}=\mathop{\sum }_{s=0}^{\infty }(-1)^{s}\frac{(\mathbf{d}\cdot \mathbf{m})^{s}}{k^{s+1}}p^{s},\end{eqnarray}$$
so that there is a function
$g^{\prime \prime }$
in
$\mathfrak{F}_{p}^{d}$
such that, for all
$\mathbf{m}$
in
$\mathbb{N}^{d}$
, we have
$$\begin{eqnarray}\frac{1}{\mathbf{d}\cdot \mathbf{m}p+k}=\frac{1}{k}+g^{\prime \prime }(\mathbf{m})p.\end{eqnarray}$$
Hence, for all
$\mathbf{a}$
in
$\{0,\ldots ,p-1\}^{d}$
, there exist a
$p$
-adic integer
$\unicode[STIX]{x1D706}_{\mathbf{a}}$
and a function
$g_{\mathbf{a}}$
in
$\mathfrak{F}_{p}^{d}$
such that, for all
$\mathbf{m}$
in
$\mathbb{N}^{d}$
, we have
$$\begin{eqnarray}\frac{\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{k=1,p\nmid k}^{\mathbf{e}_{i}\cdot \mathbf{a}}(\mathbf{e}_{i}\cdot \mathbf{m}p+k)}{\mathop{\prod }_{i=1}^{v}\mathop{\prod }_{k=1,p\nmid k}^{\mathbf{f}_{i}\cdot \mathbf{a}}(\mathbf{f}_{i}\cdot \mathbf{m}p+k)}=\unicode[STIX]{x1D706}_{\mathbf{a}}+g_{\mathbf{a}}(\mathbf{m})p.\end{eqnarray}$$
Since
$f$
is only constituted by vectors
$\mathbf{1}_{k}$
, for all
$i$
in
$\{1,\ldots ,v\}$
, we have
$\lfloor \mathbf{f}_{i}\cdot \mathbf{a}/p\rfloor =0$
. Thereby, for all
$\mathbf{a}$
in
$\{0,\ldots ,p-1\}^{d}$
, there exists a function
$h_{\mathbf{a}}$
in
$\mathbb{Z}_{p}+p\mathfrak{F}_{p}^{d}$
such that, for all
$\mathbf{m}$
in
$\mathbb{N}^{d}$
, we have
$$\begin{eqnarray}{\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{m}p)={\mathcal{Q}}_{e,f}(\mathbf{m})h_{\mathbf{a}}(\mathbf{m})p^{\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{a}/p)}\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{k=1}^{\lfloor \mathbf{e}_{i}\cdot \mathbf{a}/p\rfloor }(\mathbf{e}_{i}\cdot \mathbf{m}+k).\end{eqnarray}$$
Furthermore, we have either
$\lfloor \mathbf{e}_{i}\cdot \mathbf{a}/p\rfloor =0$
for all
$i$
, or
$\lfloor \mathbf{e}_{i}\cdot \mathbf{a}/p\rfloor \geqslant 1$
for some
$i$
and so
$\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{a}/p)\geqslant 1$
. In both cases, we obtain that
$$\begin{eqnarray}\mathbf{m}\mapsto p^{\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{a}/p)}\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{k=1}^{\lfloor \mathbf{e}_{i}\cdot \mathbf{a}/p\rfloor }(\mathbf{e}_{i}\cdot \mathbf{m}+k)\in \mathbb{Z}_{p}+p\mathfrak{F}_{p}^{d}.\end{eqnarray}$$
Let
$g$
be a function in
$\mathfrak{F}_{p}^{d}$
. For all
$\mathbf{a}$
in
$\{0,\ldots ,p-1\}^{d}$
, the function
$\mathbf{m}\mapsto g(\mathbf{a}+\mathbf{m}p)$
belongs to
$\mathbb{Z}_{p}+p\mathfrak{F}_{p}^{d}$
. For all
$\mathbf{m}$
in
$\mathbb{N}^{d}$
, we set
$$\begin{eqnarray}\unicode[STIX]{x1D70F}_{\mathbf{a}}(\mathbf{m}):=g(\mathbf{a}+\mathbf{m}p)h_{\mathbf{a}}(\mathbf{m})p^{\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{a}/p)}\mathop{\prod }_{i=1}^{u}\mathop{\prod }_{k=1}^{\lfloor \mathbf{e}_{i}\cdot \mathbf{a}/p\rfloor }(\mathbf{e}_{i}\cdot \mathbf{m}+k),\end{eqnarray}$$
so that
$\unicode[STIX]{x1D70F}_{\mathbf{a}}\in \mathbb{Z}_{p}+p\mathfrak{F}_{p}^{d}$
. Therefore, for all
$v$
in
$\{0,\ldots ,p-1\}$
and
$n$
in
$\mathbb{N}$
, we have
$$\begin{eqnarray}\displaystyle \mathfrak{S}_{e,f}^{g}(v+np) & = & \displaystyle \mathop{\sum }_{\mathbf{0}\leqslant \mathbf{a}\leqslant \mathbf{1}(p-1)}\underset{|\mathbf{a}+\mathbf{m}p|=v+np}{\mathop{\sum }_{\mathbf{m}\geqslant \mathbf{0}}}g(\mathbf{a}+\mathbf{m}p){\mathcal{Q}}_{e,f}(\mathbf{a}+\mathbf{m}p)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\sum }_{\mathbf{0}\leqslant \mathbf{a}\leqslant \mathbf{1}(p-1)}\underset{|\mathbf{a}+\mathbf{m}p|=v+np}{\mathop{\sum }_{\mathbf{m}\geqslant \mathbf{0}}}{\mathcal{Q}}_{e,f}(\mathbf{m})\unicode[STIX]{x1D70F}_{\mathbf{a}}(\mathbf{m}),\nonumber\end{eqnarray}$$
which proves (4.12).
Now, if
$|\mathbf{a}+\mathbf{m}p|=v+np$
, then we have
$|\mathbf{a}|=v+jp$
with
$$\begin{eqnarray}0\leqslant j\leqslant \min \biggl(n,\biggl\lfloor\frac{d(p-1)-v}{p}\biggr\rfloor\biggr)=:M.\end{eqnarray}$$
Furthermore, we have
$\lfloor |\mathbf{a}|/p\rfloor =j$
and there is
$k$
in
$\{1,\ldots ,d\}$
such that
$\mathbf{a}^{(k)}\geqslant (v+jp)/d$
. Since
$e$
is
$2$
-admissible, there are
$1\leqslant i_{1}<i_{2}\leqslant u$
such that
$\mathbf{e}_{i_{1}}\cdot \mathbf{a}/p\geqslant j$
and
$\mathbf{e}_{i_{2}}\cdot \mathbf{a}/p\geqslant j$
. Hence, we obtain that
$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{e,f}(\mathbf{a}/p)=\mathop{\sum }_{i=1}^{u}\biggl\lfloor\frac{\mathbf{e}_{i}\cdot \mathbf{a}}{p}\biggr\rfloor\geqslant 2j\end{eqnarray}$$
because
$f$
is constituted by vectors
$\mathbf{1}_{k}$
. In particular, there is
$\unicode[STIX]{x1D70F}_{\mathbf{a}}^{\prime }$
in
$\mathfrak{F}_{p}^{d}$
such that
$\unicode[STIX]{x1D70F}_{\mathbf{a}}=p^{2j}\unicode[STIX]{x1D70F}_{\mathbf{a}}^{\prime }$
. Hence, we have
$$\begin{eqnarray}\mathfrak{S}_{e,f}^{g}(v+np)=\underset{|\mathbf{a}|=v}{\mathop{\sum }_{\mathbf{0}\leqslant \mathbf{a}\leqslant \mathbf{1}(p-1)}}\mathop{\sum }_{|\mathbf{m}|=n}{\mathcal{Q}}_{e,f}(\mathbf{m})\unicode[STIX]{x1D70F}_{\mathbf{a}}(\mathbf{m})+\mathop{\sum }_{j=1}^{M}p^{2j}\underset{|\mathbf{a}|=v+jp}{\mathop{\sum }_{\mathbf{0}\leqslant \mathbf{a}\leqslant \mathbf{1}(p-1)}}\mathop{\sum }_{|\mathbf{m}|=n-j}{\mathcal{Q}}_{e,f}(\mathbf{m})\unicode[STIX]{x1D70F}_{\mathbf{a}}^{\prime }(\mathbf{m}).\end{eqnarray}$$
For every
$\mathbf{a}$
in
$\{0,\ldots ,p-1\}^{d}$
, we write
$\unicode[STIX]{x1D70F}_{\mathbf{a}}=\unicode[STIX]{x1D6FC}_{\mathbf{a}}+p\unicode[STIX]{x1D6FD}_{\mathbf{a}}$
, where
$\unicode[STIX]{x1D6FC}_{\mathbf{a}}$
is a constant in
$\mathbb{Z}_{p}$
and
$\unicode[STIX]{x1D6FD}_{\mathbf{a}}$
is a function in
$\mathfrak{F}_{p}^{d}$
. We set
$$\begin{eqnarray}\unicode[STIX]{x1D6FC}:=\underset{|\mathbf{a}|=v}{\mathop{\sum }_{\mathbf{0}\leqslant \mathbf{a}\leqslant \mathbf{1}(p-1)}}\unicode[STIX]{x1D6FC}_{\mathbf{a}}\in \mathbb{Z}_{p}\quad \text{and}\quad \unicode[STIX]{x1D6FD}:=\underset{|\mathbf{a}|=v}{\mathop{\sum }_{\mathbf{0}\leqslant \mathbf{a}\leqslant \mathbf{1}(p-1)}}\unicode[STIX]{x1D6FD}_{\mathbf{a}}\in \mathfrak{F}_{p}^{d}.\end{eqnarray}$$
Finally, for every
$j$
in
$\{1,\ldots ,M\}$
, we set
$$\begin{eqnarray}\unicode[STIX]{x1D6FE}_{j}:=\underset{|\mathbf{a}|=v+jp}{\mathop{\sum }_{\mathbf{0}\leqslant \mathbf{a}\leqslant \mathbf{1}(p-1)}}\unicode[STIX]{x1D70F}_{\mathbf{a}}^{\prime }\in \mathfrak{F}_{p}^{d}.\end{eqnarray}$$
Hence, we obtain that
$$\begin{eqnarray}\mathfrak{S}_{e,f}^{g}(v+np)=\unicode[STIX]{x1D6FC}\mathfrak{S}_{e,f}(n)+p\mathfrak{S}_{e,f}^{\unicode[STIX]{x1D6FD}}(n)+\mathop{\sum }_{j=1}^{M}p^{2j}\mathfrak{S}_{e,f}^{\unicode[STIX]{x1D6FE}_{j}}(n-j),\end{eqnarray}$$
where
$\unicode[STIX]{x1D6FC}\mathfrak{S}_{e,f}\in \mathfrak{A}$
,
$\mathfrak{S}_{e,f}^{\unicode[STIX]{x1D6FD}}\in \mathfrak{B}$
and
$\mathfrak{S}_{e,f}^{\unicode[STIX]{x1D6FE}_{j}}\in \mathfrak{B}$
. For every
$j$
,
$1\leqslant j\leqslant M$
, we have
$2j\geqslant j+1$
, so that there exist
$A^{\prime }$
in
$\mathfrak{A}$
and a sequence
$(B_{j})_{j\geqslant 0}$
, with
$B_{j}$
in
$\mathfrak{B}$
, such that
$$\begin{eqnarray}\mathfrak{S}_{e,f}^{g}(v+np)=A^{\prime }(n)+pB_{0}(n)+\mathop{\sum }_{j=1}^{\infty }p^{j+1}B_{j}(n-j).\end{eqnarray}$$
This shows that
$\mathfrak{S}_{e,f}$
and
$\mathfrak{B}$
satisfy Condition (a) in Proposition 1, so that Theorem 1 is proved.◻
