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The divisor function at consecutive integers

Published online by Cambridge University Press:  26 February 2010

D. R. Heath-Brown
D. R. Heath-Brown
Affiliation:
Magdalen College, Oxford, OX1 4AU
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Rather more than thirty years ago Erdős and Mirsky [2] asked whether there exist infinitely many integers n for which d(n) = d(n + 1). At one time it seemed that this might be as hard to resolve as the twin prime problem, see Vaughan [6] and Halberstam and Richert [3, pp. 268, 338]. The reasoning was roughly as follows. A natural way to arrange that d(n) = d(n + l) is to take n = 2p, where 2p + 1 = 3q, with p, q primes. However sieve methods yield only 2p + 1 = 3P2 (by the method of Chen [1]). To specify that P2 should be a prime q entails resolving the “parity problem” of sieve theory. Doing this would equally allow one to replace P2 by a prime in Chen's p + 2 = P2 result.

Type
Research Article
Information
Mathematika , Volume 31 , Issue 1 , June 1984 , pp. 141 - 149
Copyright
Copyright © University College London 1984

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References

1.Chen, J. R.. On the representation of a large even integer as the sum of a prime and the product of at most two primes. Sci. Sinica, 16 (1973), 157–176.Google Scholar
2.Erdős, P. and Mirsky, L.. The distribution of values of the divisor function d(n). Proc. London Math. Soc. (3), 2 (1952), 257–271.CrossRefGoogle Scholar
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5.Spiro, C.. Thesis (Urbana, 1981).Google Scholar
6.Vaughan, R. C.. A remark on the divisor function d(n). Glasgow Math. J., 14 (1973), 54–55.CrossRefGoogle Scholar
7.Xie, S.. On the fc-twin primes problem. Acta Math. Sinica, 26 (1983), 378–384.Google Scholar