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The divisor function at consecutive integers

Published online by Cambridge University Press:  26 February 2010

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
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), 157176.Google Scholar
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6.Vaughan, R. C.. A remark on the divisor function d(n). Glasgow Math. J., 14 (1973), 5455.CrossRefGoogle Scholar
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