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The divisor function at consecutive integers

  • D. R. Heath-Brown (a1)

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.

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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.
2.Erdős P. and Mirsky L.. The distribution of values of the divisor function d(n). Proc. London Math. Soc. (3), 2 (1952), 257271.
3.Halberstam H. and Richert H. E.. Sieve Methods (Academic Press, London, 1974).
4.Heath-Brown D. R.. A parity problem from sieve theory. Mathematika, 29 (1982), 16.
5.Spiro C.. Thesis (Urbana, 1981).
6.Vaughan R. C.. A remark on the divisor function d(n). Glasgow Math. J., 14 (1973), 5455.
7.Xie S.. On the fc-twin primes problem. Acta Math. Sinica, 26 (1983), 378384.
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  • ISSN: 0025-5793
  • EISSN: 2041-7942
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