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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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4. D. R. Heath-Brown . A parity problem from sieve theory. Mathematika, 29 (1982), 16.

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  • ISSN: 0025-5793
  • EISSN: 2041-7942
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