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    Goldston, D. A. Graham, S. W. Pintz, J. and Yildirim, C. Y. 2010. Small Gaps Between Almost Primes, the Parity Problem, and Some Conjectures of Erdos on Consecutive Integers. International Mathematics Research Notices,


    de Koninck, Jean-Marie and Luca, Florian 2009. The Product of Exponents in the Factorization of Consecutive Integers. Mathematika, Vol. 55, Issue. 1-2, p. 59.


    Schlage-Puchta, J.-C. 2003. The equation ω( n) = ω( n + 1). Mathematika, Vol. 50, Issue. 1-2, p. 99.


    Kan, Jiahai and Shan, Zun 1999. On the divisor function d( n): II. Mathematika, Vol. 46, Issue. 02, p. 419.


    Timofeev, N. M. 1999. On the difference between the number of prime divisors at consecutive integers. Mathematical Notes, Vol. 66, Issue. 4, p. 474.


    Тимофеев, Николай Михайлович and Timofeev, Nikolai Michailovich 1999. О разности числа простых делителей последовательных чисел. Математические заметки, Vol. 66, Issue. 4, p. 579.


    Gilchrist, Martin 1989. Non-commutative Noetherian Unique Factorization Domains often have stable range one. Mathematical Proceedings of the Cambridge Philosophical Society, Vol. 106, Issue. 02, p. 229.


    Hildebrand, Adolf 1989. On Integer Sets Containing Strings of Consecutive Integers. Mathematika, Vol. 36, Issue. 01, p. 60.


    1988. Elementary Theory of Numbers.


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The divisor function at consecutive integers

  • D. R. Heath-Brown (a1)
  • DOI: http://dx.doi.org/10.1112/S0025579300010743
  • Published online: 01 February 2010
Abstract

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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Mathematika
  • ISSN: 0025-5793
  • EISSN: 2041-7942
  • URL: /core/journals/mathematika
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