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%I #43 Jan 13 2024 11:46:23
%S 2,14,21,33,34,38,44,57,75,85,86,93,94,98,116,118,122,133,135,141,142,
%T 145,147,158,171,177,201,202,205,213,214,217,218,230,244,253,285,296,
%U 298,301,302,326,332,334,375,381,387,393,394,429,434,445,446,453,481
%N Numbers k with prime signature(k) = prime signature(k+1).
%C This sequence is infinite, see A189982 and Theorem 4 in Goldston-Graham-Pintz-Yıldırım. - _Charles R Greathouse IV_, Jul 17 2015
%C This is a subsequence of A005237, hence a(n) >> n sqrt(log log n) by the Erdős-Pomerance-Sárközy result cited there. - _Charles R Greathouse IV_, Jul 17 2015
%C Sequence is not the same as A280074, first deviation is at a(212): a(212) = 2041, A280074(212) = 2024. Number 2024 is the smallest number n such that A007425(n) = A007425(n+1) with different prime signatures of numbers n and n+1 (2024 = 2^3 * 11 * 23, 2025 = 3^4 * 5^2; A007425(2024) = A007425(2025) = 90). Conjecture: also numbers n such that Product_{d|n} tau(d) = Product_{d|n+1} tau(d). - _Jaroslav Krizek_, Dec 25 2016
%H T. D. Noe, <a href="/A052213/b052213.txt">Table of n, a(n) for n = 1..10000</a>
%H D. A. Goldston, S. W. Graham, J. Pintz, and C. Y. Yıldırım, <a href="http://arxiv.org/abs/0803.2636">Small gaps between almost primes, the parity problem, and some conjectures of Erdos on consecutive integers</a>, arXiv:0803.2636 [math.NT], 2008.
%H MathOverflow, <a href="http://mathoverflow.net/questions/32412">Question on consecutive integers with similar prime factorizations</a>
%H Eric Weisstein's MathWorld, <a href="http://mathworld.wolfram.com/PrimeSignature.html">Prime Signature</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Prime_signature">Prime signature</a>
%e 14 = 2^1*7^1 and 15 = 3^1*5^1, so both have prime signature {1,1}. Thus, 14 is a term.
%t pri[n_] := Sort[ Transpose[ FactorInteger[n]] [[2]]]; Select[ Range[ 2, 1000], pri[#] == pri[#+1] &]
%t Rest[SequencePosition[Table[Sort[FactorInteger[n][[All,2]]],{n,500}],{x_,x_}][[All,1]]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Aug 28 2017 *)
%o (PARI) lista(nn) = for (n=1, nn-1, if (vecsort(factor(n)[,2]) == vecsort(factor(n+1)[,2]), print1(n, ", "));); \\ _Michel Marcus_, Jun 10 2015
%o (Python)
%o from sympy import factorint
%o def aupto(limit):
%o alst, prevsig = [], [1]
%o for k in range(3, limit+2):
%o sig = sorted(factorint(k).values())
%o if sig == prevsig: alst.append(k - 1)
%o prevsig = sig
%o return alst
%o print(aupto(250)) # _Michael S. Branicky_, Sep 20 2021
%Y Cf. A005237, A189982, A260143.
%K easy,nonn
%O 1,1
%A _Erich Friedman_, Jan 29 2000
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