A052213 Numbers k with prime signature(k) = prime signature(k+1).

2, 14, 21, 33, 34, 38, 44, 57, 75, 85, 86, 93, 94, 98, 116, 118, 122, 133, 135, 141, 142, 145, 147, 158, 171, 177, 201, 202, 205, 213, 214, 217, 218, 230, 244, 253, 285, 296, 298, 301, 302, 326, 332, 334, 375, 381, 387, 393, 394, 429, 434, 445, 446, 453, 481

Offset: 1

Author

Erich Friedman, Jan 29 2000

Keywords

• easy
• nonn

Comments

This sequence is infinite, see A189982 and Theorem 4 in Goldston-Graham-Pintz-Yıldırım. - Charles R Greathouse IV, Jul 17 2015

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

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

Examples

			14 = 2^1*7^1 and 15 = 3^1*5^1, so both have prime signature {1,1}. Thus, 14 is a term.
		

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• D. A. Goldston, S. W. Graham, J. Pintz, and C. Y. Yıldırım, Small gaps between almost primes, the parity problem, and some conjectures of Erdos on consecutive integers, arXiv:0803.2636 [math.NT], 2008.
• MathOverflow, Question on consecutive integers with similar prime factorizations
• Eric Weisstein's MathWorld, Prime Signature
• Wikipedia, Prime signature

Crossrefs

Cf. A005237, A189982, A260143.

Programs

Mathematica

pri[n_] := Sort[ Transpose[ FactorInteger[n]] [[2]]]; Select[ Range[ 2, 1000], pri[#] == pri[#+1] &]
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 *)

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

Python

from sympy import factorint
def aupto(limit):
    alst, prevsig = [], [1]
    for k in range(3, limit+2):
        sig = sorted(factorint(k).values())
        if sig == prevsig: alst.append(k - 1)
        prevsig = sig
    return alst
print(aupto(250)) # Michael S. Branicky, Sep 20 2021