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

1,1

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

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 (2008)

MathOverflow, Question on consecutive integers with similar prime factorizations

Eric Weisstein's MathWorld, Prime Signature

Wikipedia, Prime signature

Example

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

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 *)

Prog

(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

Crossrefs

Cf. A005237, A189982, A260143.

Sequence in context: A138047 A005237 A140578 * A280074 A359745 A086263

Adjacent sequences: A052210 A052211 A052212 * A052214 A052215 A052216

Keyword

easy,nonn

Author

Erich Friedman, Jan 29 2000

Status

approved