A260143 -id:A260143 - cp's OEIS frontend

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

Erich Friedman, Jan 29 2000

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

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

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

Cf. A005237, A189982, A260143.

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 *)

lista(nn) = for (n=1, nn-1, if (vecsort(factor(n)[,2]) == vecsort(factor(n+1)[,2]), print1(n, ", "));); \\ Michel Marcus, Jun 10 2015

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

A346549 Runs (of length > 1) of consecutive squarefree semiprimes.

Original entry on oeis.org

14, 15, 21, 22, 33, 34, 35, 38, 39, 57, 58, 85, 86, 87, 93, 94, 95, 118, 119, 122, 123, 133, 134, 141, 142, 143, 145, 146, 158, 159, 177, 178, 201, 202, 203, 205, 206, 213, 214, 215, 217, 218, 219, 253, 254, 298, 299, 301, 302, 303, 326, 327, 334, 335, 381, 382, 393, 394, 395, 445, 446, 447, 453, 454, 481, 482

Offset: 1

Ositadima Chukwu, Sep 16 2021

nonn

Runs of length bigger than 3 are impossible as one of four consecutive numbers is divisible by 4.
The existence of consecutive pairs of such numbers is connected with primes of the form (p*q +- 1)/2 where p and q are odd primes.
From Michael S. Branicky, Sep 21 2021: (Start)
This only differs from the supersequence A038456 in the terms 26, 27 (present there but not here).
Proof. Numbers with 4 divisors are of the form p*q for p, q prime, p != q or of the form r^3 for r prime.  For two such numbers to be consecutive terms of A038456 (but not terms here) requires p*q + 1 = r^3 or p*q = r^3 + 1 for primes p, q, r, p != q.  There is no solution to either form with p, q distinct primes for r = 2. Thus, r^3 is odd and p*q must be even, so wlog p = 2. Thus, we need to solve for case 1: 2*q + 1 = r^3 for q, r prime. But r^3 - 1 = (r - 1)*(r^2 + r + 1), so r = 3 is the only prime solution producing the factor 2, leading to the pair 26, 27. Likewise, case 2: r^3 + 1 = (r + 1)*(r^2 - r + 1) has no solution with prime r producing the required factor 2. (End)

			14 and 15 are consecutive and both have prime signature {1, 1}

Wikipedia, Prime signature

Intersection of A006881 and A260143.

Subsequence of A038456.

s = Union@ Flatten@ Table[Prime[m] Prime[n], {n, Log2[#/3]}, {m, n + 1, PrimePi[#/Prime[n]]}] &[482]; s[[Flatten@ Map[Append[#, Last[#] + 1] &, Position[Differences[s], 1]]]] (* Michael De Vlieger, Oct 28 2021 *)
With[{sp=If[SquareFreeQ[#]&&PrimeOmega[#]==2,1,0]&/@Range[ 500]},DeleteDuplicates[ Flatten[SequencePosition[sp,{1,1}]]]] (* Harvey P. Dale, Jul 29 2022 *)

consecutive(n)=my(f=vecsort(factor(n)[, 2])); f==vecsort(factor(n-1)[, 2]) || f==vecsort(factor(n+1)[, 2]) \\  based on A260143
squarefree_semiprime(n)=(bigomega(n)==2&&omega(n)==2) \\ based on A006881
for(n=1, 500, if(squarefree_semiprime(n) && consecutive(n), print1(n, ", ")))

from sympy import factorint
def aupto(limit):
    aset, prevsig = set(), [1]
    for k in range(3, limit+2):
        sig = sorted(factorint(k).values())
        if sig == prevsig == [1, 1]: aset.update([k - 1, k])
        prevsig = sig
    return sorted(aset)
print(aupto(482)) # Michael S. Branicky, Sep 20 2021

cp's OEIS Frontend

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

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