

A346549


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


0



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
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OFFSET

1,1


COMMENTS

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)


LINKS

Table of n, a(n) for n=1..66.
Wikipedia, Prime signature


EXAMPLE

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


MATHEMATICA

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 *)


PROG

(PARI) consecutive(n)=my(f=vecsort(factor(n)[, 2])); f==vecsort(factor(n1)[, 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, ", ")))
(Python)
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


CROSSREFS

Intersection of A006881 and A260143.
Subsequence of A038456.
Sequence in context: A220671 A136012 A038456 * A075658 A047821 A195238
Adjacent sequences: A346546 A346547 A346548 * A346550 A346551 A346552


KEYWORD

nonn


AUTHOR

Ositadima Chukwu, Sep 16 2021


STATUS

approved



