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

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

Author

Ositadima Chukwu, Sep 16 2021

Keywords

• nonn

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)

Examples

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

Links

• Wikipedia, Prime signature

Crossrefs

Intersection of A006881 and A260143.

Subsequence of A038456.

Programs

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 *)

PARI

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, ", ")))

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