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 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 (list; graph; refs; listen; history; text; internal format)
 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 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(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 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

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Last modified January 26 16:14 EST 2023. Contains 359833 sequences. (Running on oeis4.)