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%I #38 Jul 29 2022 17:11:47
%S 14,15,21,22,33,34,35,38,39,57,58,85,86,87,93,94,95,118,119,122,123,
%T 133,134,141,142,143,145,146,158,159,177,178,201,202,203,205,206,213,
%U 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
%N Runs (of length > 1) of consecutive squarefree semiprimes.
%C Runs of length bigger than 3 are impossible as one of four consecutive numbers is divisible by 4.
%C 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.
%C From _Michael S. Branicky_, Sep 21 2021: (Start)
%C This only differs from the supersequence A038456 in the terms 26, 27 (present there but not here).
%C 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)
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Prime_signature">Prime signature</a>
%e 14 and 15 are consecutive and both have prime signature {1, 1}
%t 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 *)
%t With[{sp=If[SquareFreeQ[#]&&PrimeOmega[#]==2,1,0]&/@Range[ 500]},DeleteDuplicates[ Flatten[SequencePosition[sp,{1,1}]]]] (* _Harvey P. Dale_, Jul 29 2022 *)
%o (PARI) consecutive(n)=my(f=vecsort(factor(n)[, 2])); f==vecsort(factor(n-1)[, 2]) || f==vecsort(factor(n+1)[, 2]) \\ based on A260143
%o squarefree_semiprime(n)=(bigomega(n)==2&&omega(n)==2) \\ based on A006881
%o for(n=1, 500, if(squarefree_semiprime(n) && consecutive(n), print1(n, ", ")))
%o (Python)
%o from sympy import factorint
%o def aupto(limit):
%o aset, prevsig = set(), [1]
%o for k in range(3, limit+2):
%o sig = sorted(factorint(k).values())
%o if sig == prevsig == [1, 1]: aset.update([k - 1, k])
%o prevsig = sig
%o return sorted(aset)
%o print(aupto(482)) # _Michael S. Branicky_, Sep 20 2021
%Y Intersection of A006881 and A260143.
%Y Subsequence of A038456.
%K nonn
%O 1,1
%A _Ositadima Chukwu_, Sep 16 2021
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