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%I #28 Sep 23 2020 21:27:20
%S 2,12,18,38,79,75,153,313
%N a(n) is the least number k such that 3^k-2 and 3^k+2 are the product of n prime factors counted with multiplicity.
%C A001222(3^k-2) = A001222(3^k+2).
%C a(9) >= 521, a(10) >= 368, a(11) >= 339, a(12) >= 551. - _Chai Wah Wu_, Sep 23 2020
%e a(1) = 2 because 3^2-2 = 7 and 3^2+2 = 11 are both prime.
%t m = 6; s = Table[0, {m}]; c = 0; k = 1; While[c < m, n = PrimeOmega[3^k - 2]; If[n <= m && s[[n]] == 0, n2 = PrimeOmega[3^k + 2]; If[n2 == n, s[[n]] = k; c++]]; k++]; s (* _Amiram Eldar_, Sep 18 2020 *)
%o (PARI) a(n) = {my(k=1); while((bigomega(3^k-2) != n) || (bigomega(3^k+2) != n), k++); k;} \\ _Michel Marcus_, Sep 14 2020
%Y Cf. A000040, A001222, A001358, A014224.
%K nonn,more
%O 1,1
%A _Zak Seidov_, Sep 13 2020
%E a(8) from _Daniel Suteu_, Sep 14 2020
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