A226603 Let c(n) be the n-th number in the sequence of odd composite numbers that are not squares of primes, and let p = c(n)*2^k + 1 (with k > 0) and m be the smallest integer satisfying congruence 2^m == 1 (mod p). The number a(n) is the least k such that p is prime and c(n) does not divide m, or 0 if no such value exists.

1, 1, 2, 6, 13, 2, 9, 13, 2744, 2, 1, 93, 2, 1, 19, 15, 6, 6, 168, 6, 13, 2, 5, 1, 26, 91, 3, 6, 1, 5, 10, 18, 1, 293, 250, 11, 1, 41, 30, 5, 1, 8, 16, 4, 2, 497, 176316, 95, 4, 592, 65, 6, 3, 113, 36, 1

Offset: 1

Author

Arkadiusz Wesolowski, Jun 13 2013

Keywords

• hard
• more
• nonn

Comments

Since 78557 is a Sierpinski number, a(31513) = 0. Are there any values of n so that a(n) = 0 and c(n) is not a Sierpinski number?

Crossrefs

Cf. A226025.

Programs

Mathematica

lst = {}; Do[If[! PrimeQ[c] && ! PrimeQ@Sqrt[c], k = 1; While[True, p = c*2^k + 1; If[PrimeQ[p] && ! Divisible[MultiplicativeOrder[2, p], c], AppendTo[lst, k]; Break[]]; k++]], {c, 3, 185, 2}]; lst

Extensions

a(47)-a(56) from Arkadiusz Wesolowski, Jun 16 2013