

A226603


Let c(n) be the nth 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, 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
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OFFSET

1,3


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?


LINKS

Table of n, a(n) for n=1..56.


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


CROSSREFS

Cf. A226025.
Sequence in context: A282974 A181063 A161324 * A116534 A130533 A231384
Adjacent sequences: A226600 A226601 A226602 * A226604 A226605 A226606


KEYWORD

hard,more,nonn


AUTHOR

Arkadiusz Wesolowski, Jun 13 2013


EXTENSIONS

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


STATUS

approved



