A287145 Smallest k such that both of the consecutive Woodall numbers A003261(k) and A003261(k+1) are divisible by A014662(n), the n-th prime p with even order of 2 mod p.

4, 13, 64, 89, 83, 188, 433, 701, 449, 342, 1429, 1768, 1889, 2276, 3484, 2423, 5149, 5776, 2069, 1693, 8644, 4793, 9728, 11173, 4237, 13364, 15049, 16108, 16469, 9455, 19501, 22364, 25876, 8929, 3131, 6524, 2311, 36313, 13017, 10114, 13582, 43069, 15962

Offset: 1

Author

Amiram Eldar, May 20 2017

Keywords

• nonn

Comments

Keller proved that the occurrence of 2 consecutive Woodall numbers that are divisible by the same prime is restricted to primes p with even h(p), the order of 2 mod p, and that there are an infinity of such pairs.

Examples

			11 is the 3rd prime p with even order of 2 mod p. A003261(k)=k*2^k-1 is divisible by 11 for k = 16,48,61,64,65,73,79,100,... The first occurrence of 2 consecutive numbers is 64 and 65, thus a(3) = 64.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..350
• Wilfrid Keller, New Cullen Primes, Mathematics of Computation, Vol. 64, No. 212 (October 1995), pp. 1733-1741.

Crossrefs

Cf. A003261, A014662, A014664.

Programs

Mathematica

a = {}; For[p=0, p<=11699, p++; If[!PrimeQ[p], Continue[]]; h=MultiplicativeOrder[2, p]; If[!EvenQ[h], Continue[]]; n=(h/2+1)*p-2; a = AppendTo[a, n]]; a

Formula

a(n) = (h(p)/2 + 1)*p - 2, where p=A014662(n), and h(p) is the order of 2 modulo p (A014664).