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

1,1

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.

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.

Formula

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

Example

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.

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

Crossrefs

Cf. A003261, A014662, A014664.

Sequence in context: A356029 A213093 A135312 * A266096 A005035 A340452

Adjacent sequences: A287142 A287143 A287144 * A287146 A287147 A287148

Keyword

nonn

Author

Amiram Eldar, May 20 2017

Status

approved