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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.

%I A287145 #12 May 27 2017 05:26:48
%S A287145 4,13,64,89,83,188,433,701,449,342,1429,1768,1889,2276,3484,2423,5149,
%T A287145 5776,2069,1693,8644,4793,9728,11173,4237,13364,15049,16108,16469,
%U A287145 9455,19501,22364,25876,8929,3131,6524,2311,36313,13017,10114,13582,43069,15962
%N 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.
%C A287145 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.
%H A287145 Amiram Eldar, <a href="/A287145/b287145.txt">Table of n, a(n) for n = 1..350</a>
%H A287145 Wilfrid Keller, <a href="http://dx.doi.org/10.1090/S0025-5718-1995-1308456-3">New Cullen Primes</a>, Mathematics of Computation, Vol. 64, No. 212 (October 1995), pp. 1733-1741.
%F A287145 a(n) = (h(p)/2 + 1)*p - 2, where p=A014662(n), and h(p) is the order of 2 modulo p (A014664).
%e A287145 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.
%t A287145 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
%Y A287145 Cf. A003261, A014662, A014664.
%K A287145 nonn
%O A287145 1,1
%A A287145 _Amiram Eldar_, May 20 2017