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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.
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%I #12 May 27 2017 05:26:48

%S 4,13,64,89,83,188,433,701,449,342,1429,1768,1889,2276,3484,2423,5149,

%T 5776,2069,1693,8644,4793,9728,11173,4237,13364,15049,16108,16469,

%U 9455,19501,22364,25876,8929,3131,6524,2311,36313,13017,10114,13582,43069,15962

%N 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 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 Amiram Eldar, <a href="/A287145/b287145.txt">Table of n, a(n) for n = 1..350</a>

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

%e 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 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 Cf. A003261, A014662, A014664.

%K nonn

%O 1,1

%A _Amiram Eldar_, May 20 2017