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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.
1
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
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
KEYWORD
nonn
AUTHOR
Amiram Eldar, May 20 2017
STATUS
approved