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%I #2 Mar 31 2012 13:20:33
%S 1,3,6,5,8,7,7,11,8,9,9,12,7,11,12,11,16,8,13,12,13,16,8,7,8,8,12,6,8,
%T 14,13,5,16,13,11,19,16,8,20,19,15,11,12,13,7,9,8,9,14,6,12,11,13,20,
%U 18,13,9,12,14,13,14,11,13,14,13,13,16,13,10,10,17,20,10,13,10,20,11,19,17
%N a(n) = number of primes of the form (2n+1)!! - 2^k.
%C a(n) is the lengths of n-th row of the table below. Table of numbers k such that (2n+1)!! - 2^k is prime: {0}, {1,2,3}, {1,2,3,4,5,6}, {2,3,4,6,9}, {2,6,7,8,9,10,12,13}, {2,4,7,11,13,14,15}, {1,2,8,11,16,18,20}, {1,4,6,10,12,16,18,19,22,23,24},...
%e a(1) = 1 because there is only one prime of the form 3!! - 2^k = 3!! - 2^0 = 2.
%e a(2) = 3 because there are three primes of the form 5!! - 2^k: 5!! - 2^1 = 13, 5!! - 2^2 = 11 and 5!! - 2^3 = 7.
%t Table[Length[Select[Range[0,Floor[Log[2,(2n+1)!! ]]],PrimeQ[(2n+1)!!-2^# ]&]],{n,1,100}]
%K nonn
%O 1,2
%A _Alexander Adamchuk_, Nov 17 2006
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