%I #2 Mar 31 2012 14:39:48
%S 1,5,15,23,33,41,53,75,89,99,105,113,153,155,165,189,215,239,249,261,
%T 281,293,323,341,363,371,375,405,411,419,431,473,519,543,545,561,575,
%U 629,659,699,725,741,743,765,785,803,831,849,893,905,915,923,933,935
%N a(n)=A(n,1), the first element of each sequence A(n) defined recursively as follows. Recall that A119751 is the sequence defined recursively by a(1)=1 and a(k) is the first odd number greater than a(k-1) such that 2a(k)+1 is prime and a(k)+a(j)+1 is prime for all 1<=j<k. Let A(1)=A119751, that is, A(1,k)=A119751(k). Then A(n) is the sequence defined recursively as follows: (1) A(n,1) is the first odd number not in any A(m), 1<=m<n, such that 2A(n,1)+1 is prime. (2) A(n,k) is the first odd number greater than A(n,k-1), not in any A(m), 1<=m<n, such that 2A(n,k)+1 is prime. (3) A(n,k)+A(n,j)+1 is prime for all 1<=j<k.
%e a(1)=1 is the first element of A119751=1, 3, 9, 69, 429, 4089, 86529, 513099, ... so a(2)=5 since 5 is the first odd number not in A119751 such that 2*5+1 is prime. Furthermore, A(2)=5, 11, 35, 95, 221, 551, 1271, 5705,...
%Y Cf. A119751, A119752, A127257.
%K nonn
%O 1,2
%A _Walter Kehowski_, Jan 10 2007
|