%I #6 Apr 09 2014 10:16:43
%S 1,2,6,12,20,13,14,56,72,41,43,44,108,29,30,240,272,145,147,83,84,86,
%T 183,184,408,410,218,220,476,61,62,992,1056,545,547,291,293,613,615,
%U 1256,1320,169,171,172,364,366,751,752,1584,1586,818,820,1716,437,439
%N a(n) = the smallest positive integer m such that A152221(m) = n.
%e 12 represented in binary is 1100. A152221(12) = 3 because the binary representation of 3 (which is 11) occurs in 1100 (like so: (11)00), the binary representation of 4 (which is 100) occurs in 1100 (like so: 1(100)) and no larger pair of consecutive integers occurs within 1100. Since no positive integer m, m < 12, is such that A152221(m) = 3, then a(3) = 12.
%Y A152221
%K base,nonn
%O 0,2
%A _Leroy Quet_, Nov 29 2008
%E Extended by _Ray Chandler_, Dec 05 2008