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%I #7 Mar 11 2014 01:32:51
%S 0,0,1,0,2,0,1,0,3,0,1,1,2,1,1,0,4,0,1,2,2,0,2,1,3,2,2,0,2,1,1,0,5,0,
%T 1,3,2,1,3,2,3,1,1,1,3,0,2,1,4,3,3,0,3,1,1,1,3,2,2,1,2,1,1,0,6,0,1,4,
%U 2,2,4,3,3,0,2,2,4,1,3,2,4,2,2,1,2,0,2,2,4,1,1,2,3,0,2,1,5,4,4,0,4,1,1,2,4
%N a(n) = the minimum number of 0's that, if removed from the binary representation of n, leaves a palindrome.
%C a(2^m) = m, for all m >= 0.
%C a(2^m-1) = 0 for all m >= 0.
%C If 2^k is the largest power of 2 that divides n, then a(n) >= k.
%e 20 in binary is 10100. This is not a palindrome, so a(20) > 0. Removing one 0 gets either 1100 or 1010 (the latter in two ways). Neither of these is a palindrome, so a(20)>1. But removing the last two 0's so that we have 101 does indeed leave a palindrome. So a(20) = 2.
%K base,nonn
%O 0,5
%A _Leroy Quet_, Mar 18 2010
%E Extended by _D. S. McNeil_, May 10 2010
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