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%I #8 Jan 15 2015 11:40:19
%S 0,0,1,0,2,1,3,0,1,2,4,1,5,3,2,0,6,1,7,2,1,4,8,1,2,5,3,3,9,2,10,0,2,6,
%T 3,1,11,7,4,2,12,1,13,4,1,8,14,1,3,2,2,5,15,3,2,3,3,9,16,2,17,10,5,0,
%U 4,2,18,6,2,3,19,1,20,11,6,7,4,4,21,2,4,12,22,1,3,13,3,4,23,1,3,8,1,14,5,1,24,3,7,2,25
%N a(1) = 0; after which, a(2n) = a(n), a(2n+1) = 1 + a(A250470(n)).
%C Consider the binary tree illustrated in A252753 and A252755: If we start from any n, computing successive iterations of A253554 until 1 is reached (i.e., we are traversing level by level towards the root of the tree, starting from that vertex of the tree where n is located at), a(n) gives the number of odd numbers > 1 encountered on the path (i.e., excluding the final 1 from the count but including the starting n if it was odd).
%H Antti Karttunen, <a href="/A253556/b253556.txt">Table of n, a(n) for n = 1..8192</a>
%F a(1) = 0; after which, a(2n) = a(n), a(2n+1) = 1 + a(A250470(n)).
%F a(n) = A253555(n) - A253557(n).
%F a(n) = A253558(n) - 1.
%F a(n) = A080791(A252754(n)). [Number of nonleading 0-bits in A252754(n).]
%F Other identities. For all n >= 2:
%F a(n) = A000120(A252756(n)) - 1. [One less than the binary weight of A252756(n).]
%o (Scheme, with memoization-macro definec)
%o (definec (A253556 n) (cond ((= 1 n) 0) ((odd? n) (+ 1 (A253556 (A250470 n)))) (else (A253556 (/ n 2)))))
%Y One less than A253558.
%Y Powers of two, A000079, gives the positions of zeros.
%Y Cf. A000120, A080791, A252753, A252754, A252755, A252756, A253554, A253555, A253557, A253559.
%Y Differs from A252735 for the first time at n=21, where a(21) = 1, while A252735(21) = 3.
%K nonn
%O 1,5
%A _Antti Karttunen_, Jan 12 2015
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