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A253559 a(1) = 0; for n>1: a(n) = A253557(n) - 1.

%I A253559 #9 Jan 15 2015 11:40:48
%S A253559 0,0,0,1,0,1,0,2,1,1,0,2,0,1,1,3,0,2,0,2,2,1,0,3,1,1,1,2,0,2,0,4,2,1,
%T A253559 1,3,0,1,1,3,0,3,0,2,3,1,0,4,1,2,2,2,0,2,2,3,2,1,0,3,0,1,1,5,1,3,0,2,
%U A253559 3,2,0,4,0,1,1,2,1,2,0,4,2,1,0,4,2,1,2,3,0,4,2,2,4,1,1,5,0,2,1,3,0,3,0,3,3,1,0,3,0,3,1,4,0,3,3,2,3,1,1,4,1,1,3,2,2,2,0,6
%N A253559 a(1) = 0; for n>1: a(n) = A253557(n) - 1.
%C A253559 Consider the binary trees 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), a(n) gives the number of even numbers > 2 encountered on the path (i.e., excluding the 2 from the count but including the starting n if it was even).
%H A253559 Antti Karttunen, <a href="/A253559/b253559.txt">Table of n, a(n) for n = 1..8192</a>
%F A253559 a(n) = A080791(A252756(n)). [Number of nonleading 0-bits in A252756(n).]
%F A253559 a(1) = 0; for n>1: a(n) = A253557(n) - 1.
%F A253559 Other identities. For all n >= 2:
%F A253559 a(n) = A000120(A252754(n)) - 1. [One less than the binary weight of A252754(n).]
%F A253559 a(n) = A253555(n) - A253558(n).
%o A253559 (Scheme) (define (A253559 n) (if (= 1 n) 0 (+ -1 (A253557 n))))
%Y A253559 Essentially, one less than A253557.
%Y A253559 A008578 gives the positions of zeros.
%Y A253559 Cf. A000120, A080791, A252753, A252754, A252755, A252756, A253554, A253555, A253556, A253558.
%Y A253559 Differs from A252736 for the first time at n=21, where a(21) = 2, while A252736(21) = 1.
%K A253559 nonn
%O A253559 1,8
%A A253559 _Antti Karttunen_, Jan 12 2015