%I #7 Nov 25 2012 23:49:09
%S 1,1,1,2,2,1,2,3,2,3,3,2,3,3,3,2,2,3,3,3,2,3,4,1,2,3,3,3,2,3,4,3,3,3,
%T 5,2,3,3,3,2,3,4,3,3,3,5,3,4,4,4,5,1,2,3,3,3,2,3,4,3,3,3,5,3,4,4,4,5,
%U 3,3,3,5,5,3,5,5,3,2,3,3,3,2,3,4,3,3,3
%N Run lengths in A219642.
%C a(n) tells from how many starting values one can end to 0 in n steps, with the iterative process described in A219642 (if going around in 0->0 loop is disallowed).
%H A. Karttunen, <a href="/A219644/b219644.txt">Table of n, a(n) for n = 0..10000</a>
%F a(n) = A219643(n+1)-A219643(n). (The first differences of A219643).
%o (Scheme, with two different variants):
%o (define (A219644 n) (- (A219643 (1+ n)) (A219643 n)))
%o (define (A219644v2 n) (1+ (- (A219645 n) (A219643 n))))
%Y a(n) = 1+(A219645(n)-A219643(n)).
%Y This sequence is based on Fibonacci number system (Zeckendorf expansion): A014417. Analogous sequence for binary system: A086876, for factorial number system: A219654.
%K nonn
%O 0,4
%A _Antti Karttunen_, Nov 24 2012