%I #7 Nov 25 2012 23:56:30
%S 1,1,2,2,2,4,4,4,4,2,4,4,4,4,6,4,4,4,6,6,6,6,6,6,6,6,8,4,2,4,4,4,4,6,
%T 4,4,4,6,6,6,6,6,6,6,6,8,6,8,8,6,4,4,4,6,6,6,6,6,6,6,6,8,6,8,8,6,8,10,
%U 6,6,6,6,6,6,6,8,6,8,8,6,8,10,8,10,12,6
%N Run lengths in A219652.
%C a(n) tells from how many starting values one can end to 0 in n steps, with the iterative process described in A219652 (if going around in 0->0 loop is disallowed).
%H A. Karttunen, <a href="/A219654/b219654.txt">Table of n, a(n) for n = 0..10080</a>
%F a(n) = A219653(n+1)-A219653(n). (The first differences of A219653).
%o (Scheme with two different variants):
%o (define (A219654 n) (- (A219653 (1+ n)) (A219653 n)))
%o (define (A219654v2 n) (1+ (- (A219655 n) (A219653 n))))
%Y a(n) = 1+(A219655(n)-A219653(n)). This sequence is based on Factorial number system: A007623. Analogous sequence for binary system: A086876, for Zeckendorf expansion: A219644. Cf. A219652, A219659, A219666.
%K nonn
%O 0,3
%A _Antti Karttunen_, Nov 25 2012