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%I #11 Jul 20 2013 03:31:24
%S 1,1,1,2,1,3,2,1,4,3,2,2,1,5,4,3,4,2,3,2,1,6,5,4,6,3,6,4,2,4,3,2,2,1,
%T 7,6,5,8,4,9,6,3,8,6,4,4,2,5,4,3,4,2,3,2,1,8,7,6,10,5,12,8,4,12,9,6,6,
%U 3,10,8,6,8,4,6,4,2,6,5,4,6,3,6,4,2,4,3
%N Product of run lengths in Zeckendorf representation of n.
%C The same sequence also gives the product for the lengths of zero-runs only, as by definition, no two consecutive 1's can occur in Fibonacci number system (aka Zeckendorf representation), thus any 1's present contribute just *1 to the total product.
%H Antti Karttunen, <a href="/A227355/b227355.txt">Table of n, a(n) for n = 0..10946</a>
%F a(n) = A167489(A003714(n)) = A227350(A003714(n)).
%F a(A227352(A005408(n))) = A167489(n).
%F For n>= 3, a(A000045(n)) = n-2.
%o (Scheme) (define (A227355 n) (A167489 (A003714 n)))(define (A227355v2 n) (A227350 (A003714 n))) ;; Alternative definition.
%Y Cf. A167489, A003714, A102364, A014417, A227350.
%K nonn,base
%O 0,4
%A _Antti Karttunen_, Jul 08 2013
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