A227355 Product of run lengths in Zeckendorf representation of n.

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, 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, 3, 10, 8, 6, 8, 4, 6, 4, 2, 6, 5, 4, 6, 3, 6, 4, 2, 4, 3

Offset: 0

Author

Antti Karttunen, Jul 08 2013

Keywords

• nonn
• base

Comments

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.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10946

Crossrefs

Cf. A167489, A003714, A102364, A014417, A227350.

Programs

Scheme

(define (A227355 n) (A167489 (A003714 n)))(define (A227355v2 n) (A227350 (A003714 n))) ;; Alternative definition.

Formula

a(n) = A167489(A003714(n)) = A227350(A003714(n)).

a(A227352(A005408(n))) = A167489(n).

For n>= 3, a(A000045(n)) = n-2.