A233286 Number of trailing zeros in the factorial base representation of n-th Fibonacci number; a(n) = A230403(A000045(n)) = A233285(n)-1.

0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 5, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1

Offset

1,12

Comments

A233285 is the main entry for this topic, see comments there.

Links

Table of n, a(n) for n=1..87.

Index entries for sequences related to factorial base representation.

Formula

a(n) = A230403(A000045(n)) = A233285(n)-1.

Example

The factorial base representation (A007623(A000045(n)) of Fibonacci numbers look like this, from n=1 onward: 1, 1, 10, 11, 21, 110, 201, 311, 1120, 2101, 3221, 11000, 14221, 30221, 50120, 121011, 211201, 332220, ...
When we count the trailing zeros of each, we get 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 3, 0, 0, 1, 0, 0, 1, ..., the first terms of this sequence.

Mathematica

a[n_] := Module[{k = Fibonacci[n], m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; FirstPosition[s, _?(# > 0&)][[1]] - 1]; Array[a, 100] (* Amiram Eldar, Feb 21 2024 *)

Prog

(Scheme)
(define (A233286 n) (A230403 (A000045 n)))
;; Or alternatively:
(define (A233286 n) (- (A233285 n) 1))

Crossrefs

One less than A233285. Cf. also A007623, A000045, A230403.

Sequence in context: A307801 A186717 A344833 * A305802 A186038 A091009

Adjacent sequences: A233283 A233284 A233285 * A233287 A233288 A233289

Keyword

nonn,base

Author

Antti Karttunen, Dec 07 2013

Status

approved