A233286 Number of trailing zeros in the factorial base representation of n-th Fibonacci number.

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

Offset: 1

Author

Antti Karttunen, Dec 07 2013

Keywords

• nonn
• base

Comments

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

Examples

			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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..100000
• Index entries for sequences related to factorial base representation.

Crossrefs

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

Programs

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 *)

PARI

A233286(n) = if(n<=2, 0, my(f=fibonacci(n)); for(k=2,oo,if(f%(k!), return(k-2)))); \\ Antti Karttunen, Jan 18 2025

Scheme

(define (A233286 n) (A230403 (A000045 n)))

Formula

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

Extensions

Data section extended to a(105) by Antti Karttunen, Jan 18 2025