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A233286
Number of trailing zeros in the factorial base representation of n-th Fibonacci number; a(n) = A230403(A000045(n)) = A233285(n)-1.
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.
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 A375483 A186038
KEYWORD
nonn,base,changed
AUTHOR
Antti Karttunen, Dec 07 2013
STATUS
approved