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A280686 Largest Fibonacci proper divisor of n, a(1) = 1. 5
1, 1, 1, 2, 1, 3, 1, 2, 3, 5, 1, 3, 1, 2, 5, 8, 1, 3, 1, 5, 3, 2, 1, 8, 5, 13, 3, 2, 1, 5, 1, 8, 3, 2, 5, 3, 1, 2, 13, 8, 1, 21, 1, 2, 5, 2, 1, 8, 1, 5, 3, 13, 1, 3, 5, 8, 3, 2, 1, 5, 1, 2, 21, 8, 13, 3, 1, 34, 3, 5, 1, 8, 1, 2, 5, 2, 1, 13, 1, 8, 3, 2, 1, 21, 5, 2, 3, 8, 1, 5, 13, 2, 3, 2, 5, 8, 1, 2, 3, 5, 1, 34, 1, 13, 21, 2, 1, 3, 1, 55, 3, 8, 1, 3, 5, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

For n > 1, a(n) = greatest Fibonacci number that divides n and is less than n.

LINKS

Antti Karttunen, Table of n, a(n) for n = 1..10946

FORMULA

a(n) = n / A280687(n).

Other identities. For all n >= 1:

a(A000045(n)) = A105800(n).

a(A001690(n)) = A054494(A001690(n)).

EXAMPLE

For n=3, the greatest Fibonacci number that divides 3 and is less than 3 is A000045(1)=A000045(2)=1, thus a(3) = 1.

For n=20, the greatest Fibonacci number that divides 20 and is less than 20 is A000045(5)=5, thus a(20) = 5.

For n=21, the greatest Fibonacci number that divides 21 and is less than 21 is A000045(4)=3, thus a(21) = 3.

PROG

(Scheme)

;; A stand-alone program:

(define (A280686 n) (let loop ((f1 1) (f2 1) (lpd 1)) (cond ((>= f2 n) lpd) ((zero? (modulo n f2)) (loop f2 (+ f1 f2) f2)) (else (loop f2 (+ f1 f2) lpd)))))

(PARI) a(n)=my(r=1, lim=if(n%2, n\3, n/2), a=1, b=2); while(b<n, if(n%b==0, r=b); [a, b]=[b, a+b]); r \\ Charles R Greathouse IV, Jun 20 2017

CROSSREFS

Cf. A000045, A032742, A105800, A280687, A280696.

Cf. A001690 (gives the positions n > 1 where this sequence and A054494 obtain equal values).

Sequence in context: A128487 A056609 A014673 * A085392 A089384 A228812

Adjacent sequences:  A280683 A280684 A280685 * A280687 A280688 A280689

KEYWORD

nonn

AUTHOR

Antti Karttunen, Jan 11 2017

STATUS

approved

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Last modified September 18 16:42 EDT 2020. Contains 337170 sequences. (Running on oeis4.)