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A280686 Largest Fibonacci proper divisor of n, a(1) = 1. 6

%I #16 Jun 20 2017 01:55:34

%S 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,

%T 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,

%U 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

%N Largest Fibonacci proper divisor of n, a(1) = 1.

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

%H Antti Karttunen, <a href="/A280686/b280686.txt">Table of n, a(n) for n = 1..10946</a>

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

%F Other identities. For all n >= 1:

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

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

%e 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.

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

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

%o (Scheme)

%o ;; A stand-alone program:

%o (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)))))

%o (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

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

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

%K nonn

%O 1,4

%A _Antti Karttunen_, Jan 11 2017

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Last modified March 28 13:42 EDT 2024. Contains 371254 sequences. (Running on oeis4.)