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Largest Lucas proper divisor of n, a(1) = a(2) = 1.
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%I #15 Jan 12 2017 07:19:03

%S 1,1,1,2,1,3,1,4,3,2,1,4,1,7,3,4,1,3,1,4,7,11,1,4,1,2,3,7,1,3,1,4,11,

%T 2,7,18,1,2,3,4,1,7,1,11,3,2,1,4,7,2,3,4,1,18,11,7,3,29,1,4,1,2,7,4,1,

%U 11,1,4,3,7,1,18,1,2,3,4,11,3,1,4,3,2,1,7,1,2,29,11,1,18,7,4,3,47,1,4,1,7,11,4,1,3,1,4,7,2,1,18,1,11,3,7,1,3,1

%N Largest Lucas proper divisor of n, a(1) = a(2) = 1.

%C For n > 1, a(n) = greatest Lucas number (A000032) that divides n and is less than n.

%H Antti Karttunen, <a href="/A280696/b280696.txt">Table of n, a(n) for n = 1..15127</a>

%F a(n) = n / A280697(n).

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

%F a(A057854(n)) = A280694(A057854(n)).

%F a(A000204(n)) = A280698(n).

%o (Scheme)

%o ;; A stand-alone program:

%o (define (A280696 n) (let loop ((l1 1) (l2 3) (lpd 1)) (cond ((>= l1 n) (if (and (= 1 lpd) (even? n) (> n 2)) 2 lpd)) ((zero? (modulo n l1)) (loop l2 (+ l1 l2) l1)) (else (loop l2 (+ l1 l2) lpd)))))

%Y Cf. A000032, A000045, A000204, A280686, A280697, A280698, A280699.

%Y Cf. A057854 (gives the positions n > 1 where this sequence and A280694 obtain equal values).

%K nonn

%O 1,4

%A _Antti Karttunen_, Jan 11 2017