cp's OEIS Frontend

A280696 Largest Lucas proper divisor of n, a(1) = a(2) = 1.

%I A280696 #15 Jan 12 2017 07:19:03
%S A280696 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 A280696 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 A280696 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 A280696 Largest Lucas proper divisor of n, a(1) = a(2) = 1.
%C A280696 For n > 1, a(n) = greatest Lucas number (A000032) that divides n and is less than n.
%H A280696 Antti Karttunen, <a href="/A280696/b280696.txt">Table of n, a(n) for n = 1..15127</a>
%F A280696 a(n) = n / A280697(n).
%F A280696 Other identities. For all n >= 1:
%F A280696 a(A057854(n)) = A280694(A057854(n)).
%F A280696 a(A000204(n)) = A280698(n).
%o A280696 (Scheme)
%o A280696 ;; A stand-alone program:
%o A280696 (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 A280696 Cf. A000032, A000045, A000204, A280686, A280697, A280698, A280699.
%Y A280696 Cf. A057854 (gives the positions n > 1 where this sequence and A280694 obtain equal values).
%K A280696 nonn
%O A280696 1,4
%A A280696 _Antti Karttunen_, Jan 11 2017