A247940 Least integer m > n such that m + n divides L(m) + L(n), where L(k) refers to the Lucas number A000032(k).

5, 5, 15, 5, 19, 30, 17, 19, 15, 13, 13, 24, 35, 236, 33, 34, 31, 90, 29, 23, 27, 25, 25, 84, 47, 80, 45, 190, 43, 54, 41, 35, 39, 1216, 37, 72, 59, 212, 57, 43, 55, 66, 53, 86, 51, 76, 49, 60, 71, 53, 69, 55, 67, 222, 65, 122, 63, 112, 61, 264

Offset: 1

Author

Zhi-Wei Sun, Sep 27 2014

Keywords

• nonn

Comments

Conjecture: Let A be any integer not congruent to 3 modulo 6. Define v(0) = 2, v(1) = A, and v(n+1) = A*v(n) + v(n-1) for n > 0. Then, for any integer n > 0, there are infinitely many positive integers m such that m + n divides v(m) + v(n).

This implies that a(n) exists for any n > 0.

Examples

			 a(3) = 15 since 15 + 3 = 18 divides L(15) + L(3) = 1364 + 4 = 18*76.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000032, A247824, A247937.

Programs

Mathematica

Do[m=n+1;Label[aa];If[Mod[LucasL[m]+LucasL[n],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]