A247940 - OEIS

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

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OFFSET

1,1

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.

LINKS

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

EXAMPLE

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

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}]

CROSSREFS

Cf. A000032, A247824, A247937.

Sequence in context: A294750 A304300 A321653 * A061200 A255304 A339337

Adjacent sequences: A247937 A247938 A247939 * A247941 A247942 A247943

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Sep 27 2014

STATUS

approved