

A247940


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


11



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(n1) 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

ZhiWei 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

ZhiWei Sun, Sep 27 2014


STATUS

approved



