OFFSET
1,1
COMMENTS
Conjecture: Let A be any integer not congruent to 3 modulo 6. Define u(0) = 0, u(1) = 1, and u(n+1) = A*u(n) + u(n-1) for n > 0. Then, for any integer n > 0, there are infinitely many positive integers m such that m + n divides u(m) + u(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(2) = 22 since 22 + 2 = 24 divides F(22) + F(2) = 17711 + 1 = 17712 = 24*738.
MATHEMATICA
Do[m=n+1; Label[aa]; If[Mod[Fibonacci[m]+Fibonacci[n], m+n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 1, 60}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 27 2014
STATUS
approved