OFFSET
1,4
COMMENTS
If p is a prime congruent to 2 or 3 modulo 5, then a(p) = 1 since it is known that p divides F(p+1).
Conjecture: a(n) exists for any n > 0.
See also A297574 for a similar conjecture.
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..1000 (n = 1..120 from Zhi-Wei Sun)
Zhi-Wei Sun, Introduction to Lucas sequences, a talk give in 2017.
EXAMPLE
a(2) = 1 since 1*2 divides F(1+2) = F(3) = 2.
a(4) = 2 since 2*4 divides F(2+4) = 8.
a(5) = 14170 since 5*14170 = 70850 divides F(5+14170) = F(14175).
a(6) = 6 since 6*6 = 36 divides F(6+6) = F(12) = 144.
MATHEMATICA
Do[m=1; Label[aa]; If[Mod[Fibonacci[m+n], m*n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 1, 60}]
lpi[n_]:=Module[{k=1}, While[!Divisible[Fibonacci[k+n], k*n], k++]; k]; Array[ lpi, 60] (* Harvey P. Dale, May 05 2018 *)
PROG
(PARI) a(n) = my(m=1); while(1, if(Mod(fibonacci(m+n), m*n)==0, return(m)); m++) \\ Felix Fröhlich, Jan 01 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Jan 01 2018
STATUS
approved