A297573 Least positive integer m such that m*n divides F(m+n), where F(k) denotes the k-th Fibonacci number A000045(k).

1, 1, 1, 2, 14170, 6, 1, 136, 207, 28340, 979, 12, 1, 322, 385, 368, 1, 306, 17, 19780, 3, 68, 1, 24, 524975, 58, 2889, 92, 13, 3570, 12749, 736, 7, 2, 165, 612, 1, 34, 633, 13160, 339, 6, 1, 1846, 5355, 2, 1, 336, 8183, 509950, 21, 116, 1, 918, 4895, 184, 51, 26, 10207, 7140

Offset: 1

Author

Zhi-Wei Sun, Jan 01 2018

Keywords

• nonn

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.

Examples

			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.
		

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.

Crossrefs

Cf. A000045, A247937, A248123, A297574.

Programs

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 *)

PARI

a(n) = my(m=1); while(1, if(Mod(fibonacci(m+n), m*n)==0, return(m)); m++) \\ Felix Fröhlich, Jan 01 2018