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

1,4

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

Cf. A000045, A247937, A248123, A297574.

Sequence in context: A070832 A170994 A151599 * A346566 A159730 A264944

Adjacent sequences: A297570 A297571 A297572 * A297574 A297575 A297576

Keyword

nonn

Author

Zhi-Wei Sun, Jan 01 2018

Status

approved