

A297573


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


2



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
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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 ZhiWei Sun)
ZhiWei 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 FrÃ¶hlich, Jan 01 2018


CROSSREFS

Cf. A000045, A247937, A248123, A297574.
Sequence in context: A070832 A170994 A151599 * A159730 A264944 A242855
Adjacent sequences: A297570 A297571 A297572 * A297574 A297575 A297576


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jan 01 2018


STATUS

approved



