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%I #26 May 05 2018 15:48:53
%S 1,1,1,2,14170,6,1,136,207,28340,979,12,1,322,385,368,1,306,17,19780,
%T 3,68,1,24,524975,58,2889,92,13,3570,12749,736,7,2,165,612,1,34,633,
%U 13160,339,6,1,1846,5355,2,1,336,8183,509950,21,116,1,918,4895,184,51,26,10207,7140
%N Least positive integer m such that m*n divides F(m+n), where F(k) denotes the k-th Fibonacci number A000045(k).
%C 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).
%C Conjecture: a(n) exists for any n > 0.
%C See also A297574 for a similar conjecture.
%H Chai Wah Wu, <a href="/A297573/b297573.txt">Table of n, a(n) for n = 1..1000</a> (n = 1..120 from Zhi-Wei Sun)
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/IntroLucasSeq.pdf">Introduction to Lucas sequences</a>, a talk give in 2017.
%e a(2) = 1 since 1*2 divides F(1+2) = F(3) = 2.
%e a(4) = 2 since 2*4 divides F(2+4) = 8.
%e a(5) = 14170 since 5*14170 = 70850 divides F(5+14170) = F(14175).
%e a(6) = 6 since 6*6 = 36 divides F(6+6) = F(12) = 144.
%t 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}]
%t lpi[n_]:=Module[{k=1},While[!Divisible[Fibonacci[k+n],k*n],k++];k]; Array[ lpi,60] (* _Harvey P. Dale_, May 05 2018 *)
%o (PARI) a(n) = my(m=1); while(1, if(Mod(fibonacci(m+n), m*n)==0, return(m)); m++) \\ _Felix Fröhlich_, Jan 01 2018
%Y Cf. A000045, A247937, A248123, A297574.
%K nonn
%O 1,4
%A _Zhi-Wei Sun_, Jan 01 2018
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