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%I #18 Jan 15 2022 10:01:17
%S 10,6,84,12,16,7,27,9,144,30,28,12,8,30,14,18,57,19,342,18,20,24,66,
%T 12,9,27,144,60,112,35,16,24,60,55,20,12,40,111,24,36,88,72,80,48,10,
%U 15,72,24,224,18,50,54,270,72,54,33,224,18,28,12
%N Least positive integer m such that m + n divides F(m), where F(m) is the m-th Fibonacci number given by A000045.
%C Conjecture: a(n) exists for any n > 0. Moreover, a(n) <= n*(n-1) except for n = 1, 2, 3, 9.
%C In contrast, it is easy to show that for any integer n > 0, there is a positive integer m such that m + n divides 2^m - 1.
%C a(n) exists for any n > 0. See Bloom (1998). - _Amiram Eldar_, Jan 15 2022
%H Zhi-Wei Sun, <a href="/A248593/b248593.txt">Table of n, a(n) for n = 1..10000</a>
%H David M. Bloom, <a href="https://www.fq.math.ca/Scanned/36-1/elementary36-1.pdf">Offset Entries</a>, Solution to Problem B-830, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 36, No. 1 (1998), pp. 89-90.
%e a(1) = 10 since 10 + 1 = 11 divides F(10) = 55.
%t Do[m=1;Label[aa];If[Mod[Fibonacci[m],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
%Y Cf. A000045, A247937, A247940, A248588, A248590.
%K nonn
%O 1,1
%A _Zhi-Wei Sun_, Oct 09 2014
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