OFFSET
1,1
COMMENTS
Conjecture: a(n) exists for any n > 0. Moreover, a(n) <= n*(n-1) except for n = 1, 2, 3, 9.
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.
a(n) exists for any n > 0. See Bloom (1998). - Amiram Eldar, Jan 15 2022
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
David M. Bloom, Offset Entries, Solution to Problem B-830, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 36, No. 1 (1998), pp. 89-90.
EXAMPLE
a(1) = 10 since 10 + 1 = 11 divides F(10) = 55.
MATHEMATICA
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 09 2014
STATUS
approved