A248593 Least positive integer m such that m + n divides F(m), where F(m) is the m-th Fibonacci number given by A000045.

10, 6, 84, 12, 16, 7, 27, 9, 144, 30, 28, 12, 8, 30, 14, 18, 57, 19, 342, 18, 20, 24, 66, 12, 9, 27, 144, 60, 112, 35, 16, 24, 60, 55, 20, 12, 40, 111, 24, 36, 88, 72, 80, 48, 10, 15, 72, 24, 224, 18, 50, 54, 270, 72, 54, 33, 224, 18, 28, 12

Offset: 1

Author

Zhi-Wei Sun, Oct 09 2014

Keywords

• nonn

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

Examples

			a(1) = 10 since 10 + 1 = 11 divides F(10) = 55.
		

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.

Crossrefs

Cf. A000045, A247937, A247940, A248588, A248590.

Programs

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}]