A248125 Least positive integer m such that m + n divides C(2m,m) + C(2n,n), where C(2k,k) = (2k)!/(k!)^2.

1, 2, 5, 16, 3, 6, 2, 22, 101, 6, 21, 86, 43, 16, 15, 4, 3, 6, 21, 20, 11, 8, 49, 48, 7, 22, 29, 28, 27, 26, 25, 49, 11, 29, 133, 20, 19, 22, 71, 70, 7, 18, 13, 46, 11, 14, 25, 24, 23, 93, 45, 80, 43, 67, 29, 286, 171, 102, 97, 38

Offset: 1

Author

Zhi-Wei Sun, Oct 01 2014

Keywords

• nonn

Comments

Conjecture: a(n) exists for all n > 0. Moreover, for n > 66 we have a(n) < n except for n = 364, 408.

a(n) = n for n = 1, 2, 6, 15, 20, 28, 66, ... The next term, if it exists, is greater than 10^4. - Derek Orr, Oct 01 2014

Examples

			a(3) = 5 since 3 + 5 = 8 divides C(6,3) + C(10,5) = 20 + 252 = 272.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A247824, A247937, A247940, A248123, A248124.

Programs

Mathematica

Do[m=1;Label[aa];If[Mod[Binomial[2m,m]+Binomial[2n,n],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]

PARI

a(n)=m=1;while((binomial(2*m,m)+binomial(2*n,n))%(m+n),m++);m
vector(100,n,a(n)) \\ Derek Orr, Oct 01 2014