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

%I A248125 #21 Oct 02 2014 18:46:24
%S A248125 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,
%T A248125 29,28,27,26,25,49,11,29,133,20,19,22,71,70,7,18,13,46,11,14,25,24,23,
%U A248125 93,45,80,43,67,29,286,171,102,97,38
%N A248125 Least positive integer m such that m + n divides C(2m,m) + C(2n,n), where C(2k,k) = (2k)!/(k!)^2.
%C A248125 Conjecture: a(n) exists for all n > 0. Moreover, for n > 66 we have a(n) < n except for n = 364, 408.
%C A248125 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
%H A248125 Zhi-Wei Sun, <a href="/A248125/b248125.txt">Table of n, a(n) for n = 1..10000</a>
%e A248125 a(3) = 5 since 3 + 5 = 8 divides C(6,3) + C(10,5) = 20 + 252 = 272.
%t A248125 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}]
%o A248125 (PARI)
%o A248125 a(n)=m=1;while((binomial(2*m,m)+binomial(2*n,n))%(m+n),m++);m
%o A248125 vector(100,n,a(n)) \\ _Derek Orr_, Oct 01 2014
%Y A248125 Cf. A247824, A247937, A247940, A248123, A248124.
%K A248125 nonn
%O A248125 1,2
%A A248125 _Zhi-Wei Sun_, Oct 01 2014