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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. 8

%I #21 Oct 02 2014 18:46:24

%S 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 29,28,27,26,25,49,11,29,133,20,19,22,71,70,7,18,13,46,11,14,25,24,23,

%U 93,45,80,43,67,29,286,171,102,97,38

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

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

%C 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 Zhi-Wei Sun, <a href="/A248125/b248125.txt">Table of n, a(n) for n = 1..10000</a>

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

%t 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 (PARI)

%o a(n)=m=1;while((binomial(2*m,m)+binomial(2*n,n))%(m+n),m++);m

%o vector(100,n,a(n)) \\ _Derek Orr_, Oct 01 2014

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

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Oct 01 2014

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Last modified September 4 16:58 EDT 2024. Contains 375685 sequences. (Running on oeis4.)