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

1,2

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

Links

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

Example

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

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}]

Prog

(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

Crossrefs

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

Sequence in context: A306794 A111790 A309115 * A297473 A286381 A286382

Adjacent sequences: A248122 A248123 A248124 * A248126 A248127 A248128

Keyword

nonn

Author

Zhi-Wei Sun, Oct 01 2014

Status

approved