A118112 a(n) = binomial(3n,n) mod (n+1).

1, 0, 0, 0, 3, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 9, 0, 0, 0, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 17, 0, 0, 0, 19, 0, 0, 0, 21, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 33, 0, 0, 0, 35, 0, 0, 0, 37, 0, 0, 0, 0, 0, 0, 0, 41, 0, 0, 0, 43, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Author

Labos Elemer, Apr 13 2006

Keywords

• nonn

Comments

These divisibilities are analogous to those of Catalan numbers. For rather long sequences of consecutive integers, a(n)=0. For the first 10000 integers 9678 residues equals zero. See A118113.

If n+1 is in A061345, a(n)=0. This follows from Kummer's theorem. - Robert Israel, May 09 2018

Examples

			For n=9, binomial(27,7) = 4686825; 4686825 mod 10 = 5.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Wikipedia, Kummer's theorem

Crossrefs

Cf. A000108, A061345, A118113.

Programs

Maple

seq(binomial(3*n,n) mod (n+1), n=1..200); # Robert Israel, May 09 2018

Mathematica

Table[Mod[Binomial[3*k,k],k+1],{k,500}]

PARI

a(n) = binomial(3*n, n) % (n+1); \\ Michel Marcus, May 10 2018

Formula

a(n) = binomial(3n,n) mod (n+1).

Extensions

Mathematica program corrected by Harvey P. Dale, Dec 28 2012