A118112 - OEIS

%I #15 May 10 2018 02:48:01

%S 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,

%T 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,

%U 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

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

%C 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.

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

%H Robert Israel, <a href="/A118112/b118112.txt">Table of n, a(n) for n = 1..10000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Kummer%27s_theorem">Kummer's theorem</a>

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

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

%p seq(binomial(3*n,n) mod (n+1), n=1..200); # _Robert Israel_, May 09 2018

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

%o (PARI) a(n) = binomial(3*n, n) % (n+1); \\ _Michel Marcus_, May 10 2018

%Y Cf. A000108, A061345, A118113.

%K nonn

%O 1,5

%A _Labos Elemer_, Apr 13 2006

%E Mathematica program corrected by _Harvey P. Dale_, Dec 28 2012