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

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%I A118112 #15 May 10 2018 02:48:01
%S A118112 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 A118112 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 A118112 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 A118112 a(n) = binomial(3n,n) mod (n+1).
%C A118112 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 A118112 If n+1 is in A061345, a(n)=0. This follows from Kummer's theorem. - _Robert Israel_, May 09 2018
%H A118112 Robert Israel, <a href="/A118112/b118112.txt">Table of n, a(n) for n = 1..10000</a>
%H A118112 Wikipedia, <a href="https://en.wikipedia.org/wiki/Kummer%27s_theorem">Kummer's theorem</a>
%F A118112 a(n) = binomial(3n,n) mod (n+1).
%e A118112 For n=9, binomial(27,7) = 4686825; 4686825 mod 10 = 5.
%p A118112 seq(binomial(3*n,n) mod (n+1), n=1..200); # _Robert Israel_, May 09 2018
%t A118112 Table[Mod[Binomial[3*k,k],k+1],{k,500}]
%o A118112 (PARI) a(n) = binomial(3*n, n) % (n+1); \\ _Michel Marcus_, May 10 2018
%Y A118112 Cf. A000108, A061345, A118113.
%K A118112 nonn
%O A118112 1,5
%A A118112 _Labos Elemer_, Apr 13 2006
%E A118112 Mathematica program corrected by _Harvey P. Dale_, Dec 28 2012