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 A118112 a(n) = binomial(3n,n) mod (n+1). 3
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 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 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 Wikipedia, Kummer's theorem FORMULA a(n) = binomial(3n,n) mod (n+1). EXAMPLE For n=9, binomial(27,7) = 4686825; 4686825 mod 10 = 5. 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}] PROG (PARI) a(n) = binomial(3*n, n) % (n+1); \\ Michel Marcus, May 10 2018 CROSSREFS Cf. A000108, A061345, A118113. Sequence in context: A122480 A096133 A293381 * A245552 A195938 A184762 Adjacent sequences:  A118109 A118110 A118111 * A118113 A118114 A118115 KEYWORD nonn AUTHOR Labos Elemer, Apr 13 2006 EXTENSIONS Mathematica program corrected by Harvey P. Dale, Dec 28 2012 STATUS approved

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Last modified February 24 01:16 EST 2020. Contains 332195 sequences. (Running on oeis4.)