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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
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
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
KEYWORD
nonn
AUTHOR
Labos Elemer, Apr 13 2006
EXTENSIONS
Mathematica program corrected by Harvey P. Dale, Dec 28 2012
STATUS
approved