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A118114
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a(n) = binomial(3n,n) mod((n+1)(n+2)).
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2
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3, 3, 4, 15, 21, 28, 0, 81, 55, 99, 0, 0, 84, 120, 0, 153, 171, 285, 0, 231, 253, 0, 360, 0, 0, 0, 0, 522, 0, 496, 0, 561, 833, 945, 0, 703, 741, 156, 0, 861, 903, 1419, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2016, 1664, 2145, 2211, 3417, 0, 2415, 2485, 2556, 0
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OFFSET
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1,1
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COMMENTS
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Compared with A118112: larger nonzero value more often and in non-monotonic order.
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LINKS
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EXAMPLE
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For n=5, binomial(15,5) = 3003 = (5+1)*(5+2)*71 + 21, a(5) = 21, the residue.
Interestingly, a very large zone of zeros occurs between about n=5460 and n=7800, uninterrupted by nonzero residues.
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MAPLE
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seq(binomial(3*n, n) mod((n+1)*(n+2)), n=1..71); # Emeric Deutsch, Apr 15 2006
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MATHEMATICA
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Table[Mod[Binomial[3*k, k], (k + 1)*(k + 2)], {k, 1, 1000}]
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PROG
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(PARI) a(n) = binomial(3*n, n) % ((n+1)*(n+2)); \\ Michel Marcus, Jan 05 2017
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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