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A004782
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Numbers k such that 2*(2k-3)!/(k!*(k-1)!) is an integer.
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5
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2, 3, 7, 16, 21, 29, 43, 46, 67, 78, 89, 92, 105, 111, 127, 141, 154, 157, 171, 188, 191, 205, 210, 211, 221, 229, 232, 239, 241, 267, 277, 300, 309, 313, 316, 323, 326, 331, 346, 369, 379, 415, 421, 430, 436, 441, 443, 451, 460, 461, 465, 469, 477
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OFFSET
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1,1
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COMMENTS
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Equivalently, numbers k such that binomial(2k-3,k-1) == 0 (mod k*(k-1)/2), or: binomial(2k-2,k-1) == 0 (mod k^2-k), or: the Catalan number A000108(k-1) is divisible by k-1, i.e., a(n) = A014847(n) + 1. Indeed, 2(2k-3)!/(k!*(k-1)!) = 2(2k-2)!/(k!(k-1)!(2k-2)) = C(k-1)/(k-1). - M. F. Hasler, Nov 11 2015
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LINKS
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FORMULA
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MATHEMATICA
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PROG
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(PARI) for(n=2, 999, binomial(2*n-2, n-1)%(n^2-n)||print1(n", "))
(PARI) is_A004782(n)=!binomod(2*n-2, n-1, n^2-n) \\ Using http://home.gwu.edu/~maxal/gpscripts/binomod.gp by M. Alekseyev. - M. F. Hasler, Nov 11 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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