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A004782 Numbers k such that 2*(2k-3)!/(k!*(k-1)!) is an integer. 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Superset of A081767, as proved by Luke Pebody. Terms not in A081767 include 3, 7, 127, 511, ... - Ralf Stephan, Oct 12 2004
See A260642 for A004782 \ A081767. - M. F. Hasler, Nov 11 2015
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
LINKS
FORMULA
a(n) = A014847(n) + 1. - Enrique Pérez Herrero, Feb 03 2013
MATHEMATICA
Select[Range[500], IntegerQ[2 (2 # - 3)!/(#! (# - 1)!)] &] (* Arkadiusz Wesolowski, Sep 06 2011 *)
PROG
(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
CROSSREFS
Sequence in context: A058698 A058699 A250193 * A049956 A289844 A153056
KEYWORD
nonn
AUTHOR
EXTENSIONS
Offset corrected and initial term added by Arkadiusz Wesolowski, Sep 06 2011
STATUS
approved

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Last modified April 16 17:08 EDT 2024. Contains 371749 sequences. (Running on oeis4.)