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a(n) = binomial(4*n,n)*(2*n+1)/(3*n+1).
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%I #32 Sep 08 2022 08:45:29

%S 1,3,20,154,1260,10659,92092,807300,7152444,63882940,574221648,

%T 5188082354,47073334100,428634152730,3914819231400,35848190542920,

%U 329007937216860,3025582795190340,27872496751392496,257172019222240200,2376196095585231920,21983235825545286435

%N a(n) = binomial(4*n,n)*(2*n+1)/(3*n+1).

%C Number of standard Young tableaux of shape [3n,n]. Also the number of binary words with 3n 1's and n 0's such that for every prefix the number of 1's is >= the number of 0's. The a(1) = 3 words are: 1011, 1101, 1110. - _Alois P. Heinz_, Aug 15 2012

%H Vincenzo Librandi, <a href="/A126596/b126596.txt">Table of n, a(n) for n = 0..100</a>

%F a(n) = A039599(2*n,n).

%F a(n) = (2*n+1)*A002293(n). - _Mark van Hoeij_, Nov 17 2011

%F a(n) = A208983(2*n+1). - _Reinhard Zumkeller_, Mar 04 2012

%F a(n) = A005810(n) * A005408(n) / A016777(n). - _Reinhard Zumkeller_, Mar 04 2012

%F a(n) = [x^n] ((1 - sqrt(1 - 4*x))/(2*x))^(2*n+1). - _Ilya Gutkovskiy_, Nov 01 2017

%F Recurrence: 3*n*(3*n-1)*(3*n+1)*a(n) = 8*(2*n+1)*(4*n-3)*(4*n-1)*a(n-1). - _Vaclav Kotesovec_, Feb 03 2018

%p seq((2*n+1)*binomial(4*n,n)/(3*n+1),n=0..22); # _Emeric Deutsch_, Mar 27 2007

%t Table[(Binomial[4n,n](2n+1))/(3n+1),{n,0,30}] (* _Harvey P. Dale_, Feb 06 2016 *)

%o (Magma) [Binomial(4*n,n)*(2*n+1)/(3*n+1): n in [0..20]]; // _Vincenzo Librandi_, Nov 18 2011

%o (Haskell)

%o a126596 n = a005810 n * a005408 n `div` a016777 n

%o -- _Reinhard Zumkeller_, Mar 04 2012

%Y Column k=3 of A214776.

%K nonn,easy

%O 0,2

%A _Philippe Deléham_, Mar 13 2007

%E More terms from _Emeric Deutsch_, Mar 27 2007