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A126596 a(n) = binomial(4*n,n)*(2*n+1)/(3*n+1).

%I A126596 #35 Aug 29 2025 03:55:21
%S A126596 1,3,20,154,1260,10659,92092,807300,7152444,63882940,574221648,
%T A126596 5188082354,47073334100,428634152730,3914819231400,35848190542920,
%U A126596 329007937216860,3025582795190340,27872496751392496,257172019222240200,2376196095585231920,21983235825545286435
%N A126596 a(n) = binomial(4*n,n)*(2*n+1)/(3*n+1).
%C A126596 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 A126596 Vincenzo Librandi, <a href="/A126596/b126596.txt">Table of n, a(n) for n = 0..100</a>
%F A126596 a(n) = A039599(2*n,n).
%F A126596 a(n) = (2*n+1)*A002293(n). - _Mark van Hoeij_, Nov 17 2011
%F A126596 a(n) = A208983(2*n+1). - _Reinhard Zumkeller_, Mar 04 2012
%F A126596 a(n) = A005810(n) * A005408(n) / A016777(n). - _Reinhard Zumkeller_, Mar 04 2012
%F A126596 a(n) = [x^n] ((1 - sqrt(1 - 4*x))/(2*x))^(2*n+1). - _Ilya Gutkovskiy_, Nov 01 2017
%F A126596 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
%F A126596 a(n) ~ 2^(8*n+3/2) / (3^(3*n+3/2) * sqrt(Pi*n)). - _Amiram Eldar_, Aug 29 2025
%p A126596 seq((2*n+1)*binomial(4*n,n)/(3*n+1),n=0..22); # _Emeric Deutsch_, Mar 27 2007
%t A126596 Table[(Binomial[4n,n](2n+1))/(3n+1),{n,0,30}] (* _Harvey P. Dale_, Feb 06 2016 *)
%o A126596 (Magma) [Binomial(4*n,n)*(2*n+1)/(3*n+1): n in [0..20]]; // _Vincenzo Librandi_, Nov 18 2011
%o A126596 (Haskell)
%o A126596 a126596 n = a005810 n * a005408 n `div` a016777 n
%o A126596 -- _Reinhard Zumkeller_, Mar 04 2012
%Y A126596 Column k=3 of A214776.
%K A126596 nonn,easy,changed
%O A126596 0,2
%A A126596 _Philippe Deléham_, Mar 13 2007
%E A126596 More terms from _Emeric Deutsch_, Mar 27 2007