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%I #39 Mar 13 2023 07:22:34
%S 1,3,1,10,2,2,35,5,3,5,126,14,6,6,14,462,42,14,10,14,42,1716,132,36,
%T 20,20,36,132,6435,429,99,45,35,45,99,429,24310,1430,286,110,70,70,
%U 110,286,1430,92378,4862,858,286,154,126,154,286,858,4862
%N Array T(n,m) of super ballot numbers read along ascending antidiagonals.
%C First row is A000108. 2nd row is A007054. 3rd row and 4th column are essentially A007272.
%C 1st column is A001700. 2nd column is essentially A000108. 3rd column is A007054.
%C Main diagonal is A000984.
%H G. C. Greubel, <a href="/A135573/b135573.txt">Table of n, a(n) for the first 50 antidiagonals</a>
%H E. Allen and I. Gheorghiciuc, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Allen/gheo.html">A Weighted Interpretation for the Super Catalan Numbers</a>, J. Int. Seq. 17 (2014) # 14.10.7, Table 1.
%H Ira M. Gessel, <a href="http://dx.doi.org/10.1016/0747-7171(92)90034-2">Super ballot numbers</a>, J. Symb. Comput. vol 14, iss 2-3 (1992) pp 179-194.
%F T(n, m) = (2*n + 1)!*(2*m)! / (n!*m!*(m + n + 1)!).
%F From _Peter Luschny_, Nov 03 2021: (Start)
%F T(n, m) = (1/(2*Pi))*Integral_{x=0..4} x^m*(4 - x)^(n + 1/2)*x^(-1/2). These are integral representations of the n-th moment of a positive function on [0, 4]. The representations are unique.
%F T(n, m) = 4^(m + n)*hypergeom([1/2 + n, 1/2 - m], [3/2 + n], 1)/((2*n + 1)*Pi).
%F For fixed n and m -> oo: T(n, m) ~ (1/(2*Pi))*4^(n + m + 1)*(Gamma(3/2 + n) / m^(3/2 + n))*(1 - (2*n + 3)^2 / (8*m)) . (End)
%F T(n, m) = (-1)^m*4^(n + 1 + m)*binomial(n + 1/2, n + 1 + m)/2. - _Peter Luschny_, Nov 04 2021
%F From _Peter Bala_, Mar 12 2023: (Start)
%F T(n,m) = 2*(2*n + 1 )/(n + m + 1) * T(n-1,m) with T(0,m) = Catalan(m), where Catalan(m) = A000108(m).
%F T(n,m) = Sum_{k = 0..n} (-1)^k*4^(n-k)*binomial(n,k)*Catalan(m+k) (easily verified using Maple's sumrecursion command). Thus T(n,m) is an integer. (End)
%e Array with rows n >= 0 and columns m >= 0 starts:
%e [n\m] 0 1 2 3 4 5 6 7 8 ...
%e -------------------------------------------------------
%e [0] 1 1 2 5 14 42 132 429 1430 ... [A000108]
%e [1] 3 2 3 6 14 36 99 286 858 ... [A007054]
%e [2] 10 5 6 10 20 45 110 286 780 ... [A007272]
%e [3] 35 14 14 20 35 70 154 364 910 ... [A348893]
%e [4] 126 42 36 45 70 126 252 546 1260 ... [A348898]
%e [5] 462 132 99 110 154 252 462 924 1980 ... [A348899]
%e [6] 1716 429 286 286 364 546 924 1716 3432 ...
%e ...
%e Seen as a triangle:
%e [0] 1;
%e [1] 3, 1;
%e [2] 10, 2, 2;
%e [3] 35, 5, 3, 5;
%e [4] 126, 14, 6, 6, 14;
%e [5] 462, 42, 14, 10, 14, 42;
%e [6] 1716, 132, 36, 20, 20, 36, 132;
%e [7] 6435, 429, 99, 45, 35, 45, 99, 429.
%e .
%e T(20, 100000) = 2.442634...*10^60129. Asymptotic formula: 2.442627..*10^60129.
%p T := proc(n,m) (2*n+1)!/n!*(2*m)!/m!/(m+n+1)! ; end proc:
%p for d from 0 to 12 do for c from 0 to d do printf("%d, ",T(d-c,c)) ; od: od:
%p # Alternatively, printed as rows:
%p A135573 := (n, m) -> (1/(2*Pi))*int(x^m*(4-x)^(n+1/2)*x^(-1/2), x=0..4):
%p for n from 0 to 9 do seq(A135573(n, m), m = 0..9) od; # _Peter Luschny_, Nov 03 2021
%t T[n_, m_] := (2*n+1)!/n!*(2*m)!/m!/(m+n+1)!; Table[T[n-m, m], {n, 0, 12}, {m, 0, n}] // Flatten (* _Jean-François Alcover_, Jan 06 2014, after Maple *)
%t T[n_, m_] := 4^(m+n) Hypergeometric2F1[1/2+n, 1/2-m, 3/2+n, 1] / ((2 n + 1) Pi);
%t Table[T[n - m + 1, m], {n, 0, 9}, {m, 0, n}] // Flatten (* _Peter Luschny_, Nov 03 2021 *)
%o (Sage)
%o def T(n, m): return (-1)^m*4^(n + 1 + m)*binomial(n + 1/2, n + 1 + m)/2
%o for n in range(7): print([T(n, m) for m in range(9)]) # _Peter Luschny_, Nov 04 2021
%Y Cf. A000108, A007054, A000984, A348893, A348898, A348899.
%Y Cf. A000984 (main diagonal), A001700 (column 0), A082590 (sum of antidiagonals).
%K easy,nonn,tabl
%O 0,2
%A _R. J. Mathar_, Feb 23 2008
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