A135573 Array T(n,m) of super ballot numbers read along ascending antidiagonals.

1, 3, 1, 10, 2, 2, 35, 5, 3, 5, 126, 14, 6, 6, 14, 462, 42, 14, 10, 14, 42, 1716, 132, 36, 20, 20, 36, 132, 6435, 429, 99, 45, 35, 45, 99, 429, 24310, 1430, 286, 110, 70, 70, 110, 286, 1430, 92378, 4862, 858, 286, 154, 126, 154, 286, 858, 4862

Offset

0,2

Comments

First row is A000108. 2nd row is A007054. 3rd row and 4th column are essentially A007272.

1st column is A001700. 2nd column is essentially A000108. 3rd column is A007054.

Main diagonal is A000984.

Links

G. C. Greubel, Table of n, a(n) for the first 50 antidiagonals

E. Allen and I. Gheorghiciuc, A Weighted Interpretation for the Super Catalan Numbers, J. Int. Seq. 17 (2014) # 14.10.7, Table 1.

Ira M. Gessel, Super ballot numbers, J. Symb. Comput. vol 14, iss 2-3 (1992) pp 179-194.

Formula

T(n, m) = (2*n + 1)!*(2*m)! / (n!*m!*(m + n + 1)!).

From Peter Luschny, Nov 03 2021: (Start)

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.

T(n, m) = 4^(m + n)*hypergeom([1/2 + n, 1/2 - m], [3/2 + n], 1)/((2*n + 1)*Pi).

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)

T(n, m) = (-1)^m*4^(n + 1 + m)*binomial(n + 1/2, n + 1 + m)/2. - Peter Luschny, Nov 04 2021

From Peter Bala, Mar 12 2023: (Start)

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).

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)

Example

Array with rows n >= 0 and columns m >= 0 starts:
[n\m]  0    1    2    3    4    5    6     7     8  ...
-------------------------------------------------------
[0]    1    1    2    5   14   42  132   429  1430  ...  [A000108]
[1]    3    2    3    6   14   36   99   286   858  ...  [A007054]
[2]   10    5    6   10   20   45  110   286   780  ...  [A007272]
[3]   35   14   14   20   35   70  154   364   910  ...  [A348893]
[4]  126   42   36   45   70  126  252   546  1260  ...  [A348898]
[5]  462  132   99  110  154  252  462   924  1980  ...  [A348899]
[6] 1716  429  286  286  364  546  924  1716  3432  ...
...
Seen as a triangle:
[0] 1;
[1] 3,    1;
[2] 10,   2,   2;
[3] 35,   5,   3,  5;
[4] 126,  14,  6,  6,  14;
[5] 462,  42,  14, 10, 14, 42;
[6] 1716, 132, 36, 20, 20, 36, 132;
[7] 6435, 429, 99, 45, 35, 45, 99,  429.
.
T(20, 100000) = 2.442634...*10^60129. Asymptotic formula: 2.442627..*10^60129.

Maple

T := proc(n, m) (2*n+1)!/n!*(2*m)!/m!/(m+n+1)! ; end proc:
for d from 0 to 12 do for c from 0 to d do printf("%d, ", T(d-c, c)) ; od: od:
# Alternatively, printed as rows:
A135573 := (n, m) -> (1/(2*Pi))*int(x^m*(4-x)^(n+1/2)*x^(-1/2), x=0..4):
for n from 0 to 9 do seq(A135573(n, m), m = 0..9) od; # Peter Luschny, Nov 03 2021

Mathematica

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[n_, m_] := 4^(m+n) Hypergeometric2F1[1/2+n, 1/2-m, 3/2+n, 1] / ((2 n + 1) Pi);
Table[T[n - m + 1, m], {n, 0, 9}, {m, 0, n}] // Flatten (* Peter Luschny, Nov 03 2021 *)

Prog

(Sage)
def T(n, m): return (-1)^m*4^(n + 1 + m)*binomial(n + 1/2, n + 1 + m)/2
for n in range(7): print([T(n, m) for m in range(9)]) # Peter Luschny, Nov 04 2021

Crossrefs

Cf. A000108, A007054, A000984, A348893, A348898, A348899.

Cf. A000984 (main diagonal), A001700 (column 0), A082590 (sum of antidiagonals).

Sequence in context: A331155 A008299 A016478 * A257254 A126953 A134284

Adjacent sequences: A135570 A135571 A135572 * A135574 A135575 A135576

Keyword

easy,nonn,tabl

Author

R. J. Mathar, Feb 23 2008

Status

approved