A361032 Square array read by ascending antidiagonals: T(n,k) = F(n) * (4*k)!/(k!*(k + n + 1)!^3), where F(n) = (1/8)*(4*n + 4)!/(n + 1)!; n, k >= 0.

3, 315, 9, 46200, 280, 280, 7882875, 17325, 3675, 17325, 1466593128, 1513512, 116424, 116424, 1513512, 288592936632, 162954792, 5885880, 2134440, 5885880, 162954792, 59064793444800, 20193091776, 399072960, 67953600, 67953600, 399072960, 20193091776, 12445136556298875

Offset: 0

Author

Peter Bala, Mar 04 2023

Keywords

• nonn
• tabl
• easy

Comments

The Catalan numbers A000108 are given by the formula Catalan(k) = (2*k)!/(k!*(k + 1)!). Gessel (1992) considered generalized Catalan numbers defined by Catalan(n,k) = J(n) * (2*k)!/(k!*(k + n + 1)!), where J(n) = (2*n + 2)!/(2*(n + 1)!) = (2^n)*Product_{j = 0..n} (2*j + 1) is chosen so that these numbers are always integers. Gessel's generalized Catalan numbers are particular cases of super ballot numbers. See A135573 for a table of these generalized Catalan numbers.

For this table we carry out an analogous construction using the numbers B(k) = (4*k)!/k!^4 = A008977(k) in place of the central binomial numbers (2*k)!/k!^2. We define sequences {B(n,k) : k >= 0}, n = 0, 1, 2, ..., by B(n,k) = F(n) * (4*k)!/(k!*(k + n + 1)!^3), where choosing F(n) = (4*n + 4)!/(8*(n + 1)!) = (1/2)*(4^n)*Product_{j = 0..n} (4*j + 1)*(4*j + 2)*(4*j + 3) appears to produce integer values for these quantities. The rows of the square array below are the sequences {B(0,k)}, {B(1,k)}, {B(2,k)}, ....

An alternative expression is B(n,k) = G(n,k) * B(n+k+1), where G(n,k) = (1/8)*Product_{j = 0..n} ( (4*j + 1)*(4*j + 2)*(4*j + 3)/((4*k + 4*j + 1)*(4*k + 4*j + 2)*(4*k + 4*j + 3)) ).

Examples

			The square array with rows n >= 0 and columns k >= 0 begins:
  n\k|          0          1          2           3           4 ...
  ----------------------------------------------------------------------
   0 |          3          9        280       17325     1513512 ...
   1 |        315        280       3675      116424     5885880 ...
   2 |      46200      17325     116424     2134440    67953600 ...
   3 |    7882875    1513512    5885880    67953600  1449322875 ...
   4 | 1466593128  162954792  399072960  3086579925 46235189000 ...
   5 |  ...
  ...
As a triangle:
 Row
  0 |             3
  1 |           315          9
  2 |         46200        280      280
  3 |       7882875      17325     3675    17325
  4 |    1466593128    1513512   116424   116424  1513512
  5 |  288592936632  162954792  5885880  2134440  5885880  162954792
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1325 (first 51 antidiagonals)
• Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194.

Crossrefs

A361033 (row 0), A361034 (row 2), A361035 (row 3).

Cf. A008977, A135573, A361027.

Programs

Maple

# as a square array
T := proc (n, k) (-1)^k*(1/8)*256^(n+1+k)*binomial(n+1/4, n+1+k)*binomial(n+2/4, n+1+k)* binomial(n+3/4, n+1+k); end proc:
for n from 0 to 10 do seq(T(n, k), k = 0..10); end do;
# as a triangle
T := proc (n, k) (-1)^k*(1/8)*256^(n+1+k)*binomial(n+1/4, n+1+k)*binomial(n+2/4, n+1+k)* binomial(n+3/4, n+1+k); end proc:
for n from 0 to 10 do seq(T(n-k, k), k = 0..n); end do;

PARI

T(n,k) = (-1)^k*(1/8)*256^(n+k+1)*binomial(n+1/4, n+k+1)*binomial(n+1/2, n+k+1)* binomial(n+3/4, n+k+1) \\ Andrew Howroyd, Jan 05 2024

Formula

T(n,k) = (-1)^k*(1/8)*256^(n+k+1)*binomial(n+1/4, n+k+1)*binomial(n+1/2, n+k+1)* binomial(n+3/4, n+k+1).

P-recursive: (n + k + 1)^3*T(n,k) = 4*(4*k - 1)*(4*k - 2)*(4*k - 3)*T(n,k-1) with T(n,0) = (1/8)*(4*n + 4)!/(n + 1)!^4 = (1/8)*A008977(n+1).

(n + k + 1)^3*T(n,k) = 4*(4*n + 1)*(4*n + 2)*(4*n + 3)*T(n-1,k) with T(0,k) = 3*(4*k)!/(k!*(k+1)!^3) = A361033(k).

T(n,k) = (1/2) * (1/(2*Pi))^3 * 256^(n+k+1) * Integral_{x = 0..1} (1 - x)^(n+1/4)*x^(k-1/4) dx * Integral_{x = 0..1} (1 - x)^(n+1/2)*x^(k-1/2) dx * Integral_{x = 0..1} (1 - x)^(n+3/4)*x^(k-3/4) dx.