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%I #16 Jan 05 2024 14:27:22
%S 3,315,9,46200,280,280,7882875,17325,3675,17325,1466593128,1513512,
%T 116424,116424,1513512,288592936632,162954792,5885880,2134440,5885880,
%U 162954792,59064793444800,20193091776,399072960,67953600,67953600,399072960,20193091776,12445136556298875
%N 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.
%C 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.
%C 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)}, ....
%C 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)) ).
%H Andrew Howroyd, <a href="/A361032/b361032.txt">Table of n, a(n) for n = 0..1325</a> (first 51 antidiagonals)
%H Ira M. Gessel, <a href="http://people.brandeis.edu/~gessel/homepage/papers/superballot.pdf">Super ballot numbers</a>, J. Symbolic Comp., 14 (1992), 179-194.
%F 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).
%F 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).
%F (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).
%F 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.
%e The square array with rows n >= 0 and columns k >= 0 begins:
%e n\k| 0 1 2 3 4 ...
%e ----------------------------------------------------------------------
%e 0 | 3 9 280 17325 1513512 ...
%e 1 | 315 280 3675 116424 5885880 ...
%e 2 | 46200 17325 116424 2134440 67953600 ...
%e 3 | 7882875 1513512 5885880 67953600 1449322875 ...
%e 4 | 1466593128 162954792 399072960 3086579925 46235189000 ...
%e 5 | ...
%e ...
%e As a triangle:
%e Row
%e 0 | 3
%e 1 | 315 9
%e 2 | 46200 280 280
%e 3 | 7882875 17325 3675 17325
%e 4 | 1466593128 1513512 116424 116424 1513512
%e 5 | 288592936632 162954792 5885880 2134440 5885880 162954792
%p # as a square array
%p 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:
%p for n from 0 to 10 do seq(T(n, k), k = 0..10); end do;
%p # as a triangle
%p 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:
%p for n from 0 to 10 do seq(T(n-k, k), k = 0..n); end do;
%o (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
%Y A361033 (row 0), A361034 (row 2), A361035 (row 3).
%Y Cf. A008977, A135573, A361027.
%K nonn,tabl,easy
%O 0,1
%A _Peter Bala_, Mar 04 2023