A361035 -id:A361035 - cp's OEIS frontend

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

Peter Bala, Mar 04 2023

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)) ).

			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

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.

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

Cf. A008977, A135573, A361027.

# 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;

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

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.

A361033 a(n) = 3(4n)!/(n!*(n+1)!^3).

Original entry on oeis.org

3, 9, 280, 17325, 1513512, 162954792, 20193091776, 2768662192725, 409716429837000, 64358256798795960, 10605621798062141760, 1817833036248401270280, 321997225483126007438400, 58649494641569379926280000, 10941649720331183519046796800, 2084191938036600263793119045925

Offset: 0

Peter Bala, Mar 01 2023

Row 0 of A361032.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that A000984(n) is divisible by n + 1 and the result (2*n)!/(n!*(n+1)!) is the n-th Catalan number A000108(n). Similarly, the  numbers A008977(n) = (4*n)!/n!^4 appear to have the property that 3*A008977(n) is divisible by (n + 1)^3, leading to the present sequence. Cf. A361028. Do these numbers have a combinatorial interpretation?
Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0.

Cf. A000108, A007226, A007228, A008977, A361028, A361032, A361034, A361035.

Maple
```
seq(3*(4*n)!/(n!*(n+1)!^3), n = 0..20);
```

Table[3 (4n)!/(n! ((n+1)!)^3),{n,0,15}] (* Harvey P. Dale, Jul 30 2024 *)

a(n) = 3*A008977(n)/(n+1)^3.
a(n) = (3/4)*A008977(n+1)/((4*n+1)*(4*n+2)*(4*n+3)).
a(n) = (1/2)*A007228(n)*A007226(n)*A000108(n).
P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+1)^3 * a(n-1) with a(0) = 3.
The o.g.f. A(x) satisfies the differential equation
x^3*(1 - 256*x)*A(x)''' + x^2*(6 - 1152*x)*A(x)'' + x*(7 - 816*x)*A(x)' + (1 - 24*x)*A(x) - 3 = 0 with A(0) = 3, A'(0) = 9 and A''(0) = 560.
a(n) ~ 3*sqrt(1/(2*Pi^3)) * 2^(8*n)/n^(9/2).

A361034 a(n) = 2520(4n)!/(n!*(n+2)!^3).

Original entry on oeis.org

315, 280, 3675, 116424, 5885880, 399072960, 33129291195, 3190228041000, 344161801063080, 40616781150254400, 5155510596280207800, 695029472211496161600, 98570579229528369624000, 14597207555235045670540800, 2243893009052293495117018875, 356344642367340570239409729000

Offset: 0

Peter Bala, Mar 01 2023

Row 1 of A361032.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 6*A000984(n) is divisible by (n + 1)*(n + 2) and the result (2*n)!/(n!*(n+2)!) is the super ballot number A007054(n). Similarly, the  numbers A008977(n) = (4*n)!/n!^4 appear to have the property that 2520*A008977(n) is divisible by ((n + 1)*(n + 2))^3, leading to the present sequence. Cf. A361029.
Conjecture: a(n) is odd iff n = 2^k - 2 for some k >= 1.

Cf. A007054, A008977, A361029, A361032, A361033, A361035.

seq(2520*(4*n)!/(n!*(n+2)!^3), n = 0..20);

a(n) = 2520*A008977(n)/((n+1)*(n+2))^3.
a(n) = (315/2)*A008977(n+2)/((4*n+1)*(4*n+2)*(4*n+3)*(4*n+5)*(4*n+6)*(4*n+7)).
P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+2)^3 * a(n-1) with a(0) = 315.
The o.g.f. A(x) satisfies the differential equation
x^3*(1 - 256*x)*A(x)''' + x^2*(9 - 1152*x)*A(x)'' + x*(19 - 816*x)*A(x)' + (8 - 24*x)*A(x) - 2520 = 0 with A(0) = 315, A'(0) = 280 and A''(0) = 7350.
a(n) ~ 630*sqrt(8/Pi^3) * 2^(8*n)/n^(15/2).

cp's OEIS Frontend

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

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A361033 a(n) = 3(4n)!/(n!*(n+1)!^3).

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A361034 a(n) = 2520(4n)!/(n!*(n+2)!^3).

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cp's OEIS Frontend

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.

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A361033 a(n) = 3*(4*n)!/(n!*(n+1)!^3).

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Maple

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A361034 a(n) = 2520*(4*n)!/(n!*(n+2)!^3).

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Author

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Comments

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Maple

Formula

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

A361033 a(n) = 3(4n)!/(n!*(n+1)!^3).

A361034 a(n) = 2520(4n)!/(n!*(n+2)!^3).