A303987 Triangle read by rows: T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0, k = 0..n.

1, 1, 4, 1, 36, 36, 1, 144, 900, 400, 1, 400, 8100, 19600, 4900, 1, 900, 44100, 313600, 396900, 63504, 1, 1764, 176400, 2822400, 9922500, 7683984, 853776, 1, 3136, 571536, 17640000, 133402500, 276623424, 144288144, 11778624, 1, 5184, 1587600, 85377600, 1200622500, 5194373184, 7070119056, 2650190400, 165636900

Offset: 0

Author

Wolfdieter Lang, May 14 2018

Keywords

• nonn
• easy
• tabl

Comments

The row sums of this triangle are b(n) = A005259(n), for n >= 0. This sequence b was used in R. Apéry's 1979 proof of the irrationality of Zeta(3). See A005259 for references and links.

Row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], x), hence b(n) = hypergeometric([-n, -n, n+1, n+1], [1, 1, 1], 1) (see the formula in A005259 given by K. A. Penson. This is the solution to Exercise 2.14 of the Koepf reference given there, p. 29).

Examples

			The triangle T begins:
n\k  0    1       2        3          4          5          6          7 ...
0:   1
1:   1    4
2:   1   36      36
3:   1  144     900      400
4:   1  400    8100    19600       4900
5:   1  900   44100   313600     396900      63504
6:   1 1764  176400  2822400    9922500    7683984     853776
7:   1 3136  571536 17640000  133402500  276623424  144288144   11778624
----------------------------------------------------------------------------
row n = 8:   1 5184 1587600 85377600 1200622500 5194373184 7070119056 2650190400 165636900,
row n = 9: 1 8100 3920400 341510400 8116208100 63631071504 176752976400 169612185600 47869064100 2363904400,
row n = 10: 1 12100 8820900 1177862400 44188244100 572679643536 2828047622400 5446435737600 3877394192100 853369488400 34134779536.
...
		

Crossrefs

Cf. A000984, A002894, A005259, A008459, A063007.

The column sequences (without zeros) are A000012, A035287(n+1) = 4*A000217(n)^2, 36*A288876, 400*A000579(n+6)^2, 4900*A000581(n+8)^2, 63504*A001287(n+10)^2, ...

Programs

GAP

Flat(List([0..10],n->List([0..n],k->(Binomial(n,k)*Binomial(n+k,k))^2))); # Muniru A Asiru, May 15 2018

Mathematica

T[n_, k_] := (Gamma[k + n + 1]/(Gamma[k + 1]^2*Gamma[-k + n + 1]))^2;
Flatten[Table[T[n, k], {n, 0, 8}, {k, 0, n}]] (* Peter Luschny, May 14 2018 *)

Formula

T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0 and k = 0..n.

T(n, k) = (binomial(n+k, 2*k)*cbi(k))^2, with cbi(k) = A000984(k) = binomial(2*k, k), and cbi(k)^2 = A002894(k).

G.f. for column sequences (without leading zeros):

cbi(k)^2*P2(2*k, x)/(1 - x)^(4*k+1), with the row polynomials of A008459 (Pascal entries squared) P2(2*k, x) = Sum_{j=0..2*k} A008459(2*k, j)*x^j. For a proof see the general comment in A288876 on the diagonals and columns of A008459.