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%I #18 Dec 29 2018 23:55:28
%S 1,1,4,1,36,36,1,144,900,400,1,400,8100,19600,4900,1,900,44100,313600,
%T 396900,63504,1,1764,176400,2822400,9922500,7683984,853776,1,3136,
%U 571536,17640000,133402500,276623424,144288144,11778624,1,5184,1587600,85377600,1200622500,5194373184,7070119056,2650190400,165636900
%N 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.
%C 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.
%C 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).
%F T(n, k) = (binomial(n,k)*binomial(n+k,k))^2 = A063007(n, k)^2, for n >= 0 and k = 0..n.
%F 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).
%F G.f. for column sequences (without leading zeros):
%F 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.
%e The triangle T begins:
%e n\k 0 1 2 3 4 5 6 7 ...
%e 0: 1
%e 1: 1 4
%e 2: 1 36 36
%e 3: 1 144 900 400
%e 4: 1 400 8100 19600 4900
%e 5: 1 900 44100 313600 396900 63504
%e 6: 1 1764 176400 2822400 9922500 7683984 853776
%e 7: 1 3136 571536 17640000 133402500 276623424 144288144 11778624
%e ----------------------------------------------------------------------------
%e row n = 8: 1 5184 1587600 85377600 1200622500 5194373184 7070119056 2650190400 165636900,
%e row n = 9: 1 8100 3920400 341510400 8116208100 63631071504 176752976400 169612185600 47869064100 2363904400,
%e row n = 10: 1 12100 8820900 1177862400 44188244100 572679643536 2828047622400 5446435737600 3877394192100 853369488400 34134779536.
%e ...
%t T[n_, k_] := (Gamma[k + n + 1]/(Gamma[k + 1]^2*Gamma[-k + n + 1]))^2;
%t Flatten[Table[T[n, k], {n, 0, 8}, {k, 0, n}]] (* _Peter Luschny_, May 14 2018 *)
%o (GAP) Flat(List([0..10],n->List([0..n],k->(Binomial(n,k)*Binomial(n+k,k))^2))); # _Muniru A Asiru_, May 15 2018
%Y Cf. A000984, A002894, A005259, A008459, A063007.
%Y 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, ...
%K nonn,easy,tabl
%O 0,3
%A _Wolfdieter Lang_, May 14 2018
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