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%I #19 Sep 08 2022 08:45:24
%S 1,2,1,6,5,1,20,22,8,1,70,93,47,11,1,252,386,244,81,14,1,924,1586,
%T 1186,500,124,17,1,3432,6476,5536,2794,888,176,20,1,12870,26333,25147,
%U 14649,5615,1435,237,23,1,48620,106762,112028,73489,32714,10135,2168,307,26,1
%N Riordan array (1/sqrt(1-4*x), (1/sqrt(1-4*x)-1)/2).
%C Row sums are A007854. Diagonal sums are A116396.
%C Triangle T(n,k), 0 <= k <= n, read by rows given by [2,1,1,1,1,1,1,...] DELTA [1,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Jun 05 2007
%C Inverse of Riordan array (1/(1+2*x), x*(1+x)/(1+2*x)^2) (see A123876). - _Philippe Deléham_, Oct 25 2007
%H G. C. Greubel, <a href="/A116395/b116395.txt">Rows n = 0..100 of triangle, flattened</a>
%H Joseph Pappe, Digjoy Paul and Anne Schilling, <a href="https://arxiv.org/abs/2109.06300">An area-depth symmetric q,t-Catalan polynomial</a>, arXiv:2109.06300 [math.CO], 2021. See Remark 2.4 p. 4.
%F Number triangle T(n,k) = (4^n/2^k)*Sum_{j=0..k} C(k,j)*C(n+(j-1)/2,n)*(-1)^(k-j).
%F Sum_{k=0..n} (-1)^k*T(n,k) = A000108(n), Catalan numbers. - _Philippe Deléham_, Nov 07 2006
%F T(n,k) = Sum_{j>=0} A039599(n,j)*binomial(j,k). - _Philippe Deléham_, Mar 30 2007
%F Sum_{k=0..n} T(n,k)*x^k = A127053(n), A126985(n), A127016(n), A127017(n), A126987(n), A126986(n), A126982(n), A126984(n), A126983(n), A000007(n), A000108(n), A000984(n), A007854(n), A076035(n), A076036(n), A127628(n), A126694(n), A115970(n) for x = -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively. - _Philippe Deléham_, Oct 25 2007
%e Triangle begins:
%e 1;
%e 2, 1;
%e 6, 5, 1;
%e 20, 22, 8, 1;
%e 70, 93, 47, 11, 1;
%e 252, 386, 244, 81, 14, 1;
%t T[n_, k_]:= (4^n/2^k)*Sum[(-1)^(k-j)*Binomial[k, j]*Binomial[n+(j-1)/2, n], {j, 0, k}]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* _G. C. Greubel_, May 28 2019 *)
%o (PARI) {T(n,k) = (4^n/2^k)*sum(j=0, k, (-1)^(k-j)*binomial(k, j)* binomial(n+(j-1)/2, n))}; \\ _G. C. Greubel_, May 28 2019
%o (Magma) [[ Round((4^n/2^k)*(&+[ (-1)^(k-j)*Binomial(k, j)*Gamma(n+(j+1)/2)/(Factorial(n)*Gamma((j+1)/2)) : j in [0..k]])) : k in [0..n]]: n in [0..12]]; // _G. C. Greubel_, May 28 2019
%o (Sage) [[(4^n/2^k)*sum( (-1)^(k-j)*binomial(k, j)* binomial(n+(j-1)/2, n) for j in (0..k)) for k in (0..n)] for n in (0..12)] # _G. C. Greubel_, May 28 2019
%K easy,nonn,tabl
%O 0,2
%A _Paul Barry_, Feb 12 2006
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