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%I #24 Dec 02 2020 08:28:30
%S 1,1,1,2,2,1,5,5,3,1,14,15,10,4,1,42,51,37,17,5,1,132,188,150,77,26,6,
%T 1,429,731,654,371,141,37,7,1,1430,2950,3012,1890,798,235,50,8,1,4862,
%U 12235,14445,10095,4706,1539,365,65,9,1,16796,51822,71398,56040,28820,10392,2726,537,82,10,1
%N A(n, k) is the n-th binomial transform of the Catalan sequence (A000108) evaluated at k. Array read by descending antidiagonals for n >= 0 and k >= 0.
%C Interestingly, the determinant of the n X n array of entries of the form A(i,j) is equal to the (n-1)-th superfactorial number (see A000178).
%C As indicated in A104455, the k-th binomial transform of A000108 will have:
%C o.g.f.: (1-sqrt((1-(k+4)*x)/(1-k*x)))/(2*x),
%C e.g.f.: exp((k+2)*x)*(BesselI(0,2x) - BesselI(1,2x)) and
%C a(n) = Sum_{i=0..n} binomial(n, i)*CatalanNumber(i)*k^(n-i).
%C The columns of this array are polynomial integer sequences. The successive polynomials corresponding to the columns of this array are: p0(n) = 1, p1(n) = n + 1, p2(n) = n^2 + 2n + 2, p3(n) = n^3 + 3*n^2 + 6*n + 5, p4(n) = n^4 + 4*n^3 + 12*n^2 + 20*n + 14, and so forth. The coefficients of these successive polynomials form a number triangle, which is given by A098474.
%F A(0,j) = A000108(j).
%F A(i,j) = Sum_{k=0..j} binomial(j,k)*A(i-1,k) for i >= 1.
%F A(i,j) = Sum_{k=0..j} binomial(j,k)*A000108(k)*i^(j-k).
%F From _Peter Luschny_, Jan 27 2020: (Start)
%F A(n,k) = n^k*hypergeom([1/2, -k], [2], -4/n) for n >= 1.
%F A(n,k) = (2/Pi)*Integral_{x=-1..1}(k + 4*x^2)^(n - k)*sqrt(1 - x^2). (End)
%e The array given by integers of the form A(n,k) is illustrated below:
%e [0] 1, 1, 2, 5, 14, 42, 132, 429, 1430, ...
%e [1] 1, 2, 5, 15, 51, 188, 731, 2950, 12235, ...
%e [2] 1, 3, 10, 37, 150, 654, 3012, 14445, 71398, ...
%e [3] 1, 4, 17, 77, 371, 1890, 10095, 56040, 320795, ...
%e [4] 1, 5, 26, 141, 798, 4706, 28820, 182461, 1188406, ...
%e [5] 1, 6, 37, 235, 1539, 10392, 72267, 516474, 3783115, ...
%e [6] 1, 7, 50, 365, 2726, 20838, 162996, 1303485, 10642310, ...
%e [7] 1, 8, 65, 537, 4515, 38654, 337007, 2991340, 27013723, ...
%e [8] 1, 9, 82, 757, 7086, 67290, 648420, 6340365, 62893270, ...
%e [9] 1, 10, 101, 1031, 10643, 111156, 1174875, 12568686, 136080971, ...
%e Seen as a triangle:
%e 1
%e 1, 1
%e 2, 2, 1
%e 5, 5, 3, 1
%e 14, 15, 10, 4, 1
%e 42, 51, 37, 17, 5, 1
%e 132, 188, 150, 77, 26, 6, 1
%e 429, 731, 654, 371, 141, 37, 7, 1
%e 1430, 2950, 3012, 1890, 798, 235, 50, 8, 1
%p A := (n, k) -> (2/Pi)*int((k+4*x^2)^(n-k)*sqrt(1 - x^2), x=-1..1):
%p for n from 0 to 9 do seq(A(n,k), k=0..n) od; # _Peter Luschny_, Jan 27 2020
%t A000108[n_]:= Binomial[2*n,n]/(n+1) ;
%t T[i_,j_]: Sum[Binomial(j,k)*A000108(k)*i^(j-k), {k,0,j}] ;
%t A[0, k_] := CatalanNumber[k]; A[n_, k_] := n^k*Hypergeometric2F1[1/2, -k, 2, -4/n];
%t Table[A[n, k], {n, 0, 6}, {k, 0, 8}] (* _Peter Luschny_, Jan 27 2020 *)
%o (Sage) def A000108(n): return binomial(2*n,n)/(n+1) ;
%o def T(i,j): return sum(binomial(j,k)*A000108(k)*i^(j-k) for k in range(j+1))
%Y Cf. A098474, A000178. Rows 0-5: A000108, A007317, A064613, A104455, A104498, A154623. Columns 0-3: A000012, A000027, A002522, A005491.
%K nonn,tabl,easy
%O 0,4
%A _John M. Campbell_, Mar 28 2016
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