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A271025 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.

%I A271025 #24 Dec 02 2020 08:28:30
%S A271025 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 A271025 1,429,731,654,371,141,37,7,1,1430,2950,3012,1890,798,235,50,8,1,4862,
%U A271025 12235,14445,10095,4706,1539,365,65,9,1,16796,51822,71398,56040,28820,10392,2726,537,82,10,1
%N A271025 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 A271025 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 A271025 As indicated in A104455, the k-th binomial transform of A000108 will have:
%C A271025    o.g.f.: (1-sqrt((1-(k+4)*x)/(1-k*x)))/(2*x),
%C A271025    e.g.f.: exp((k+2)*x)*(BesselI(0,2x) - BesselI(1,2x)) and
%C A271025    a(n) = Sum_{i=0..n} binomial(n, i)*CatalanNumber(i)*k^(n-i).
%C A271025 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 A271025 A(0,j) = A000108(j).
%F A271025 A(i,j) = Sum_{k=0..j} binomial(j,k)*A(i-1,k) for i >= 1.
%F A271025 A(i,j) = Sum_{k=0..j} binomial(j,k)*A000108(k)*i^(j-k).
%F A271025 From _Peter Luschny_, Jan 27 2020: (Start)
%F A271025 A(n,k) = n^k*hypergeom([1/2, -k], [2], -4/n) for n >= 1.
%F A271025 A(n,k) = (2/Pi)*Integral_{x=-1..1}(k + 4*x^2)^(n - k)*sqrt(1 - x^2). (End)
%e A271025 The array given by integers of the form A(n,k) is illustrated below:
%e A271025 [0] 1, 1,  2,   5,    14,    42,     132,     429,      1430, ...
%e A271025 [1] 1, 2,  5,   15,   51,    188,    731,     2950,     12235, ...
%e A271025 [2] 1, 3,  10,  37,   150,   654,    3012,    14445,    71398, ...
%e A271025 [3] 1, 4,  17,  77,   371,   1890,   10095,   56040,    320795, ...
%e A271025 [4] 1, 5,  26,  141,  798,   4706,   28820,   182461,   1188406, ...
%e A271025 [5] 1, 6,  37,  235,  1539,  10392,  72267,   516474,   3783115, ...
%e A271025 [6] 1, 7,  50,  365,  2726,  20838,  162996,  1303485,  10642310, ...
%e A271025 [7] 1, 8,  65,  537,  4515,  38654,  337007,  2991340,  27013723, ...
%e A271025 [8] 1, 9,  82,  757,  7086,  67290,  648420,  6340365,  62893270, ...
%e A271025 [9] 1, 10, 101, 1031, 10643, 111156, 1174875, 12568686, 136080971, ...
%e A271025 Seen as a triangle:
%e A271025                           1
%e A271025                          1, 1
%e A271025                        2, 2, 1
%e A271025                       5, 5, 3, 1
%e A271025                    14, 15, 10, 4, 1
%e A271025                  42, 51, 37, 17, 5, 1
%e A271025              132, 188, 150, 77, 26, 6, 1
%e A271025           429, 731, 654, 371, 141, 37, 7, 1
%e A271025       1430, 2950, 3012, 1890, 798, 235, 50, 8, 1
%p A271025 A := (n, k) -> (2/Pi)*int((k+4*x^2)^(n-k)*sqrt(1 - x^2), x=-1..1):
%p A271025 for n from 0 to 9 do seq(A(n,k), k=0..n) od; # _Peter Luschny_, Jan 27 2020
%t A271025 A000108[n_]:= Binomial[2*n,n]/(n+1) ;
%t A271025 T[i_,j_]: Sum[Binomial(j,k)*A000108(k)*i^(j-k), {k,0,j}] ;
%t A271025 A[0, k_] := CatalanNumber[k]; A[n_, k_] := n^k*Hypergeometric2F1[1/2, -k, 2, -4/n];
%t A271025 Table[A[n, k], {n, 0, 6}, {k, 0, 8}] (* _Peter Luschny_, Jan 27 2020 *)
%o A271025 (Sage) def A000108(n): return binomial(2*n,n)/(n+1) ;
%o A271025 def T(i,j): return sum(binomial(j,k)*A000108(k)*i^(j-k) for k in range(j+1))
%Y A271025 Cf. A098474, A000178. Rows 0-5: A000108, A007317, A064613, A104455, A104498, A154623. Columns 0-3: A000012, A000027, A002522, A005491.
%K A271025 nonn,tabl,easy
%O A271025 0,4
%A A271025 _John M. Campbell_, Mar 28 2016