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%I #20 Nov 24 2021 03:55:21
%S 1,1,0,1,1,0,1,2,2,0,1,3,8,5,0,1,4,18,40,14,0,1,5,32,135,224,42,0,1,6,
%T 50,320,1134,1344,132,0,1,7,72,625,3584,10206,8448,429,0,1,8,98,1080,
%U 8750,43008,96228,54912,1430,0,1,9,128,1715,18144,131250,540672,938223,366080,4862,0
%N Square array A(n,k), n>=0, k>=0, read by antidiagonals, where column k is the expansion of 2/(1 + sqrt(1 - 4*k*x)).
%C Number of 2n-length strings of balanced parentheses of at most k different types. Also number of binary trees with n inner nodes of at most k different dimensions. - _Alois P. Heinz_, Oct 28 2019
%H Alois P. Heinz, <a href="/A290605/b290605.txt">Antidiagonals n = 0..140, flattened</a>
%F A(n,k) = k^n*(2*n)!/(n!*(n + 1)!).
%F A(n,k) = k^n*A000108(n).
%F G.f. of column k: 2/(1 + sqrt(1 - 4*k*x)).
%F G.f. of column k: 1/(1 - k*x/(1 - k*x/(1 - k*x/(1 - k*x/(1 - k*x/(1 - ...)))))), a continued fraction.
%F E.g.f. of column k: (BesselI(0,2*k*x) - BesselI(1,2*k*x))*exp(2*k*x).
%F If g.f. = 2/(1 + sqrt(1 - 4*k*x)), then a(n) ~ k^n*4^n/(sqrt(Pi)*n^(3/2)).
%F A(n,k) = Sum_{i=0..k} binomial(k,i) * A256061(n,k-i). - _Alois P. Heinz_, Oct 28 2019
%F For fixed k >= 1, Sum_{n>=0} 1/A(n,k) = 2*k*(8*k + 1) / (4*k - 1)^2 + 24 * k^2 * arcsin(1/(2*sqrt(k))) / (4*k - 1)^(5/2). - _Vaclav Kotesovec_, Nov 23 2021
%F For fixed k >= 1, Sum_{n>=0} (-1)^n / A(n,k) = 2*k*(8*k - 1) / (4*k + 1)^2 - 24 * k^2 * log((1 + sqrt(4*k + 1))/(2*sqrt(k))) / (4*k + 1)^(5/2). - _Vaclav Kotesovec_, Nov 24 2021
%e G.f. of column k: A(x) = 1 + k*x + 2*k^2*x^2 + 5*k^3*x^3 + 14*k^4*x^4 + 42*k^5*x^5 + 132*k^6*x^6 + ...
%e Square array begins:
%e 1, 1, 1, 1, 1, 1, ...
%e 0, 1, 2, 3, 4, 5, ...
%e 0, 2, 8, 18, 32, 50, ...
%e 0, 5, 40, 135, 320, 625, ...
%e 0, 14, 224, 1134, 3584, 8750, ...
%e 0, 42, 1344, 10206, 43008, 131250, ...
%p ctln:= proc(n) option remember; binomial(2*n, n)/(n+1) end:
%p A:= proc(n, k) option remember; k^n*ctln(n) end:
%p seq(seq(A(n, d-n), n=0..d), d=0..10); # _Alois P. Heinz_, Oct 28 2019
%t Table[Function[k, SeriesCoefficient[2/(1 + Sqrt[1 - 4 k x]), {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten
%t Table[Function[k, SeriesCoefficient[1/(1 + ContinuedFractionK[-k x, 1, {i, 1, n}]), {x, 0, n}]][j - n], {j, 0, 10}, {n, 0, j}] // Flatten
%Y Columns k=0-10 give: A000007, A000108, A151374, A005159, A151403, A156058, A156128, A156266, A156270, A156273, A156275.
%Y Rows n=0-2 give: A000012, A001477, A001105.
%Y Main diagonal gives A291699.
%Y Cf. A000108, A253180, A256061.
%K nonn,tabl
%O 0,8
%A _Ilya Gutkovskiy_, Aug 07 2017
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