%I #36 Dec 12 2023 18:48:14
%S 1,0,1,2,-1,1,3,3,-2,1,11,0,5,-3,1,31,11,-5,8,-4,1,101,20,16,-13,12,
%T -5,1,328,81,4,29,-25,17,-6,1,1102,247,77,-25,54,-42,23,-7,1,3760,855,
%U 170,102,-79,96,-65,30,-8,1,13036,2905,685,68,181,-175,161,-95,38,-9,1,45750,10131,2220,617,-113,356,-336,256,-133,47,-10,1
%N Triangle read by rows of coefficients of polynomials C_n(x) = Sum_{k=0..n} (2*k)!*(x - 1)^(n-k)/((k + 1)!*k!).
%C The polynomials C_n(x) have generating function G(x,t) = (1 - sqrt(1 - 4*t))/(2*t*(1 + t - x*t)) = 1 + x*t + (x^2 - x + 2)*t^2 + (x^3 - 2*x^2 + 3*x + 3)*t^3 + ...
%C C_n(x) can be defined by the recurrence relation C_n(x) = (x - 1)*C_(n-1)(x) + (2n)!/((n + 1)!*n!), C_0(x) = 1 or the equivalent form C_n(x) = (x - 1)*C_(n-1)(x) + C_n(1), C_0(x) = 1.
%C C_n(x) can be defined as convolution of Catalan numbers and powers of (x - 1).
%C Discriminants of C_n(x) gives the sequence: 1, 1, -7, -543, 533489, 7080307052, -1318026434480736, -3526797951451513832247, 137992774365121594001729513153, ...
%C C_n(0) = A032357(n).
%C C_n(1) = C_n(x) - (x - 1)*C_(n-1)(x) = A000108(n).
%C C_n(2) = Sum_{m=0..n} C_1(m) = A014137(n).
%C C_n(3) = A014318(n).
%C C_n(5) = A000346(n).
%C C_n(6) = A046714(n).
%H G. C. Greubel, <a href="/A271453/b271453.txt">Rows n=0..100 of triangle, flattened</a>
%H Ilya Gutkovskiy, <a href="/A271453/a271453_2.pdf">Polynomials C_n(x)</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CatalanNumber.html">Catalan Number</a>
%F For triangle: T(n,n)=1, T(n,0) = Sum_{k=0..n} (-1)^(n-k)*(2*k)!/(k! * (k+1)!), T(n, k) = T(n-1, k-1) - T(n-1, k). - _G. C. Greubel_, Nov 04 2018
%e Triangle begins:
%e 1;
%e 0, 1;
%e 2, -1, 1;
%e 3, 3, -2, 1;
%e 11, 0, 5, -3, 1;
%e 31, 11, -5, 8, -4, 1;
%e ...
%e The first few polynomials are:
%e C_0(x) = 1;
%e C_1(x) = x;
%e C_2(x) = x^2 - x + 2;
%e C_3(x) = x^3 - 2*x^2 + 3*x + 3;
%e C_4(x) = x^4 - 3*x^3 + 5*x^2 + 11;
%e C_5(x) = x^5 - 4*x^4 + 8*x^3 - 5*x^2 + 11*x + 31;
%e ...
%t CoefficientList[RecurrenceTable[{c[0] == 1, c[n] == (x - 1) c[n - 1] + CatalanNumber[n]}, c, {n, 11}], x]
%t T[n_, n_]:= 1; T[n_, 0]:= (-1)^n*Sum[CatalanNumber[k]*(-1)^k, {k, 0, n}]; T[n_, k_]:= T[n - 1, k - 1] - T[n - 1, k]; Table[T[n, k], {n, 0, 5}, {k, 0, n}]//Flatten (* _G. C. Greubel_, Nov 04 2018 *)
%o (PARI) {T(n, k) = if(k==n, 1, if(k==0, sum(j=0,n, (-1)^(n-j)*(2*j)!/(j!*(j+1)!)), T(n-1, k-1) - T(n-1, k))) };
%o for(n=0, 10, for(k=0, n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Nov 04 2018
%Y Cf. A000108, A130595.
%K sign,tabl,easy
%O 0,4
%A _Ilya Gutkovskiy_, Apr 09 2016