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A319093 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1, k) + T(n-1, k-1) - T(n-1, k-2) + 2*T(n-1, k-3) + T(n-1, k-4) for k = 0..4*n; T(n,k)=0 for n or k < 0.

%I A319093 #33 Nov 24 2018 18:11:29
%S A319093 1,1,1,-1,2,1,1,2,-1,2,7,-2,2,4,1,1,3,0,1,15,3,-4,24,6,-1,9,6,1,1,4,2,
%T A319093 0,23,20,-14,48,55,-24,46,52,2,12,20,8,1,1,5,5,0,30,51,-15,60,180,-25,
%U A319093 49,280,15,30,180,72,15,45,35,10,1,1,6,9,2,36,96,11,54,387,116,-51,774,376,-162,804,532
%N A319093 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1, k) + T(n-1, k-1) - T(n-1, k-2) + 2*T(n-1, k-3) + T(n-1, k-4) for k = 0..4*n; T(n,k)=0 for n or k < 0.
%C A319093 Row n gives the coefficients in the expansion of (1 + x - x^2 + 2*x^3 + x^4)^n, where n is a nonnegative integer. The row sum at row n is s(n) = 4^n. In the center-justified triangle, the sum of numbers along "first layer" skew diagonals pointing top-right are the coefficients in the expansion of 1/(1 - x - x^2 + x^3 - 2*x^4 - x^5) and the sum of numbers along "first layer" skew diagonals pointing top-left are the coefficients in the expansion of 1/(1 - x - 2*x^2 + x^3 - x^4 - x^5), see links. The central coefficients are given in A319096.
%D A319093 Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.
%H A319093 Shara Lalo, <a href="/A319093/a319093.pdf">Triangle of coefficients in expansions of (1 + x - x^2 + 2*x^3 + x^4)^n.</a>
%H A319093 Shara Lalo, <a href="/A319093/a319093_1.pdf">First layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + x - x^2 + 2*x^3 + x^4)^n.</a>
%H A319093 Shara Lalo, <a href="/A319093/a319093_4.pdf">Formulas for Coefficients in Expansion of (1 + x - x^2 + 2*x^3 + x^4)^n.</a>
%F A319093 T(n,k) = Sum_{i=0..k} Sum_{j=2*i..k} Sum_{q=3*i..k}(f) for k = 0..4*n; f= (-1)^(i + q - 2*j)*2^(j - 2*i)*n!)/((n - k + q)!*(k + j - 2*q)!*(i + q - 2*j)!*(j - 2*i)!*i!); f=0 for (n - k + q)<0 or (k + j - 2*q)<0 or (i + q - 2*j) <0 or (j - 2*i) <0. A novel formula proven by Shara Lalo and Zagros Lalo. Also see formula in Links section.
%F A319093 G.f.: 1/(1 - t*x - t*x^2 + t*x^3 - 2*t*x^4 - t*x^5).
%e A319093 Triangle begins:
%e A319093 1;
%e A319093 1, 1, -1, 2,  1;
%e A319093 1, 2, -1, 2,  7, -2,  2,   4,   1;
%e A319093 1, 3,  0, 1, 15,  3, -4,  24,   6,  -1,  9,   6,  1;
%e A319093 1, 4,  2, 0, 23, 20, -14, 48,  55, -24, 46,  52,  2, 12,  20,  8,  1;
%e A319093 1, 5,  5, 0, 30, 51, -15, 60, 180, -25, 49, 280, 15, 30, 180, 72, 15, 45, 35, 10, 1;
%e A319093 ...
%t A319093 Clear[t, n, k]; t[n_, k_] := t[n, k] = Sum[((-1)^(i + q - 2*j)*2^(j - 2*i)*n!)/((n - k + q)!*(k + j - 2*q)!*(i + q - 2*j)!*(j - 2*i)!*i!), {i, 0, k}, {j, 2*i, k}, {q, 3*i, k}]; Flatten[Table[t[n, k], {n, 0, 7}, {k, 0, 4*n}]]
%t A319093 Clear[t, n, k]; t[0, 0] = 1; t[n_, k_] :=  t[n, k] =   If[n < 0 || k < 0, 0,    t[n - 1, k] + t[n - 1, k - 1] - t[n - 1, k - 2] + 2 t[n - 1, k - 3] + t[n - 1, k - 4]]; Table[t[n, k], {n, 0, 6}, {k, 0, 4*n}  ]  // Flatten
%o A319093 (PARI) row(n) = Vecrev((1 + x - x^2 + 2*x^3 + x^4)^n);
%o A319093 tabl(nn) = for (n=0, nn, print(row(n))); \\ _Michel Marcus_, Oct 15 2018
%Y A319093 Cf. A319096.
%K A319093 tabf,sign,easy
%O A319093 0,5
%A A319093 _Shara Lalo_, Oct 01 2018