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A319092 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1, k) + 2*T(n-1, k-1) + 3*T(n-1, k-2) + 4*T(n-1, k-3) + 5*T(n-1, k-4) + 6*T(n-1, k-5) for k = 0..5*n; T(n,k)=0 for n or k < 0.

%I A319092 #37 Dec 10 2023 18:05:57
%S A319092 1,1,2,3,4,5,6,1,4,10,20,35,56,70,76,73,60,36,1,6,21,56,126,252,441,
%T A319092 684,954,1204,1365,1344,1169,882,540,216,1,8,36,120,330,792,1688,3232,
%U A319092 5619,8944,13088,17568,21642,24456,25236,23528,19489,14232,8856,4320,1296,1,10,55,220,715,2002,4970
%N A319092 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1, k) + 2*T(n-1, k-1) + 3*T(n-1, k-2) + 4*T(n-1, k-3) + 5*T(n-1, k-4) + 6*T(n-1, k-5) for k = 0..5*n; T(n,k)=0 for n or k < 0.
%C A319092 Row n gives the coefficients in the expansion of (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5)^n, where n is a nonnegative integer.
%C A319092 The row sum is s(n)=21^n (see A009965).
%C A319092 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 - 2*x^2 -3*x^3 - 4*x^4 - 5*x^5 - 6*x^6) and the sum of numbers along "first layer" skew diagonals pointing top-left are the coefficients in the expansion of 1/(1 - 6*x - 5*x^2 - 4*x^3 - 3*x^4 - 2*x^5 - x^6), see links.
%D A319092 Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.
%H A319092 Shara Lalo, <a href="/A319092/a319092.pdf">Triangle of coefficients in expansions of (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5)^n.</a>
%H A319092 Shara Lalo, <a href="/A319092/a319092_1.pdf">First layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5)^n.</a>
%H A319092 Shara Lalo, <a href="/A319092/a319092_3.pdf">A formula for the coefficients in expansions of (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 )^n.</a>
%F A319092 T(n,k) = Sum_{i=0..k} Sum_{j=2*i..k} Sum_{q=3*i..k} Sum_{r=4*i..k}(f) for k=0..5*n; f=((2^(k + q - 2*r)*3^(j + r - 2*q)*4^(i + q - 2*j)*5^(j - 2*i)*6^i*n!)/((n - k + r)!*(k + q - 2*r)!*(j + r - 2*q)!*(i + q - 2*j)!*(j - 2*i)!*i!) ); f=0 for (n - k + r)<0 or (k + q - 2*r)<0; (j + r - 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 A319092 G.f.: 1/(1 - x*t- 2*x^2*t - 3*x^3*t - 4*x^4*t - 5*x^5*t - 6*x^6*t).
%e A319092 Triangle begins:
%e A319092 1;
%e A319092 1, 2,  3,  4,   5,   6;
%e A319092 1, 4, 10, 20,  35,  56,  70,  76,  73,  60,    36;
%e A319092 1, 6, 21, 56, 126, 252, 441, 684, 954, 1204, 1365, 1344, 1169, 882, 540, 216;
%e A319092 ...
%t A319092 t[n_, k_] := t[n, k] = Sum[(2^(k + q - 2*r)*3^(j + r - 2*q)*4^(i + q - 2*j)*5^(j - 2*i)*6^i*n!)/((n - k + r)!*(k + q - 2*r)!*(j + r - 2*q)!*(i + q -2*j)!*(j - 2*i)!*i!), {i, 0, k}, {j, 2*i, k}, {q, 3*i, k}, {r, 4*i, k}]; Flatten[Table[t[n, k], {n, 0, 5}, {k, 0, 5 n}]]
%t A319092 t[0, 0] = 1; t[n_, k_] :=  t[n, k] =   If[n < 0 || k < 0, 0, t[n - 1, k] + 2 t[n - 1, k - 1] + 3 t[n - 1, k - 2] + 4 t[n - 1, k - 3] + 5 t[n - 1, k - 4] + 6 t[n - 1, k - 5]]; Table[t[n, k], {n, 0, 5}, {k, 0, 5 n}  ]  // Flatten
%o A319092 (PARI) row(n) = Vecrev((1+2*x+3*x^2+4*x^3+5*x^4+6*x^5)^n);
%o A319092 tabl(nn) = for (n=0, nn, print(row(n))); \\ _Michel Marcus_, Oct 15 2018
%Y A319092 Cf. A009965, A319093, A319094, A319095.
%K A319092 tabf,nonn,easy
%O A319092 0,3
%A A319092 _Shara Lalo_, Oct 01 2018