login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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.
2

%I #37 Dec 10 2023 18:05:57

%S 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 684,954,1204,1365,1344,1169,882,540,216,1,8,36,120,330,792,1688,3232,

%U 5619,8944,13088,17568,21642,24456,25236,23528,19489,14232,8856,4320,1296,1,10,55,220,715,2002,4970

%N 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 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 The row sum is s(n)=21^n (see A009965).

%C 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 Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

%H 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 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 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 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 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 Triangle begins:

%e 1;

%e 1, 2, 3, 4, 5, 6;

%e 1, 4, 10, 20, 35, 56, 70, 76, 73, 60, 36;

%e 1, 6, 21, 56, 126, 252, 441, 684, 954, 1204, 1365, 1344, 1169, 882, 540, 216;

%e ...

%t 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 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 (PARI) row(n) = Vecrev((1+2*x+3*x^2+4*x^3+5*x^4+6*x^5)^n);

%o tabl(nn) = for (n=0, nn, print(row(n))); \\ _Michel Marcus_, Oct 15 2018

%Y Cf. A009965, A319093, A319094, A319095.

%K tabf,nonn,easy

%O 0,3

%A _Shara Lalo_, Oct 01 2018