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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.
3
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, 0, 23, 20, -14, 48, 55, -24, 46, 52, 2, 12, 20, 8, 1, 1, 5, 5, 0, 30, 51, -15, 60, 180, -25, 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
OFFSET
0,5
COMMENTS
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.
REFERENCES
Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.
FORMULA
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.
G.f.: 1/(1 - t*x - t*x^2 + t*x^3 - 2*t*x^4 - t*x^5).
EXAMPLE
Triangle begins:
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, 0, 23, 20, -14, 48, 55, -24, 46, 52, 2, 12, 20, 8, 1;
1, 5, 5, 0, 30, 51, -15, 60, 180, -25, 49, 280, 15, 30, 180, 72, 15, 45, 35, 10, 1;
...
MATHEMATICA
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}]]
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
PROG
(PARI) row(n) = Vecrev((1 + x - x^2 + 2*x^3 + x^4)^n);
tabl(nn) = for (n=0, nn, print(row(n))); \\ Michel Marcus, Oct 15 2018
CROSSREFS
Cf. A319096.
Sequence in context: A185286 A153905 A327812 * A228726 A346175 A165357
KEYWORD
tabf,sign,easy
AUTHOR
Shara Lalo, Oct 01 2018
STATUS
approved