OFFSET
0,4
COMMENTS
Row n gives coefficients in expansion of (1 - 2*x^2 + x^3)^n. Row sum s(n) = 1 when n = 0 and s(n) = 0 when n > 0, see link. In the center-justified triangle, the sum of numbers along "first layer" skew diagonals pointing top-right are the coefficients in expansion of 1/(1 - x + 2 x^3 - x^4) and the sum of numbers along "first layer" skew diagonals pointing top-left are the coefficients in expansion of 1/(1 - x + 2*x^2 - x^4), see links.
REFERENCES
Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.
LINKS
FORMULA
T(0,0) = 1; T(n,k) = T(n-1,k) - 2*T(n-1,k-2) + T(n-1,k-3) for k = 0..3n; T(n,k)=0 for n or k < 0.
G.f.: 1/(1 - t + 2*t x^2 - t*x^3).
EXAMPLE
Triangle begins:
1;
1, 0, -2, 1;
1, 0, -4, 2, 4, -4, 1;
1, 0, -6, 3, 12, -12, -5, 12, -6, 1;
1, 0, -8, 4, 24, -24, -26, 48, -8, -28, 24, -8, 1;
1, 0, -10, 5, 40, -40, -70, 120, 20, -150, 88, 40, -75, 40, -10, 1;
1, 0, -12, 6, 60, -60, -145, 240, 120, -460, 168, 360, -401, 48, 180, -154, 60, -12, 1;
...
MATHEMATICA
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 - 2] + t[n - 1, k - 3]]; Table[t[n, k], {n, 0, 7}, {k, 0, 3 n} ] // Flatten
CROSSREFS
KEYWORD
tabf,sign,easy
AUTHOR
Shara Lalo, Sep 06 2018
STATUS
approved