OFFSET
0,2
COMMENTS
The numbers in rows of the triangle are along "second layer" skew diagonals pointing top-left in center-justified triangle given in A303901 ((3-2*x)^n) and along "second layer" skew diagonals pointing top-right in center-justified triangle given in A317498 ((-2+3x)^n), see links. (Note: First layer skew diagonals in center-justified triangles of coefficients in expansions of (3-2*x)^n and (-2+3x)^n are given in A303941 and A302747 respectively.) The coefficients in the expansion of 1/(1 + 2x - 3x^3) are given by the sequence generated by the row sums. The row sums give A317499.
REFERENCES
Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 136, 396, 397.
LINKS
FORMULA
T(n,k) = (-2)^(n - 3k) * 3^k / ((n - 3k)! k!) * (n - 2k)! where n is a nonnegative integer and k = 0..floor(n/3).
EXAMPLE
Triangle begins:
1;
-2;
4;
-8, 3;
16, -12;
-32, 36;
64, -96, 9;
-128, 240, -54;
256, -576, 216;
-512, 1344, -720, 27;
1024, -3072, 2160, -216;
-2048, 6912, -6048, 1080;
4096, -15360, 16128, -4320, 81;
-8192, 33792, -41472, 15120, -810;
16384, -73728, 103680, -48384, 4860;
-32768, 159744, -253440, 145152, -22680, 243;
65536, -344064, 608256, -414720, 90720, -2916;
MATHEMATICA
t[n_, k_] := t[n, k] = (-2)^(n - 3k) * 3^k/((n - 3 k)! k!) * (n - 2 k)!; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/3]} ] // Flatten
t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, -2 * t[n - 1, k] + 3 * t[n - 3, k - 1]]; Table[t[n, k], {n, 0, 15}, {k, 0, Floor[n/3]}] // Flatten
CROSSREFS
KEYWORD
tabf,sign,easy
AUTHOR
Shara Lalo, Aug 02 2018
STATUS
approved