A318775 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 2 * T(n-5,k-1) for k = 0..floor(n/5); T(n,k)=0 for n or k < 0.

1, 1, 1, 1, 1, 1, 2, 1, 4, 1, 6, 1, 8, 1, 10, 1, 12, 4, 1, 14, 12, 1, 16, 24, 1, 18, 40, 1, 20, 60, 1, 22, 84, 8, 1, 24, 112, 32, 1, 26, 144, 80, 1, 28, 180, 160, 1, 30, 220, 280, 1, 32, 264, 448, 16, 1, 34, 312, 672, 80, 1, 36, 364, 960, 240, 1, 38, 420, 1320, 560, 1, 40, 480, 1760, 1120

Offset: 0

Author

Zagros Lalo, Sep 04 2018

Keywords

• tabf
• nonn
• easy

Comments

The numbers in rows of the triangle are along a "fourth layer" skew diagonals pointing top-right in center-justified triangle given in A013609 ((1+2*x)^n) and along a "fourth layer" skew diagonals pointing top-left in center-justified triangle given in A038207 ((2+x)^n), see links. (Note: First layer skew diagonals in center-justified triangles of coefficients in expansions of (1+2*x)^n and (2+x)^n are given in A128099 and A207538 respectively.)

The coefficients in the expansion of 1/(1-x-2*x^5) are given by the sequence generated by the row sums.

The row sums give A318777.

If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 1.4510850920547191..., when n approaches infinity.

Examples

			Triangle begins:
  1;
  1;
  1;
  1;
  1;
  1,  2;
  1,  4;
  1,  6;
  1,  8;
  1, 10;
  1, 12,   4;
  1, 14,  12;
  1, 16,  24;
  1, 18,  40;
  1, 20,  60;
  1, 22,  84,    8;
  1, 24, 112,   32;
  1, 26, 144,   80;
  1, 28, 180,  160;
  1, 30, 220,  280;
  1, 32, 264,  448,  16;
  1, 34, 312,  672,  80;
  1, 36, 364,  960, 240;
  1, 38, 420, 1320, 560;
  ...
		

References

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

Links

• Zagros Lalo, Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n
• Zagros Lalo, Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n

Crossrefs

Row sums give A318777.

Cf. A013609, A038207, A128099, A207538.

Programs

Mathematica

t[n_, k_] := t[n, k] = 2^k/((n - 5 k)! k!) (n - 4 k)!; Table[t[n, k], {n, 0, 24}, {k, 0, Floor[n/5]} ] // Flatten
t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, t[n - 1, k] + 2 t[n - 5, k - 1]]; Table[t[n, k], {n, 0, 24}, {k, 0, Floor[n/5]}] // Flatten

Formula

T(n,k) = 2^k / ((n - 5*k)! k!) * (n - 4*k)! where n >= 0 and 0 <= k <= floor(n/5).