A317501 Triangle read by rows: T(0,0) = 1; T(n,k) = 2 T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4); T(n,k)=0 for n or k < 0.

1, 2, 4, 8, 16, 1, 32, 4, 64, 12, 128, 32, 256, 80, 1, 512, 192, 6, 1024, 448, 24, 2048, 1024, 80, 4096, 2304, 240, 1, 8192, 5120, 672, 8, 16384, 11264, 1792, 40, 32768, 24576, 4608, 160, 65536, 53248, 11520, 560, 1, 131072, 114688, 28160, 1792, 10, 262144, 245760, 67584, 5376, 60

Offset

0,2

Comments

Unsigned version of the triangle in A317506.

The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-left in center-justified triangle given in A013609 ((1+2*x)^n) and along a "third layer" skew diagonals pointing top-right 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+2x)^n and (2+x)^n are given in A128099 and A207538 respectively.)

The coefficients in the expansion of 1/(1-2x-x^4) are given by the sequence generated by the row sums.

The row sums give A008999.

If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.106919340376..., when n approaches infinity.

References

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

Links

Table of n, a(n) for n=0..54.

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

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

Formula

T(n,k) = 2^(n - 4*k) / ((n - 4*k)! k!) * (n - 3*k)! where n >= 0 and 0 <= k <= floor(n/4).

Example

Triangle begins:
       1;
       2;
       4;
       8;
      16,      1;
      32,      4;
      64,     12;
     128,     32;
     256,     80,     1;
     512,    192,     6;
    1024,    448,    24;
    2048,   1024,    80;
    4096,   2304,   240,    1;
    8192,   5120,   672,    8;
   16384,  11264,  1792,   40;
   32768,  24576,  4608,  160;
   65536,  53248, 11520,  560,  1;
  131072, 114688, 28160, 1792, 10;
  262144, 245760, 67584, 5376, 60;

Mathematica

t[n_, k_] := t[n, k] = 2^(n - 4 k)/((n - 4 k)! k!) (n - 3 k)!; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/4]} ]  // Flatten.
t[0, 0] = 1; t[n_, k_] := t[n, k] = If[n < 0 || k < 0, 0, 2 t[n - 1, k] + t[n - 4, k - 1]]; Table[t[n, k], {n, 0, 18}, {k, 0, Floor[n/4]}] // Flatten.

Crossrefs

Row sums give A008999.

Cf. A013609, A038207, A128099, A207538, A317506.

Sequence in context: A010745 A269266 A317506 * A097777 A089738 A110333

Adjacent sequences: A317498 A317499 A317500 * A317502 A317503 A317504

Keyword

tabf,nonn,easy

Author

Zagros Lalo, Sep 03 2018

Status

approved