A318773 Triangle T(n,k) = 3*T(n-1,k) + T(n-4,k-1) for k = 0..floor(n/4), with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

1, 3, 9, 27, 81, 1, 243, 6, 729, 27, 2187, 108, 6561, 405, 1, 19683, 1458, 9, 59049, 5103, 54, 177147, 17496, 270, 531441, 59049, 1215, 1, 1594323, 196830, 5103, 12, 4782969, 649539, 20412, 90, 14348907, 2125764, 78732, 540, 43046721, 6908733, 295245, 2835, 1, 129140163, 22320522, 1082565, 13608, 15

Offset: 0

Author

Zagros Lalo, Sep 04 2018

Keywords

• tabf
• nonn
• easy

Comments

The numbers in rows of the triangle are along a "third layer" skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n) and along a "third layer" skew diagonals pointing top-right in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in center-justified triangles of coefficients in expansions of (1+3*x)^n and (3+x)^n are given in A304236 and A304249 respectively.)

The coefficients in the expansion of 1/(1-3*x-x^4) are given by the sequence generated by the row sums.

If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.035744112294..., when n approaches infinity.

Examples

			Triangle begins:
          1;
          3;
          9;
         27;
         81,        1;
        243,        6;
        729,       27;
       2187,      108;
       6561,      405,       1;
      19683,     1458,       9;
      59049,     5103,      54;
     177147,    17496,     270;
     531441,    59049,    1215,     1;
    1594323,   196830,    5103,    12;
    4782969,   649539,   20412,    90;
   14348907,  2125764,   78732,   540;
   43046721,  6908733,  295245,  2835,   1;
  129140163, 22320522, 1082565, 13608,  15;
  387420489, 71744535, 3897234, 61236, 135;
  ...
		

References

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

Links

• G. C. Greubel, Rows n = 0..120 of the irregular triangle, flattened
• Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3x)^n
• Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n

Crossrefs

Row sums give A052917.

Cf. A013610, A027465.

Cf. A304236, A304249

Cf. A317496, A317497.

Cf. A000244 (column 0), A027471 (column 1), A027472 (column 2), A036216 (column 3).

Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), A304249 (q=2), A317497 (q=3), this sequence (q=4).

Programs

Magma

[3^(n-4*k)*Binomial(n-3*k,k): k in [0..Floor(n/4)], n in [0..24]]; // G. C. Greubel, May 12 2021

Mathematica

T[n_, k_]:= T[n, k] = 3^(n-4k)*(n-3k)!/((n-4k)! k!); Table[T[n, k], {n, 0, 17}, {k, 0, Floor[n/4]} ]//Flatten
T[0, 0] = 1; T[n_, k_]:= T[n, k] = If[n<0 || k<0, 0, 3T[n-1, k] + T[n-4, k-1]]; Table[T[n, k], {n, 0, 17}, {k, 0, Floor[n/4]}]//Flatten

Sage

flatten([[3^(n-4*k)*binomial(n-3*k,k) for k in (0..n//4)] for n in (0..24)]) # G. C. Greubel, May 12 2021

Formula

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