A317497 Triangle T(n,k) = 3*T(n-1,k) + T(n-3,k-1) for k = 0..floor(n/3) with T(0,0) = 1 and T(n,k) = 0 for n or k < 0, read by rows.

1, 3, 9, 27, 1, 81, 6, 243, 27, 729, 108, 1, 2187, 405, 9, 6561, 1458, 54, 19683, 5103, 270, 1, 59049, 17496, 1215, 12, 177147, 59049, 5103, 90, 531441, 196830, 20412, 540, 1, 1594323, 649539, 78732, 2835, 15, 4782969, 2125764, 295245, 13608, 135, 14348907, 6908733, 1082565, 61236, 945, 1

Offset: 0

Author

Zagros Lalo, Jul 31 2018

Keywords

• tabf
• nonn
• easy

Comments

The numbers in rows of the triangle are along a "second layer" of skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n) and along a "second layer" of skew diagonals pointing top-right in center-justified triangle given in A027465 ((3+x)^n), see links. (Note: First layer of skew diagonals in 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/(1-3x-x^3) are given by the sequence generated by the row sums.

The row sums give A052541.

If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.1038034027355..., when n approaches infinity.

Examples

			Triangle begins:
         1;
         3;
         9;
        27,        1;
        81,        6;
       243,       27;
       729,      108,       1;
      2187,      405,       9;
      6561,     1458,      54;
     19683,     5103,     270,      1;
     59049,    17496,    1215,     12;
    177147,    59049,    5103,     90;
    531441,   196830,   20412,    540,    1;
   1594323,   649539,   78732,   2835,   15;
   4782969,  2125764,  295245,  13608,  135;
  14348907,  6908733, 1082565,  61236,  945,  1;
  43046721, 22320522, 3897234, 262440, 5670, 18;
		

References

• Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 364-366

Links

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

Crossrefs

Row sums give A052541.

Cf. A013610, A027465, A317496, A304236, A304249.

Cf. A000244 (column 0), A027471 (column 1), A027472 (column 2), A036216 (column 3), A036217 (column 4).

Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), A304249 (q=2), this sequence (q=3), A318773 (q=4).

Programs

GAP

Flat(List([0..16],n->List([0..Int(n/3)],k->3^(n-3*k)/(Factorial(n-3*k)*Factorial(k))*Factorial(n-2*k)))); # Muniru A Asiru, Aug 01 2018

Magma

[3^(n-3*k)*Binomial(n-2*k,k): k in [0..Floor(n/3)], n in [0..24]]; // G. C. Greubel, May 12 2021

Mathematica

T[n_, k_]:= T[n, k] = 3^(n-3k)(n-2k)!/((n-3k)! 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, 3 T[n-1, k] + T[n-3, k-1]]; Table[T[n, k], {n, 0, 15}, {k, 0, Floor[n/3]}]//Flatten

Sage

flatten([[3^(n-3*k)*binomial(n-2*k,k) for k in (0..n//3)] for n in (0..24)]) # G. C. Greubel, May 12 2021

Formula

T(n,k) = 3^(n-3*k) * (n-2*k)!/(k! * (n-3*k)!) where n is a nonnegative integer and k = 0..floor(n/3).