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

0,2

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.

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

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).

Example

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;

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

Prog

(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
(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

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).

Sequence in context: A148923 A058143 A126025 * A114181 A036134 A317502

Adjacent sequences: A317494 A317495 A317496 * A317498 A317499 A317500

Keyword

tabf,nonn,easy

Author

Zagros Lalo, Jul 31 2018

Status

approved