A304249 Triangle T(n,k) = 3*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2), with T(0,0) = 1 and T(n,k) = 0 for n < 0 or k < 0, read by rows.

1, 3, 9, 1, 27, 6, 81, 27, 1, 243, 108, 9, 729, 405, 54, 1, 2187, 1458, 270, 12, 6561, 5103, 1215, 90, 1, 19683, 17496, 5103, 540, 15, 59049, 59049, 20412, 2835, 135, 1, 177147, 196830, 78732, 13608, 945, 18, 531441, 649539, 295245, 61236, 5670, 189, 1

Offset: 0

Author

Zagros Lalo, May 08 2018

Keywords

• tabf
• nonn
• easy

Comments

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013610 ((1+3*x)^n), cf. formula.

The coefficients in the expansion of 1/(1-3x-x^2) are given by the sequence generated by the row sums.

The sequence of the row sums are the "Bronze Fibonacci numbers" A006190, and the limit of their ratio is 3.30277563773... (Bronze ratio), see A098316.

Examples

			Triangle begins:
          1;
          3;
          9,         1;
         27,         6;
         81,        27,         1;
        243,       108,         9;
        729,       405,        54,        1;
       2187,      1458,       270,       12;
       6561,      5103,      1215,       90,        1;
      19683,     17496,      5103,      540,       15;
      59049,     59049,     20412,     2835,      135,       1;
     177147,    196830,     78732,    13608,      945,      18;
     531441,    649539,    295245,    61236,     5670,     189,      1;
    1594323,   2125764,   1082565,   262440,    30618,    1512,     21;
    4782969,   6908733,   3897234,  1082565,   153090,   10206,    252,    1;
   14348907,  22320522,  13817466,  4330260,   721710,   61236,   2268,   24;
   43046721,  71744535,  48361131, 16888014,  3247695,  336798,  17010,  324,  1;
  129140163, 229582512, 167403915, 64481508, 14073345, 1732104, 112266, 3240, 27;
		

References

• Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 86, 363.

Links

• Zagros Lalo, Left justified triangle
• Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1+3x)^n

Crossrefs

Row sums give A006190.

Cf. A013610, A098316.

Sequences of the form 3^(n-q*k)*binomial(n-(q-1)*k, k): A027465 (q=1), this sequence (q=2), A317497 (q=3), A318773 (q=4).

Programs

Magma

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

Mathematica

T[0, 0] = 1; T[n_, k_]:= If[n<0 || k<0, 0, 3T[n-1, k] + T[n-2, k-1]];
Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}]//Flatten
With[{q=2}, Table[3^(n-q*k)*Binomial[n-(q-1)*k, k], {n,0,24}, {k,0,Floor[n/q]}] ]//Flatten (* G. C. Greubel, May 12 2021 *)

PARI

T(n,k)=if( n>0 && k>0, 3*T(n-1, k) + T(n-2, k-1), !n && !k)
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n,k), ", ")); print); \\ Michel Marcus, May 10 2018

Sage

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

Formula

T(n,k) = A013610(n-k, n-2k). - M. F. Hasler, Jun 01 2018