A317051 Triangle read by rows: T(0,0) = 1; T(n,k) = T(n-1,k) + 9 * T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

1, 1, 1, 9, 1, 18, 1, 27, 81, 1, 36, 243, 1, 45, 486, 729, 1, 54, 810, 2916, 1, 63, 1215, 7290, 6561, 1, 72, 1701, 14580, 32805, 1, 81, 2268, 25515, 98415, 59049, 1, 90, 2916, 40824, 229635, 354294, 1, 99, 3645, 61236, 459270, 1240029, 531441, 1, 108, 4455, 87480, 826686, 3306744, 3720087

Offset: 0

Author

Zagros Lalo, Jul 20 2018

Keywords

• tabf
• nonn
• easy

Comments

The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A013616 ((1+9*x)^n) and along skew diagonals pointing top-left in center-justified triangle given in A038291 ((9+x)^n).

The coefficients in the expansion of 1/(1-x-9*x^2) are given by the sequence generated by the row sums.

The row sums are Generalized Fibonacci numbers (see A015445).

If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 3.5413812651491... ((1+sqrt(37))/2), when n approaches infinity.

Examples

			Triangle begins:
  1;
  1;
  1, 9;
  1, 18;
  1, 27, 81;
  1, 36, 243;
  1, 45, 486, 729;
  1, 54, 810, 2916;
  1, 63, 1215, 7290, 6561;
  1, 72, 1701, 14580, 32805;
  1, 81, 2268, 25515, 98415, 59049;
  1, 90, 2916, 40824, 229635, 354294;
  1, 99, 3645, 61236, 459270, 1240029, 531441;
  1, 108, 4455, 87480, 826686, 3306744, 3720087;
		

References

• Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 100

Links

• Zagros Lalo, Left-justified triangle
• Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (1 + 9x)^n
• Zagros Lalo, Skew diagonals in center-justified triangle of coefficients in expansion of (9 + x)^n

Crossrefs

Row sums give A015445.

Cf. A013616, A038291.

Programs

GAP

Flat(List([0..13],n->List([0..Int(n/2)],k->9^k*Binomial(n-k,k)))); # Muniru A Asiru, Jul 20 2018

Magma

/* As triangle */ [[9^k*Binomial(n-k,k): k in [0..Floor(n/2)]]: n in [0.. 15]]; // Vincenzo Librandi, Sep 05 2018

Mathematica

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, t[n - 1, k] + 9 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten
Table[9^k Binomial[n - k, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten

PARI

T(n, k) = 9^k*binomial(n-k,k);
tabf(nn) = for (n=0, nn, for (k=0, n\2, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 20 2018

Formula

T(n,k) = 9^k*binomial(n-k,k), n >= 0, 0 <= k <= floor(n/2).