A143929 Eigentriangle by rows, termwise products of the natural numbers decrescendo and the bisected Fibonacci series.

1, 2, 1, 3, 2, 3, 4, 3, 6, 8, 5, 4, 9, 16, 21, 6, 5, 12, 24, 42, 55, 7, 6, 15, 32, 63, 110, 144, 8, 7, 18, 40, 84, 165, 288, 377, 9, 8, 21, 48, 105, 220, 432, 754, 987, 10, 9, 24, 56, 126, 275, 576, 1131, 1974, 2584

Offset: 1

Author

Gary W. Adamson, Sep 05 2008

Keywords

• nonn
• easy
• tabl

Comments

Row sums = even-indexed Fibonacci terms A001906.

Sum of n-th row terms = rightmost term of next row.

Examples

			First rows of the triangle T(n, m), n >= 1, m = 1..n:
  1;
  2, 1;
  3, 2,  3;
  4, 3,  6,  8;
  5, 4,  9, 16,  21;
  6, 5, 12, 24,  42,  55;
  7, 6, 15, 32,  63, 110, 144;
  8, 7, 18, 40,  84, 165, 288, 377;
  9, 8, 21, 48, 105, 220, 432, 754, 987;
  ...
Example: row 4 = (4, 3, 6, 8) = termwise product of (4, 3, 2, 1) and (1, 1, 3, 8).
		

Crossrefs

Cf. A000045, A001906, A004736, A088305.

Formula

Given A004736: (1; 2,1; 3,2,1; 4,3,2,1; ...), we apply the termwise products of the sequence {A088305(n-1)}_{n>=1} starting (1, 1, 3, 8, 21, ...).

From Wolfdieter Lang, Jan 07 2021: (Start)

T(n, m) = 0 if n < m; T(n, 1) = n, for n >= 1, and T(n, m) = F(2*(m-1))*(n-m+1) for n >= m >= 2, with F = A000045 (Fibonacci).

G.f. column m: G(1, x) = x/(1-x)^2, G(m, x) = F(2*(m-1))*x^m/(1-x)^2, for m >= 2. (End)

With offset 0: g.f. of row polynomials R(n, x) := Sum_{m=0..n} t(n, m)*x^m, that is, g.f. of triangle t(n,m) = T(n+1, m+1):

G(z, x) = (1 - x*z)^2 / ((1 - z)^2*(1 - 3*x*z + (x*z)^2)). - Wolfdieter Lang, Apr 09 2021