OFFSET
0,5
COMMENTS
Number of lattice paths from (0,0) to (n,k) using steps (1,0), (1,1), (2,2). - Joerg Arndt, Jul 01 2011
The n-th diagonal D(n) = {T(n,0), T(n+1,1), ..., T(n+m,m), ...} of the triangle has generating function F(x) = 1/(1 - x - x^2)^(n+1) for n = 0,1,2,.... - L. Edson Jeffery, Mar 20 2011
Let p(n,x) denote the Fibonacci polynomial, defined by p(1,x) = 1, p(2,x) = x, and p(n,x) = x*p(n-1,x) + p(n-2,x) for n >= 3. Let q(n,x) be the numerator polynomial of the rational function p(n, 1 + 1/x). The coefficients of the polynomial q(n,x) are given by the (n-1)-th row of T(n,k). E.g., p(5,x) = 1 + 3*x^2 + x^4 gives q(5,x) = 1 + 4*x + 9*x^2 + 10*x^2 + 5*x^4. - Clark Kimberling, Nov 04 2013
LINKS
Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Pieter Moree, Convoluted convolved Fibonacci numbers, arXiv:math/0311205 [math.CO], 2003.
Pieter Moree, Convoluted convolved Fibonacci numbers, Journal of Integer Sequences, Vol. 7 (2004), Article 04.2.2.
Eric Weisstein's World of Mathematics, Fibonacci polynomial.
FORMULA
From Paul Barry, Oct 24 2005: (Start)
G.f.: 1/(1 - x - x*y - x^2*y^2).
T(n,k) = Sum_{j=0..n} C((n+j)/2, j) * (1 + (-1)^(n+j)) * C(j, n-k)/2. (End)
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k-2), T(n,k) = 0 if n < 0 or if n < k, and T(0,0) = 1. - Philippe Deléham, Nov 30 2006
Sum_{k=0..n} (-1)^k*T(n,k) = A059841(n). - Philippe Deléham, Nov 30 2006
T(n,k) = A208336(n+1,k).- Philippe Deléham, Apr 05 2012
EXAMPLE
Triangle T(n,k) (with rows n >= 0 and columns 0 <= k <= n) begins
1;
1, 1;
1, 2, 2;
1, 3, 5, 3;
1, 4, 9, 10, 5;
1, 5, 14, 22, 20, 8;
1, 6, 20, 40, 51, 38, 13;
1, 7, 27, 65, 105, 111, 71, 21;
...
PROG
(Haskell)
a038137 n k = a038137_tabl !! n !! k
a038137_row n = a038137_tabl !! n
a038137_tabl = map reverse a037027_tabl
-- Reinhard Zumkeller, Jul 08 2012
CROSSREFS
KEYWORD
AUTHOR
EXTENSIONS
Title corrected by L. Edson Jeffery, Apr 23 2011
Corrected by Philippe Deléham, Apr 05 2012
STATUS
approved