OFFSET
0,2
COMMENTS
The diagonals d>=0 (d=0: main diagonal) give convolutions of Lucas numbers L(n+1) := A000204(n+1), n>=0, with those with d-shifted index: a(d+n,d) = Sum_{k=0..n} L(k+1)*L(d+n+1-k).
The row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) are generated by A(x*z)*(A(z)-x*A(x*z))/(1-x), with A(x) := (1+2*x)/(1-x-x^2) (g.f. Lucas L(n+1), n>=0).
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..10000
FORMULA
a(n, m) = Sum_{k=0..m} L(k+1)*L(n-k+1), n>=m>=0, else 0.
a(n, m) = (m+1)*L(n-m+1)*F(m) + ((m+1)*L(n-m+1) + m*L(n-m))*F(m+1), n>=m>=0, with F(n) := A000045(n) (Fibonacci) and L(n) := A000032(n) (Lucas).
G.f. for diagonals d= n-m>=0: (x^d)*(L(d+1)+L(d)*x)*(1-2*x)/(1-x-x^2)^2.
a(n, m) = -(-1)^m*F(n-2*m-1) + m*L(n+2)+F(n+3), with F(-n) = (-1)^(n+1) * F(n), hence a(n, m) = -5*A067330(n, m)+2*(m+1)*L(n+2), n>=m>=0. - Ehren Metcalfe, Apr 11 2016
EXAMPLE
Triangle begins:
{1};
{3,6};
{4,13,17}; p(2,x) = 4+13*x+17*x^2
{7,19,31,38};
...
MATHEMATICA
Table[Sum[LucasL[k + 1] LucasL[n - k + 1], {k, 0, m}], {n, 0, 10}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 11 2016 *)
PROG
(PARI) for(n=0, 10, for(k=0, n, print1(sum(m=0, k, (fibonacci(m+2) + fibonacci(m))*(fibonacci(n-m+2) + fibonacci(n-m))), ", "))) \\ G. C. Greubel, Dec 17 2017
CROSSREFS
KEYWORD
AUTHOR
Wolfdieter Lang, Feb 15 2002
STATUS
approved