A067330 Triangle read by rows of incomplete convolutions of Fibonacci numbers F(n+1) = A000045(n+1), n>=0.

1, 1, 2, 2, 3, 5, 3, 5, 7, 10, 5, 8, 12, 15, 20, 8, 13, 19, 25, 30, 38, 13, 21, 31, 40, 50, 58, 71, 21, 34, 50, 65, 80, 96, 109, 130, 34, 55, 81, 105, 130, 154, 180, 201, 235, 55, 89, 131, 170, 210, 250, 289, 331, 365, 420, 89, 144, 212, 275, 340, 404, 469, 532, 600, 655, 744, 144, 233, 343, 445

Offset: 0

Author

Wolfdieter Lang, Feb 15 2002

Keywords

• nonn
• easy
• tabl

Comments

The diagonals d>=0 (d=0: main diagonal) give convolutions of Fibonacci numbers F(n+1), n>=0, with those with d-shifted index: a(d+n,d)=sum(F(k+1)*F(d+n+1-k),k=0..n), n>=0.

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/(1-x-x^2) (g.f. Fibonacci F(n+1), n>=0).

The diagonals give A001629(n+2), A023610, A067331-4, A067430-1, A067977-8 for d= n-m= 0..9, respectively.

A row with n terms = the dot product of vectors with n terms: (1,1,2,3,...)dot(...3,2,1,1) with carryovers; such that (3, 5, 7, 10) = (1*3=3), (1*2+3=5), (2*1+5=7), (3*1+7=10).

Examples

			{1}; {1,2}; {2,3,5}; {3,5,7,10}; ...; p(2,n)= 2+3*x+5*x^2.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A067418 (triangle with rows read backwards).

Programs

Mathematica

Table[Sum[Fibonacci[k + 1] Fibonacci[n - k + 1], {k, 0, m}], {n, 0, 11}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 11 2016 *)

Formula

a(n, m)= sum(F(k+1)*F(n-k+1), k=0..m), n>=m>=0, else 0.

a(n, m)= (((3*m+5)*F(m+1)+(m+1)*F(m))*F(n-m+1)+(m*F(m+1)+2*(m+1)*F(m))*F(n-m))/5.

G.f. for diagonals d=n-m>=0: (x^d)*(F(d+1)+F(d)*x)/(1-x-x^2)^2, with F(n) := A000045(n) (Fibonacci).

a(n, m) = ((-1)^m*F(n-2*m-1)+m*L(n+2)+5*F(n)+4*F(n-1))/5, with F(-n) = (-1)^(n+1)*F(n), hence a(n, m) = (2*(m+1)*L(n+2)-A067979(n, m))/5, n>=m>=0. - Ehren Metcalfe, Apr 11 2016