cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Original entry on oeis.org

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

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Author

Wolfdieter Lang, Feb 15 2002

Keywords

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.
		

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

A067418 Triangle A067330 with rows read backwards.

Original entry on oeis.org

1, 2, 1, 5, 3, 2, 10, 7, 5, 3, 20, 15, 12, 8, 5, 38, 30, 25, 19, 13, 8, 71, 58, 50, 40, 31, 21, 13, 130, 109, 96, 80, 65, 50, 34, 21, 235, 201, 180, 154, 130, 105, 81, 55, 34, 420, 365, 331, 289, 250, 210, 170, 131, 89, 55, 744, 655, 600, 532, 469, 404, 340, 275, 212, 144, 89, 1308, 1164, 1075, 965
Offset: 0

Views

Author

Wolfdieter Lang, Feb 15 2002

Keywords

Comments

The column m (without leading 0's) gives the convolution of Fibonacci numbers F(n+1) := A000045(n+1), n>=0, with those with m-shifted index: a(n+m,m)=sum(F(k+1)*F(m+n+1-k),k=0..n), n>=0, m=0,1,...
The row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) are generated by A(z)*(A(z)-x*A(x*z))/(1-x), with A(x) := 1/(1-x-x^2) (g.f. for Fibonacci F(n+1), n>=0).
The columns give A001629(n+2), A023610, A067331-4, A067430-1, A067977-8 for m= 0..9, respectively. Row sums give A067988.

Examples

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

Programs

  • Mathematica
    Reverse /@ 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)=A067330(n, n-m), n>=m>=0, else 0.
a(n, m)= (((3*(n-m)+5)*F(n-m+1)+(n-m+1)*F(n-m))*F(m+1)+((n-m)*F(n-m+1)+2*(n-m+1)*F(n-m))*F(m))/5.
G.f. for column m=0, 1, ...: (x^m)*(F(m+1)+F(m)*x)/(1-x-x^2)^2, with F(m) := A000045(m) (Fibonacci).
a(n, m) = ((-1)^m*F(n-2*m+1)-m*L(n+2)+n*L(n+2)+5*F(n)+4*F(n-1))/5, with F(-n) = (-1)^(n+1)*F(n), hence a(n, m) = (2*(n-m+1)*L(n+2)-A067990(n, m))/5, n>=m>=0. - Ehren Metcalfe, Apr 11 2016
Showing 1-2 of 2 results.