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.

A180663 Mirror image of the Golden Triangle: T(n,k) = A001654(n-k) for n>=0 and 0<=k<=n.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 6, 2, 1, 0, 15, 6, 2, 1, 0, 40, 15, 6, 2, 1, 0, 104, 40, 15, 6, 2, 1, 0, 273, 104, 40, 15, 6, 2, 1, 0, 714, 273, 104, 40, 15, 6, 2, 1, 0, 1870, 714, 273, 104, 40, 15, 6, 2, 1, 0, 4895, 1870, 714, 273, 104, 40, 15, 6, 2, 1, 0
Offset: 0

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Author

Johannes W. Meijer, Sep 21 2010

Keywords

Comments

This triangle is the mirror image of the Golden Triangle A180662. The terms in the n-th row of the triangle are the first (n+1) golden rectangle numbers in reversed order. The golden rectangle numbers are A001654(n)=F(n)*F(n+1), with F(n) the Fibonacci numbers.
The chess sums, see A180662 for their definitions, mirror those of the Golden Triangle: Row1 & Row1; Row 2 & Row2; Kn1 and Kn2; Kn3 and Kn4; Fi1 and Fi2; Ca1 and Ca2; Ca3 and Ca4; Gi1 and Gi2; Gi3 and Gi4; Ze1 and Ze2; Ze3 and Ze4.

Examples

			The first few rows of this triangle are:
0;
1, 0;
2, 1, 0;
6, 2, 1, 0;
15, 6, 2, 1, 0;
40, 15, 6, 2, 1, 0;
		

Crossrefs

Cf. A180662 (Golden Triangle), A001654 (Golden Rectangle numbers), A000045 (F(n)).
The triangle sums lead to: A064831 (Row1, Kn21, Kn22, Kn3, Ca2, Ca3, Gi2, Gi3), A077916 (Row2), A180664 (Kn23), A180665 (Kn11, Kn12, Kn13, Fi1, Ze1), A180665(2*n) (Kn4, Fi2, Ze4), A115730(n+1) (Ca1, Ze3), A115730(3*n+1) (Ca4, Ze2), A180666 (Gi1), A180666(4*n) (Gi4).

Programs

  • Haskell
    a180663 n k = a180663_tabl !! n !! k
    a180663_row n = a180663_tabl !! n
    a180663_tabl = map reverse a180662_tabl
    -- Reinhard Zumkeller, Jun 08 2013
  • Maple
    F:= combinat[fibonacci]:
    T:= (n, k)-> F(n-k)*F(n-k+1):
    seq(seq(T(n,k), k=0..n), n=0..10); # revised Johannes W. Meijer, Sep 13 2012
  • Mathematica
    Module[{nn=20,fb},fb=Times@@@Partition[Fibonacci[Range[0,(nn(nn+1))/2]],2,1];Table[ Reverse[Take[fb,n]],{n,nn}]]//Flatten (* Harvey P. Dale, Jan 30 2023 *)

Formula

T(n,k) = F(n-k)*F(n-k+1) with F(n) = A000045(n), for n>=0 and 0<=k<=n.