A180662 The Golden Triangle: T(n,k) = A001654(k) for n>=0 and 0<=k<=n.
0, 0, 1, 0, 1, 2, 0, 1, 2, 6, 0, 1, 2, 6, 15, 0, 1, 2, 6, 15, 40, 0, 1, 2, 6, 15, 40, 104, 0, 1, 2, 6, 15, 40, 104, 273, 0, 1, 2, 6, 15, 40, 104, 273, 714, 0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895
Offset: 0
Examples
The first few rows of the Golden Triangle are: 0; 0, 1; 0, 1, 2; 0, 1, 2, 6; 0, 1, 2, 6, 15; 0, 1, 2, 6, 15, 40;
References
- David Hooper and Kenneth Whyld, The Oxford Companion to Chess, p. 221, 1992.
Links
- Reinhard Zumkeller, Rows n = 0..120 of triangle, flattened
- Verner E. Hoggatt, Jr., A new angle on Pascal’s triangle, The Fibonacci Quarterly, Vol. 6, Number 4, pp. 228-230, Oct. 1968.
- Edouard Lucas, Recherches sur plusieurs ouvrages de Léonard de Pise, Ch. 1, pp. 12-14, 1877.
- Johannes W. Meijer, Famous numbers on a chessboard, Acta Nova, Volume 4, No.4, December 2010; pp. 589-598.
- Johannes W. Meijer, Illustrations of the triangle sums, Mar 07 2013.
- S. Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.
Crossrefs
Programs
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Haskell
import Data.List (inits) a180662 n k = a180662_tabl !! n !! k a180662_row n = a180662_tabl !! n a180662_tabl = tail $ inits a001654_list -- Reinhard Zumkeller, Jun 08 2013
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Magma
[Fibonacci(k)*Fibonacci(k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, May 25 2021
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Maple
F:= combinat[fibonacci]: T:= (n, k)-> F(k)*F(k+1): seq(seq(T(n, k), k=0..n), n=0..10); # revised Johannes W. Meijer, Sep 13 2012
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Mathematica
Table[Times @@ Fibonacci@ {k, k + 1}, {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Aug 18 2016 *) Module[{nn=20,f},f=Times@@@Partition[Fibonacci[Range[0,nn]],2,1];Table[Take[f,n],{n,nn}]]//Flatten (* Harvey P. Dale, Nov 26 2022 *)
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PARI
T(n,k)=fibonacci(k)*fibonacci(k+1) \\ Charles R Greathouse IV, Nov 07 2016
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Sage
flatten([[fibonacci(k)*fibonacci(k+1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 25 2021
Formula
T(n, k) = F(k)*F(k+1) with F(n) = A000045(n), for n>=0 and 0<=k<=n.
From Johannes W. Meijer, Jun 22 2015: (Start)
Kn1p(n) = Sum_{k=0..floor(n/2)} T(n-k+p-1, k+p-1), p >= 1.
Kn1p(n) = Kn11(n+2*p-2) - Sum_{k=0..p-2} T(n-k+2*p-2, k), p >= 2.
Kn2p(n) = Sum_{k=0..floor(n/2)} T(n-k+p-1, n-2*k), p >= 1.
Kn2p(n) = Kn21(n+2*p-2) - Sum_{k=0..p-2} T(n+k+p, n+2*k+2), p >= 2. (End)
G.f. as triangle: xy/((1-x)(1+xy)(1-3xy+x^2 y^2)). - Robert Israel, Sep 06 2015
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