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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.

A060959 Table by antidiagonals of generalized Fibonacci numbers: T(n,k) = T(n,k-1) + n*T(n,k-2) with T(n,0)=0 and T(n,1)=1.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 3, 3, 1, 1, 0, 1, 5, 5, 4, 1, 1, 0, 1, 8, 11, 7, 5, 1, 1, 0, 1, 13, 21, 19, 9, 6, 1, 1, 0, 1, 21, 43, 40, 29, 11, 7, 1, 1, 0, 1, 34, 85, 97, 65, 41, 13, 8, 1, 1, 0, 1, 55, 171, 217, 181, 96, 55, 15, 9, 1, 1, 0, 1, 89, 341, 508, 441, 301, 133, 71, 17, 10, 1, 1, 0
Offset: 0

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Author

Henry Bottomley, May 10 2001

Keywords

Examples

			Square array begins as:
  0, 1, 1, 1,  1,  1,  1, ...
  0, 1, 1, 2,  3,  5,  8, ...
  0, 1, 1, 3,  5, 11, 21, ...
  0, 1, 1, 4,  7, 19, 40, ...
  0, 1, 1, 5,  9, 29, 65, ...
  0, 1, 1, 6, 11, 41, 96, ...
		

Crossrefs

Programs

  • GAP
    Flat(List([0..12], n-> List([0..n], k-> (((1+Sqrt(1+4*k))/2)^(n-k) - ((1-Sqrt(1+4*k))/2)^(n-k))/Sqrt(1+4*k) ))); # G. C. Greubel, Jan 15 2020
  • Magma
    [Round( (((1+Sqrt(1+4*k))/2)^(n-k) - ((1-Sqrt(1+4*k))/2)^(n-k) )/Sqrt(1+4*k) ): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 15 2020
    
  • Maple
    seq(seq( round((((1+sqrt(1+4*k))/2)^(n-k) - ((1-sqrt(1+4*k))/2)^(n-k) )/sqrt(1+4*k)), k=0..n), n=0..12); # G. C. Greubel, Jan 15 2020
  • Mathematica
    T[n_, k_]:= If[n==k==0, 0, Round[(((1+Sqrt[1+4n])/2)^k - ((1-Sqrt[1+4n])/2)^k)/Sqrt[1+4n]]]; Table[T[k, n-k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 15 2020 *)
  • PARI
    T(n,k) = ( ((1+sqrt(1+4*n))/2)^k - ((1-sqrt(1+4*n))/2)^k )/sqrt(1+4*n);
    for(n=0,12, for(k=0,n, print1( round(T(k,n-k)), ", "))) \\ G. C. Greubel, Jan 15 2020
    
  • Sage
    [[ round( (((1+sqrt(1+4*k))/2)^(n-k) - ((1-sqrt(1+4*k))/2)^(n-k) )/sqrt(1+4*k) ) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Jan 15 2020
    

Formula

T(n, k) = ( ((1+sqrt(1+4*n))/2)^k - ((1-sqrt(1+4*n))/2)^k )/sqrt(1+4*n).