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.

A337129 Triangular array read by rows: T(n,0) = 2^n, T(n,k) = Sum_{i=n-k..n, j=0..i-n+k, i<>n or j<>k} T(i,j) for k > 0.

Original entry on oeis.org

1, 2, 3, 4, 6, 16, 8, 12, 32, 84, 16, 24, 64, 168, 440, 32, 48, 128, 336, 880, 2304, 64, 96, 256, 672, 1760, 4608, 12064, 128, 192, 512, 1344, 3520, 9216, 24128, 63168, 256, 384, 1024, 2688, 7040, 18432, 48256, 126336, 330752, 512, 768, 2048, 5376, 14080, 36864, 96512, 252672, 661504, 1731840
Offset: 0

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Author

Oboifeng Dira, Sep 14 2020

Keywords

Examples

			The triangle  T(n,k) begins:
   n\k  0    1    2    3    4    5
   0:   1
   1:   2    3
   2:   4    6    16
   3:   8    12   32  84
   4:   16   24   64  168  440
   5:   32   48   128 336  880  2304
   ...
T(3,2) = ((3+sqrt(5))^3-(3-sqrt(5))^3)*(2)/(4*sqrt(5)) = (64*sqrt(5))/(2*sqrt(5)) = 32.
		

Crossrefs

Cf. A000079 (1st column), A069429 (diagonal), A018903 (row sums), A001906, A004171.

Programs

  • Maple
    T := proc (n, k) if k = 0 and 0 <= n then 2^n elif 1 <= k and k <= n then round((((3+sqrt(5))^(k+1)-(3-sqrt(5))^(k+1))*(2^(n-k))/(4*sqrt(5)))) else 0 end if end proc:seq(print(seq(T(n, k), k=0..n)), n=0..9);
  • Mathematica
    T[n_, 0] := 2^n;
    T[n_, n_] := 2^(n-1) Fibonacci[2n+2];
    T[n_, k_] /; 0Jean-François Alcover, Nov 13 2020 *)
  • PARI
    T(n,k) = if (k == 0, 2^n, my(w=quadgen(5, 'w)); ((2*w+2)^(k+1)-(4-2*w)^(k+1))*(2^(n-k))/(4*(2*w-1))); \\ Michel Marcus, Sep 14 2020
    
  • PARI
    Row(n)={Vecrev(polcoef((1-x*y)*(1-2*x*y)/((1-6*x*y+4*x^2*y^2)*(1-2*x)) + O(x*x^n), n))} \\ Andrew Howroyd, Sep 23 2020

Formula

T(n,0) = 2^n.
T(n,k) = ((3+sqrt(5))^(k+1)-(3-sqrt(5))^(k+1))*(2^(n-k))/(4*sqrt(5)) for 1<=k<=n.
T(n+1,n) = 2*T(n,n).
T(n+m,n) = 2^m*T(n,n), for m>=1.
T(n,n) = A069429(n) = 2^(n-1)*A001906(n+1) for n>=1.
T(2*n,n) = (1/2)*A099157(n+1) = A004171(n-1)*A001906(n+1) for n>=1.
G.f.: (1 - x*y)*(1 - 2*x*y)/((1 - 6*x*y + 4*x^2*y^2)*(1 - 2*x)). - Andrew Howroyd, Sep 23 2020