A068913 Square array read by antidiagonals of number of k step walks (each step +-1 starting from 0) which are never more than n or less than -n.
1, 0, 1, 0, 2, 1, 0, 2, 2, 1, 0, 4, 4, 2, 1, 0, 4, 6, 4, 2, 1, 0, 8, 12, 8, 4, 2, 1, 0, 8, 18, 14, 8, 4, 2, 1, 0, 16, 36, 28, 16, 8, 4, 2, 1, 0, 16, 54, 48, 30, 16, 8, 4, 2, 1, 0, 32, 108, 96, 60, 32, 16, 8, 4, 2, 1, 0, 32, 162, 164, 110, 62, 32, 16, 8, 4, 2, 1, 0, 64, 324, 328, 220, 124, 64, 32, 16, 8, 4, 2, 1
Offset: 0
Examples
Rows start: 1, 0, 0, 0, 0, ... 1, 2, 2, 4, 4, ... 1, 2, 4, 6, 12, ... 1, 2, 4, 8, 14, ... ...
Links
- Alois P. Heinz, Antidiagonals n = 0..200, flattened
Crossrefs
Programs
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Mathematica
T[n_,0]=1; T[n_,k_]:=2^k/(n+1) Sum[(-1)^r Cos[(Pi (2r-1))/(2 (n+1))]^k Cot[(Pi (1-2r))/(4 (n+1))],{r,1,n+1}]; Table[T[r,n-r],{n,0,20},{r,0,n}]//Round//Flatten (* Herbert Kociemba, Sep 23 2020 *)
Formula
Starting with T(n, 0) = 1, if (k-n) is negative or even then T(n, k) = 2*T(n, k-1), otherwise T(n, k) = 2*T(n, k-1) - A061897(n+1, (k-n-1)/2). So for n>=k, T(n, k) = 2^k. [Corrected by Sean A. Irvine, Mar 23 2024]
T(n,0) = 1, T(n,k) = (2^k/(n+1))*Sum_{r=1..n+1} (-1)^r*cos((Pi*(2*r-1))/(2*(n+1)))^k*cot((Pi*(1-2*r))/(4*(n+1))). - Herbert Kociemba, Sep 23 2020