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

0,5

Links

Alois P. Heinz, Antidiagonals n = 0..200, flattened

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

Example

Rows start:
  1,  0,  0,  0,  0, ...
  1,  2,  2,  4,  4, ...
  1,  2,  4,  6, 12, ...
  1,  2,  4,  8, 14, ...
  ...

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 *)

Crossrefs

Cf. early rows: A000007, A016116 (without initial term), A068911, A068912, A216212, A216241, A235701.

Central and lower diagonals are A000079, higher diagonals include A000918, A028399.

Sequence in context: A289281 A212957 A035393 * A128306 A372626 A305152

Adjacent sequences: A068910 A068911 A068912 * A068914 A068915 A068916

Keyword

nonn,tabl

Author

Henry Bottomley, Mar 06 2002

Status

approved