A064045 Square array read by antidiagonals of number of length 2k walks on an n-dimensional hypercubic lattice starting and finishing at the origin and staying in the nonnegative part.
1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 5, 10, 3, 1, 0, 14, 70, 24, 4, 1, 0, 42, 588, 285, 44, 5, 1, 0, 132, 5544, 4242, 740, 70, 6, 1, 0, 429, 56628, 73206, 16016, 1525, 102, 7, 1, 0, 1430, 613470, 1403028, 410928, 43470, 2730, 140, 8, 1, 0, 4862, 6952660, 29082339, 11925672, 1491210, 96684, 4445, 184, 9, 1
Offset: 0
Examples
Rows start: 1, 0, 0, 0, 0, 0, 0, ... 1, 1, 2, 5, 14, 42, 132, ... 1, 2, 10, 70, 588, 5544, 56628, ... 1, 3, 24, 285, 4242, 73206, 1403028, ...
Links
- Alois P. Heinz, Antidiagonals n = 0..140, flattened
- R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6
Crossrefs
Programs
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Maple
a:= proc(n, k) option remember; `if`(n=0, `if`(k=0, 1, 0), add(binomial(2*k, 2*j)*binomial(2*j, j)/ (j+1)*a(n-1, k-j), j=0..k)) end: seq(seq(a(n, d-n), n=0..d), d=0..12); # Alois P. Heinz, May 06 2014
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Mathematica
a[n_, k_] := a[n, k] = If[n == 0, If[k == 0, 1, 0], Sum[Binomial[2*k, 2*j]* Binomial[2*j, j]/(j+1)*a[n-1, k-j], {j, 0, k}]]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Feb 26 2015, after Alois P. Heinz *)
Formula
a(n,k) = Sum_{j=0..k} C(2k,2j) c(j) a(n-1,k-j) where c(j) = C(2j,j)/(j+1) = A000108(j) with a(0,0) = 1 and a(0,k) = 0 for k>0.