OFFSET
1,1
COMMENTS
Apparently the sequence gives the even-order terms in the diagonal of the rational function R(x,y,z) = 1/(1 - (x^2 + y^2 + z^2 - x*y - y*z + x*z)), i.e., a(n) = [(x*y*z)^(2*n)] R(x,y,z), n >= 1. - Gheorghe Coserea, Aug 09 2018
This is because those even-order terms would be the same if y were replaced by -y, giving S(x,y,z) = 1/(1 - (x^2 + y^2 + z^2 + x*y + y*z + x*z)) = Sum_{i>=0} (x^2 + y^2 + z^2 + x*y + y*z + x*z)^i, and the terms for (x*y*z)^(2*k) come from i=3*k in this sum. - Robert Israel, Jan 15 2023
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..33
MAPLE
f:= proc(n) coeftayl((x^2+y^2+z^2+x*y+y*z+x*z)^(3*n), [x, y, z]=[0, 0, 0], [2*n, 2*n, 2*n]) end proc:
map(f, [$1..30]); # Robert Israel, Jan 15 2023
MATHEMATICA
a[n_] := SeriesCoefficient[(x^2 + y^2 + z^2 + x*y + y*z + x*z)^(3n), {x, 0, 2n}, {y, 0, 2n}, {z, 0, 2n}];
Table[a[n], {n, 1, 12}] (* Jean-François Alcover, May 17 2023, after Robert Israel *)
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, Feb 06 2010
STATUS
approved