A097513 Number of ways to label the vertices of the octahedron (or faces of the cube) with nonnegative integers summing to n, where labelings that differ only by rotation or reflection are considered the same.

1, 1, 3, 5, 10, 15, 27, 38, 60, 84, 122, 164, 229, 298, 398, 509, 658, 823, 1041, 1278, 1582, 1917, 2331, 2786, 3343, 3948, 4676, 5471, 6408, 7428, 8622, 9912, 11406, 13023, 14871, 16866, 19135, 21571, 24321, 27275, 30580, 34122, 38070, 42284, 46956, 51942

Offset: 0

Author

Isabel C. Lugo (izzycat(AT)gmail.com), Aug 26 2004

Keywords

• easy
• nonn

Examples

			a(3) = 5 because we can label the faces of the cube with nonnegative integers summing to three in five ways: 3 on one face, 2 on one face and 1 on an adjacent face, 2 on one face and 1 on the opposite face, 1 on three faces sharing a corner, 1 on three faces not sharing a corner.
		

Links

• Index entries for two-way infinite sequences
• Index entries for linear recurrences with constant coefficients, signature (2,0,-1,0,-2,3,-2,1,1,-2,3,-2,0,-1,0,2,-1).

Crossrefs

Cf. A006381.

Programs

Maple

a:= n-> (Matrix([[1, 0$8, -1$2, -3, -5, -10, -15, -27, -38]]).Matrix(17, (i,j)-> if (i=j-1) then 1 elif j=1 then [2, 0, -1, 0, -2, 3, -2, 1, 1, -2, 3, -2, 0, -1, 0, 2, -1][i] else 0 fi)^n)[1,1]; seq(a(n), n=0..50); # Alois P. Heinz, Jul 31 2008

Formula

G.f.: (q^8-q^7+q^6+q^4+q^2-q+1)/((-1+q)^6*(q+1)^3*(q^2+q+1)^2*(q^2-q+1)*(q^2+1)).

a(n) is asymptotically equal to n^5/5760. - Isabel C. Lugo (izzycat(AT)gmail.com), Aug 31 2004