Table 4.

Magic formulas for the cuboctahedron

	fcc cuboctahedron n=4
	Atoms	$\frac{10}{3}{n}^3+5n^2+\frac{11}{3}n+1,n\ge 1$
	Bonds	20n3+12n2+4n, n≥0
	cn=5	12, n≥1
	cn=7	24n−24, n≥1
	cn=8	6n2−12n+6, n≥1
	cn=9	4n2−12n+8, n≥1
	cn=12	$\frac{10}{3}{n}^3-5n^2+\frac{11}{3}n-1,n\ge 1$