A371515 Centered truncated octahedral numbers: the number of integer triples (x,y,z) such that max(|x|,|y|,|z|) <= 2n and |x|+|y|+|z| <= 3n.

1, 57, 341, 1045, 2361, 4481, 7597, 11901, 17585, 24841, 33861, 44837, 57961, 73425, 91421, 112141, 135777, 162521, 192565, 226101, 263321, 304417, 349581, 399005, 452881, 511401, 574757, 643141, 716745, 795761, 880381, 970797, 1067201, 1169785, 1278741

Offset: 0

Author

Peter Kagey, Mar 25 2024

Keywords

• nonn
• easy

Examples

			The a(1) = 57 lattice points are all permutations of the points (0,0,0),(0,0,1),(0,0,2),(0,1,1),(0,1,2),(1,1,1), where any number of the coordinates can also be made negative (e.g., (1,-1,0)).
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..10000
• Eric Weisstein's World of Mathematics, Figurate Number.
• Eric Weisstein's World of Mathematics, Truncated Octahedron.
• Wikipedia, Ehrhart polynomial.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A001845, A005900.

Programs

Mathematica

Array[32*#^3 + 18*#^2 + 6*# + 1 &, 50, 0] (* or *)
LinearRecurrence[{4, -6, 4, -1}, {1, 57, 341, 1045}, 50] (* Paolo Xausa, Apr 02 2024 *)

Python

def A371515(n): return n*(n*(32*n+18)+6)+1 # Chai Wah Wu, Apr 02 2024

Formula

a(n) = 32*n^3 + 18*n^2 + 6*n + 1.

a(n) = A001845(3*n) - 6*A005900(n).

G.f.: (19*x^3 + 119*x^2 + 53*x + 1) / (x-1)^4. - Paolo Xausa, Apr 02 2024

E.g.f.: exp(x)*(1 + 56*x + 114*x^2 + 32*x^3). - Stefano Spezia, Apr 02 2024