A365486 a(n) is equal to the number of black 1 X 1 X 1 cubes in a certain coloring of the n X n X n cube (see comments for precise definition).

1, 8, 7, 32, 57, 112, 159, 256, 353, 504, 647, 864, 1081, 1376, 1663, 2048, 2433, 2920, 3399, 4000, 4601, 5328, 6047, 6912, 7777, 8792, 9799, 10976, 12153, 13504, 14847, 16384, 17921, 19656, 21383, 23328, 25273, 27440, 29599, 32000, 34401, 37048, 39687

Offset

1,2

Comments

The coloring of the n X n X n cube is carried out as follows. First, the cubes of the central cross are painted black. For odd cubes, the ends of the central cross are 1 X 1 squares; for even cubes, the ends of the central cross are 2 X 2 squares. Then all the cubes adjacent to black are painted white so that a layer is obtained with all white cubes. At the next step, all cubes adjacent to white are painted black so that a layer of black cubes is obtained, and so on, until all the cubes of the n X n X n cube are painted.

Links

Table of n, a(n) for n=1..43.

Nicolay Avilov, Illustration of initial terms

Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Formula

Let r = n (mod 4), then a(n) = (n^3)/2 if r = 0; (n^3 - 3*n)/2 +2 if r = 1; (n^3)/2 + 4 if r = 2; (n^3 - 3*n)/2 - 2 if r = 3.

G.f.: x*(1 + 6*x - 9*x^2 + 20*x^3 + 7*x^4 - 2*x^5 + x^6)/((-1 + x)^4*(1 + x)^2*(1 + x^2)). - Thomas Scheuerle, Sep 06 2023

Example

a(3) = 7;
a(5) = 13 + 44 = 57;
a(9) = 25 +236 + 92 = 353.

Mathematica

LinearRecurrence[{2, 0, -2, 2, -2, 0, 2, -1}, {1, 8, 7, 32, 57, 112, 159, 256}, 50] (* Amiram Eldar, Sep 05 2023 *)

Prog

(PARI) Vec(x*(1 + 6*x - 9*x^2 + 20*x^3 + 7*x^4 - 2*x^5 + x^6)/((-1 + x)^4*(1 + x)^2*(1 + x^2)) + O(x^50)) \\ Thomas Scheuerle, Sep 06 2023

Crossrefs

Cf. A266725.

Sequence in context: A303311 A267093 A090099 * A138809 A286460 A317231

Adjacent sequences: A365483 A365484 A365485 * A365487 A365488 A365489

Keyword

nonn,easy

Author

Nicolay Avilov, Sep 05 2023

Status

approved