Centered cube number

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title: "Centered cube number" type: doc version: 1 created: 2026-02-28 author: "Wikipedia contributors" status: active scope: public tags: ["figurate-numbers"] description: "Centered figurate number that counts points in a three-dimensional pattern" topic_path: "general/figurate-numbers" source: "https://en.wikipedia.org/wiki/Centered_cube_number" license: "CC BY-SA 4.0" wikipedia_page_id: 0 wikipedia_revision_id: 0

::summary Centered figurate number that counts points in a three-dimensional pattern ::

::data[format=table title="Infobox integer sequence"]

Field	Value
image	Body centered cubic 35 balls.svg
image_size	200px
caption	35 points in a body-centered cubic lattice, forming two cubical layers around a central point.
number	Infinity
parentsequence	Polyhedral numbers
formula	n^3 + (n + 1)^3
first_terms	1, 9, 35, 91, 189, 341, 559
OEIS	A005898
OEIS_name	Centered cube
::

| image = Body centered cubic 35 balls.svg | image_size = 200px | alt = | caption = 35 points in a body-centered cubic lattice, forming two cubical layers around a central point. | number = Infinity | parentsequence = Polyhedral numbers | formula = n^3 + (n + 1)^3 | first_terms = 1, 9, 35, 91, 189, 341, 559 | OEIS = A005898 | OEIS_name = Centered cube A centered cube number is a centered figurate number that counts the points in a three-dimensional pattern formed by a point surrounded by concentric cubical layers of points, with i2 points on the square faces of the ith layer. Equivalently, it is the number of points in a body-centered cubic pattern within a cube that has n + 1 points along each of its edges.

The first few centered cube numbers are

:1, 9, 35, 91, 189, 341, 559, 855, 1241, 1729, 2331, 3059, 3925, 4941, 6119, 7471, 9009, ... .

Formulas

The centered cube number for a pattern with n concentric layers around the central point is given by the formula

:n^3 + (n + 1)^3 = (2n+1)\left(n^2+n+1\right).

The same number can also be expressed as a trapezoidal number (difference of two triangular numbers), or a sum of consecutive numbers, as :\binom{(n+1)^2+1}{2}-\binom{n^2+1}{2} = (n^2+1)+(n^2+2)+\cdots+(n+1)^2.

Properties

Because of the factorization (2n + 1)(n2 + n + 1), it is impossible for a centered cube number to be a prime number. The only centered cube numbers which are also the square numbers are 1 and 9,{{citation | last = Stroeker | first = R. J. | title = On the sum of consecutive cubes being a perfect square | journal = Compositio Mathematica | volume = 97 | year = 1995 | issue = 1–2 | pages = 295–307 | mr = 1355130 | url = http://www.numdam.org/item?id=CM_1995__97_1-2_295_0}}. which can be shown by solving , the only integer solutions being (x,y) from {(0,0), (1,2), (3,6), (12,42)}, By substituting a=(x-1)/2 and b=y/2, we obtain x^2=2y^3+3y^2+3y+1. This gives only (a,b) from {(-1/2,0), (0,1), (1,3), (11/2,21)} where a,b are half-integers.

References

References

1. (2012). "Figurate Numbers". World Scientific.
2. Lanski, Charles. (2005). "Concepts in Abstract Algebra". American Mathematical Society.
3. {{Cite OEIS. A005898
4. (2007). "The Magic Numbers of the Professor". Mathematical Association of America.

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