Centered cube number

35 points in a body-centered cubic lattice, forming two cubical layers around a central point.
Totalno. of terms	Infinity
Subsequence of	Polyhedral numbers
Formula	${\displaystyle n^3+(n+1{)}^3}$
First terms	1,9,35,91,189,341,559
OEIS index	A005898 Centered cube

Acentered cube number is acenteredfigurate number that counts the points in a three-dimensional pattern formed by a point surrounded by concentriccubical layers of points, withi² points on the square faces of theith layer. Equivalently, it is the number of points in abody-centered cubic pattern within a cube that hasn + 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, ... (sequenceA005898 in theOEIS).

The centered cube number for a pattern withn concentric layers around the central point is given by the formula^[1]

${\displaystyle n^3+(n+1{)}^3=(2n+1)\left(n^2+n+1\right).}$

The same number can also be expressed as atrapezoidal number (difference of twotriangular numbers), or a sum of consecutive numbers, as^[2]

${\displaystyle (\genfrac{}{}{0ex}{}{(n+1{)}^2+1}{2})-(\genfrac{}{}{0ex}{}{n^2+1}{2})=(n^2+1)+(n^2+2)+\cdots +(n+1{)}^2.}$

Because of the factorization(2n + 1)(n² +n + 1), it is impossible for a centered cube number to be aprime number.^[3]The only centered cube numbers which are also thesquare numbers are 1 and 9,^[4][5] which can be shown by solvingx² =y³ + 3y, 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.

• Cube number

• Weisstein, Eric W."Centered Cube Number".MathWorld.

• Figurate numbers

• Articles with short description
• Short description is different from Wikidata
• Use American English from March 2021
• All Wikipedia articles written in American English
• Use mdy dates from March 2021