- Square pyramidal number
In

mathematics , a**pyramid number**, or**square pyramidal number**, is afigurate number that represents apyramid with a base and four sides. These numbers can be expressed in a formula as:$sum\_\{k=1\}^nk^2=\{n(n\; +\; 1)(2n\; +\; 1)\; over\; 6\}=\{2n^3\; +\; 3n^2\; +\; n\; over\; 6\}$

that is, by adding up the squares of the first "n"

integer s, or by multiplying the "n"thpronic number by the "n"th odd number. Bymathematical induction it is possible to derive one formula from the other. An equivalent formula is given inFibonacci 'sLiber Abaci (1202, ch. II.12).This is a special case of

Faulhaber's formula .The first few pyramid numbers are:

1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819

OEIS|id=A000330.

Pyramid numbers can be modelled in physical space with a given number of balls and a square frame that hold in place the number of balls forming the base, that is, "n"

^{2}. They also solve the problem of counting the number of squares in an "n" × "n" grid.**Relations to other figurate numbers**The pyramid numbers can also be expressed as sums of

binomial coefficient s (or of two consecutivetetrahedral number s) thus::$$n + 2} choose 3} + n + 1} choose 3}.

We can derive another relation between square pyramidal numbers and tetrahedral numbers: if we let $P\_n$ be the "n"th square pyramid number then

:$P\_n=\{n(n+1)(2n\; +\; 1)\; over\; 6\}=\{2n(2n+2)(2n+1)over\; \{4*6$=(1/4)T_{2n}

where $T\_n$ is the "n"th tetrahedral number.

The sum of two consecutive square pyramidal numbers is an

octahedral number .Besides 1, there is only one other number that is both a square and a pyramid number, 4900, the 70th square number and the 24th square pyramidal number. This fact was proven by

G. N. Watson in 1918.**quares in a square**A common

mathematical puzzle involves finding the number of squares in a large "n" by "n" square grid. This number can be derived as follows:*The number of 1×1 boxes found in the grid is $n^2$.

*The number of 2×2 boxes found in the grid is $(n-1)^2$. These can be counted by counting all of the possible upper-left corners of 2×2 boxes.

*The number of "k"×"k" boxes (1 ≤ "k" ≤ "n") found in the grid is $(n-k+1)^2$. These can be counted by counting all of the possible upper-left corners of "k"×"k" boxes.It follows that the number of squares in an "n" by "n" square grid is::$x\; =\; n^2\; +\; (n-1)^2\; +\; (n-2)^2\; +\; (n-3)^2\; +\; ldots\; +\; 1^2$or::$x\; =\; n(n+1)(2n+1)/6$That is, the solution to the puzzle is given by the square pyramidal numbers.

**ee also***

Tetrahedral number

*Squared triangular number **References***cite book

author = Abramowitz, M.; Stegun, I. A. (Eds.)

title = Handbook of Mathematical Functions

publisher = National Bureau of Standards, Applied Math. Series 55

year = 1964

pages = 813

id = ISBN 0486612724*cite book

author = Beiler, A. H.

title = Recreations in the Theory of Numbers

publisher = Dover

year = 1964

pages = 194

id = ISBN 0486210960*cite book

title = Fibonacci's Liber Abaci

author = Sigler, Laurence E. (trans.)

publisher = Springer-Verlag

year = 2002

id = ISBN 0-387-95419-8

pages = 260–261**External links***MathWorld | urlname = SquarePyramidalNumber | title = Square Pyramidal Number

*Wikimedia Foundation.
2010.*