Square pyramidal number

Geometric representation of the square pyramidal number 1 + 4 + 9 + 16 = 30.

A pyramid of cannonballs in the Musée historique de Strasbourg. The number of balls in the pyramid can be calculated as the fifth square pyramidal number, 55.

In mathematics, a pyramid number, or square pyramidal number, is a figurate number that represents the number of stacked spheres in a pyramid with a square base. Square pyramidal numbers also solve the problem of counting the number of squares in an n × n grid.

Formula

The first few square pyramidal numbers are:

1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819 (sequence A000330 in the OEIS).

These numbers can be expressed in a formula as

$P_{n}=\sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {2n^{3}+3n^{2}+n}{6}}.$

This is a special case of Faulhaber's formula, and may be proved by a straightforward mathematical induction. An equivalent formula is given in Fibonacci's Liber Abaci (1202, ch. II.12).

In modern mathematics, figurate numbers are formalized by the Ehrhart polynomials. The Ehrhart polynomial L(P,t) of a polyhedron P is a polynomial that counts the number of integer points in a copy of P that is expanded by multiplying all its coordinates by the number t. The Ehrhart polynomial of a pyramid whose base is a unit square with integer coordinates, and whose apex is an integer point at height one above the base plane, is (t + 1)(t + 2)(2t + 3)/6 = P_t + 1.

proof outline

It is possible to write down this formula:

$\sum _{k=0}^{n}k^{2}=\sum _{k=0}^{n-1}\sum _{i=k+1}^{n}i$

This is an alternative way to group the sum of the squares: to quote an example, we assume n=5:

$0^{2}+1^{2}+2^{2}+3^{2}+4^{2}+5^{2}=$

$1+2+3+4+5+$

$2+3+4+5+$

$3+4+5+$

$4+5+$

$5$

In this representation the number five is summed 5 times, the number 4 is summed 4 times, and so on, until zero. After that, it is possible to use this result to solve the summation:

$\sum _{i=k}^{n}i={\frac {n(n+1)}{2}}-{\frac {k(k-1)}{2}}$

The main formula is equivalent to

$\sum _{k=0}^{n}k^{2}=\sum _{k=0}^{n-1}[{\frac {n(n+1)}{2}}-{\frac {k(k+1)}{2}}]$

After some agebraic manipulation it is possible to obtain:

${\frac {3}{2}}\sum _{k=0}^{n}k^{2}={\frac {n^{2}(n+1)}{2}}-{\frac {n(n-1)}{4}}+{\frac {n^{2}}{2}}$

This is equivalent to the desired formula.

Relations to other figurate numbers

The square pyramidal numbers can also be expressed as sums of binomial coefficients:

$P_{n}={{n+2} \choose 3}+{{n+1} \choose 3}.$

The binomial coefficients occurring in this representation are tetrahedral numbers, and this formula expresses a square pyramidal number as the sum of two tetrahedral numbers in the same way as square numbers are the sums of two consecutive triangular numbers. In this sum, one of the two tetrahedral numbers counts the number of balls in a stacked pyramid that are directly above or to one side of a diagonal of the base square, and the other tetrahedral number in the sum counts the number of balls that are to the other side of the diagonal. Square pyramidal numbers are also related to tetrahedral numbers in a different way:

$P_{n}={\frac {1}{4}}{\binom {2n+2}{3}}.$

The sum of two consecutive square pyramidal numbers is an octahedral number.

Augmenting a pyramid whose base edge has n balls by adding to one of its triangular faces a tetrahedron whose base edge has n − 1 balls produces a triangular prism. Equivalently, a pyramid can be expressed as the result of subtracting a tetrahedron from a prism. This geometric dissection leads to another relation:

$P_{n}=n{\binom {n+1}{2}}-{\binom {n+1}{3}}.$

Besides 1, there is only one other number that is both a square and a pyramid number: 4900, which is both the 70th square number and the 24th square pyramidal number. This fact was proven by G. N. Watson in 1918.

In the same way that the square pyramidal numbers can be defined as a sum of consecutive squares, the squared triangular numbers can be defined as a sum of consecutive cubes.

Squares in a square

A 5 by 5 square grid, with three of its 55 squares highlighted.

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:

$n^{2}+(n-1)^{2}+(n-2)^{2}+(n-3)^{2}+\ldots +1^{2}={\frac {n(n+1)(2n+1)}{6}}.$

That is, the solution to the puzzle is given by the square pyramidal numbers.

The number of rectangles in a square grid is given by the squared triangular numbers.

References

• Abramowitz, M.; Stegun, I. A. (Eds.) (1964). Handbook of Mathematical Functions. National Bureau of Standards, Applied Math. Series 55. p. 813. ISBN 0486612724.{{cite book}}: CS1 maint: multiple names: authors list (link)
• Beiler, A. H. (1964). Recreations in the Theory of Numbers. Dover. p. 194. ISBN 0486210960.
• Goldoni, G. (2002). "A visual proof for the sum of the first n squares and for the sum of the first n factorials of order two". The Mathematical Intelligencer. 24 (4): 67–69.
• Sigler, Laurence E. (trans.) (2002). Fibonacci's Liber Abaci. Springer-Verlag. pp. 260–261. ISBN 0-387-95419-8.

External links

• Weisstein, Eric W. "Square Pyramidal Number". MathWorld.