# On Pentagonal Numbers

Hypsicles of Alexandria (c.190 – c.120 B.C.) gave the following definition of polygonal numbers :

If there are as many numbers as we please beginning from 1 and increasing by the same common difference, then, when the common difference is 1, the sum of all the numbers is a triangular number; when 2 a square; when 3, a pentagonal number [and so on]. And the number of angles is called after the number which exceeds the common difference by 2, and the side after the number of terms including 1.

Consider the following sequence :

$$1, 4, 7, 10, 13, 16, 19, 22, 25, 28, …$$

The terms have a common difference of $$3$$ beginning from $$1$$.

When they are summed, we have the following sequence :

$$1, 5, 12, 22, 35, 51, 70, 92, 117, 145, …$$

These are the pentagonal numbers.

An arithmetic sequence with first term $$a_1$$ and common difference $$d$$ can be specified by the following recurrence system :

$$a_1 = a_1, \,\, a_{n+1} = a_n + d, \, n = 1, 2, 3, …$$

It has the closed form :

$$a_n = a_1 + (n-1)d, \, n = 1, 2, 3, …$$

The first sequence has $$a_1 = 1$$ and $$d = 3$$, so

$$a_n = 1 + 3(n-1) = 1 + 3n-3 = 3n-2, \, n =1, 2, 3, …$$

We may used the closed form to sum the first sequence using the following standard results for the natural numbers and a constant :

$$\displaystyle \sum_{k=1}^n ck = \frac {cn(n + 1)}{2} \text {and} \sum_{k=1}^n c = cn$$,

where $$c$$ is an arbitrary constant.

Thus :

\displaystyle \begin{align} \sum_{k=1}^n 3k-2 = \sum_{k=1}^n 3k -\sum_{k=1}^n 2 \\ & = \frac {3n(n + 1)}{2} -2n \\ & = \frac {3n^2 + 3n}{2} – \frac {4n}{2} \\ & = \frac {3n^2-n}{2} \\ & = \frac {n(3n-1)}{2}\end{align}

Hence, the pentagonal numbers are given by :

$$\displaystyle P_n = \frac {n(3n-1)}{2}$$,  where $$n = 1, 2, 3, …$$

There are other ways to derive the formula.

$$\begin{array}{c|cccccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10\\ \hline T_n & 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & 45 & 55\\ S_n & 1 & 4 & 9 & 16 & 25 & 36 & 49 & 64 & 81 & 100\\ P_n & 1 & 5 & 12 & 22 & 35 & 51 & 70 & 92 & 117 & 145 \end{array}$$

We can see by inspection of the above that :

$$P_n = T_n + 2T_{n-1} = S_n + T_{n-1}$$,

and show it algebraically :

$$\displaystyle P_n = \frac{n(n +1)}{2} + \frac {2(n-1)n}{2} = n^2 + \frac {(n-1)n}{2} = \frac {n(3n-1)}{2}$$

For example :

$$P_6 = 21 + 2\cdot15 = 36 +15 = (6\cdot17)/2 = 51$$

Pentagonal numbers have the property that they are the sums of consecutive numbers :

$$\begin{array}{c|l|c} P_n & Run & Sum\\ \hline P_1 & 1 & 1\\ P_2 & 2 + 3 & 5\\ P_3 & 3 + 4 + 5 & 12\\ P_4 & 4 + 5 + 6 + 7 & 22\\ P_5 & 5 + 6 + 7 + 8 + 9 & 35\\ P_6 & 6 + 7 + 8 + 9 + 10 + 11 & 51 \end{array}$$

Euler generalised the pentagonal numbers in his Pentagonal Number Theorem which leads to a way of calculating the number of partitions of $$n$$.

© OldTrout $$2019$$

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