# Generalised Pentagonal Numbers We considered the pentagonal numbers which begin :

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

That is :

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

Now consider the following sequence :

$$2, 5, 8, 11, 14, 17, 20, 23, 26, 29, …$$

which begins at $$2$$ and has a common difference of $$3$$ between the terms.

For any arithmetic sequence, we have :

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

So :

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

Now we can sum this using standard results :

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

This gives us the sequence :

$$2, 7, 15, 26, 40, 57, 77, 100, 126, 155, …$$

Let us call this sequence $$Q_n$$ :

$$\displaystyle Q_n = \frac {n(3n + 1)}{2}$$ for $$n = 1, 2, 3, …$$

What happens when we consider the negative integers?

If,

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

Then,

$$P_n = 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, …$$

If,

$$\displaystyle Q_n = \frac {n(3n + 1)}{2}$$ for $$n = -1, -2, -3, …$$

Then,

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

The two sequences can be interlaced together using the formula for $$P_n$$ with $$n$$ taking all values in the order $$0, 1, -1, 2, -2, 3, -3, …$$ which produces the following sequence :

$$0, 1, 2, 5, 7, 12, 15, 22, 26, 35, …$$

This sequence gives the generalised pentagonal numbers, $$G_n$$.

The formulas for $$P_n$$ and $$Q_n$$ and the sequence $$G_n$$ appear in Euler’s Pentagonal Number Theorem.

Note that the sequence $$Q_n$$ is the number of internal dots in a pentagonal number diagram. © OldTrout $$2019$$

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