Generalised Pentagonal Numbers

OldTrout, Going Postal

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, Going Postal
 

© OldTrout \(2019\)
 

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