Sequence of Differences on Generalized Pentagonal Numbers

Theorem

Recall the generalised pentagonal numbers $GP_n$ for $n = 0, 1, 2, \ldots$

Consider the sequence defined as $\Delta_n = GP_{n + 1} - GP_n$:

$1, 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 13, 7, 15, 8, \ldots$

Then:

The values of $\Delta_n$ for odd $n$ consist of the odd numbers
The values of $\Delta_n$ for even $n$ consist of the natural numbers.

Proof

Recall the definition of the generalised pentagonal numbers $GP_n$ for $n = 0, 1, 2, \ldots$

$GP_n = \begin{cases} \dfrac {m \paren {3 m + 1} } 2 & : n = 2 m \\ \dfrac {m \paren {3 m - 1} } 2 & : n = 2 m - 1 \end{cases}$

for $n = 0, 1, 2, \ldots$

Hence:

 $\displaystyle \Delta_{2 n - 1}$ $=$ $\displaystyle GP_{2 n} - GP_{2 n - 1}$ $\displaystyle$ $=$ $\displaystyle \dfrac {n \paren {3 n + 1} } 2 - \dfrac {n \paren {3 n - 1} } 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {3 n^2 + n - 3 n^2 + n} 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {2 n} 2$ $\displaystyle$ $=$ $\displaystyle n$

which defines the sequence of natural numbers.

Then:

 $\displaystyle \Delta_{2 n}$ $=$ $\displaystyle GP_{2 n + 1} - GP_{2 n}$ $\displaystyle$ $=$ $\displaystyle GP_{2 \paren {n + 1} - 1} - GP_{2 n}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\paren {n + 1} \paren {3 \paren {n + 1} - 1} } 2 - \dfrac {n \paren {3 n + 1} } 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {\paren {3 n^2 + 6 n + 3} - \paren {n + 1} - 3 n^2 - n} 2$ $\displaystyle$ $=$ $\displaystyle \dfrac {4 n + 2} 2$ $\displaystyle$ $=$ $\displaystyle 2 n + 1$

which defines the sequence of odd numbers.

$\blacksquare$