Equivalence of Definitions of Pentagonal Number

Definition 1 implies Definition 2
Let $P_n$ be a pentagonal number by definition 1.

Let $n = 0$.

By definition:
 * $P_0 = 0$

By vacuous summation:
 * $\displaystyle P_0 = \sum_{i \mathop = 1}^0 \left({3 i - 2}\right) = 0$

By definition of summation:

and so:

Thus $P_n$ is a pentagonal number by definition 2.

Definition 2 implies Definition 1
Let $P_n$ be a pentagonal number by definition 2.

Then:

Then:
 * $\displaystyle P_0 = \sum_{i \mathop = 1}^0 \left({3 n - 2}\right)$

is a vacuous summation and so:


 * $P_0 = 0$

Thus $P_n$ is a pentagonal number by definition 1.

Definition 1 equivalent to Definition 3
We have by definition that $P_n = 0 = P \left({5, n}\right)$.

Then:

Thus $P \left({5, n}\right)$ and $P_n$ are generated by the same recurrence relation.