# When is a factorial of a number equal to its triangular number?

Consider the set of all natural numbers $n$ for which the following proposition is true.

$$\sum_{k=1}^{n} k = \prod_{k=1}^{n} k$$

Here's an example:

$$\sum_{k=1}^{3}k = 1+2+3 = 6 = 1\cdot 2\cdot 3=\prod_{k=1}^{3}k$$

Therefore, $3$ is in this set.

Does this set include more than just $3$? If so, is this set finite or infinite? Furthermore, can this set be described by a rule or formula?

[Just a tidbit: This question indicates the triangular number $1+2+3+\cdots+n$ is called the termial of $n$ and is denoted $n?$. I'm all for it; let's see if it catches on.]

[Another tidbit: the factorial of $n$, written $n!$ and called "$n$-factorial," is abbreviated "$n$-bang" in spoken word.]

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I've never heard of termial. I would perhaps say summorial, informially (?) at least. –  Dan Brumleve Nov 16 '12 at 8:15

The only members of the set are $1$ and $3$: by actual computation $2$ is not in it, and by an easy induction argument the inequality $$\binom{n+1}2=\sum_{k=1}^nk<\prod_{k=1}^nk=n!$$ holds for all $n>3$.

Clearly $\binom{4+1}2=10<24=4!$. Suppose that $\binom{n+1}2<n!$ for some $n>3$. Then

\begin{align*} \sum_{k=1}^{n+1}k&=n+1+\sum_{k=1}^nk\\ &<n+1+n!&&\text{induction hypothesis}\\ &<2n!&&\text{since }n+1<n!\text{ for }n\ge3\\ &<(n+1)n!\\ &=(n+1)!\;, \end{align*}

and the result follows by induction.

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You are asking for which natural numbers $$\frac{n(n+1)}2=n!$$

This is true for $n=1$; let us assume $n>1$ from now on.

Notice that whenever $\frac{n+1}2<n-1$ then you have $\frac{n(n+1)}2<n(n-1)\le n!$. So there are no solutions such that \begin{align*} \frac{n+1}2&<n-1\\ n+1&<2(n-1)\\ n+1&<2n-2\\ 3&<n \end{align*}

Therefore you only have to try $n=1,2,3$ and you find out that $1$ and $3$ are the only solutions.

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The other answers are fine but it shouldn't be necessary to actually carry out the induction to see that the solution set is finite: the triangular numbers grow quadratically and factorials grow super-exponentially. A more interesting problem would be: how many solutions $(m,n)$ are there to $\sum_{k=1}^{m}{k} = \prod_{k=1}^{n}{k}$? In addition to $(1,1)$ and $(3,3)$ we also have $(15,5)$.

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I asked it here. –  Dan Brumleve Nov 16 '12 at 7:51