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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.2012-11-16

3 Answers 3

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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 for some $n>3$. Then

$\begin{align*} \sum_{k=1}^{n+1}k&=n+1+\sum_{k=1}^nk\\ &

and the result follows by induction.

7

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 then you have $\frac{n(n+1)}2. So there are no solutions such that $ \begin{align*} \frac{n+1}2&

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](http://math.stackexchange.com/questions/238546/when-is-the-factorial-of-a-number-triangular).2012-11-16