Has this conjecture been proven/disproven/found before?

Question

My 11-year-old son has made the following conjecture, and wants to know if it's already known to be true or false:

if n is a whole number, define an operation #(n) = 1+2+3…..+n (like factorial but with addition) The conjecture: if n is odd, then #(n-1) = X*n (where X may be any natural number). That is, #(n-1) should be evenly divisible by n.

Answer 1

Hint: $1+2+...+n=\sum\limits_{k=1}^nk=\dfrac{n(n+1)}{2}$

Answer 2

$2\#(n)=(1+2+\cdots+n)+(n+n-1+\cdots+1)=(1+n)+(2+n-1)+\cdots+(n+1)=(n+1)+(n+1)+\cdots+(n+1)=n(n+1)$

So $\#(n)=\frac{1}{2}n(n+1)$.

Now $\#(n-1)=\frac{1}{2}(n-1)n$

This is divible by $n$ if $\frac{1}{2}(n-1)$ is an integer.

If $n$ is odd then $n-1$ is even so $\frac{1}{2}(n-1)$ is indeed an integer, therefore the conjecture is true!

Answer 3

That's a great observation for an 11 year old to make!

I'm not sure this is a proof (perhaps someone can check) but I think I can justify it.

Because $n$ is odd there are an even number of numbers in the sum you're considering.

This means they can always be paired up.

It remains to show that the sum is divisible by $n$ to do this I choose to pair them up outside in.

Notice that each pair totals $n$.

(It always will be if we step inwards by the same amounts from both ends)

Here is a specific example $n=7$ $$1+2+3+4+5+6+7$$

Pair up $$1+6 \\ 2+5 \\ 3+4$$ All the pairs sum to $7$