# Given $n+1\mid2\sum_{k=1}^{n}{a_k}$, find $a_k$.

Let $m$ be a positive integer. There are only 2 finite sequences of positive integers like $a_1,a_2,...,a_m$ such that $$(\forall n \leq m)\left(n+1\mid2\sum_{k=1}^{n}{a_k}, \quad a_n\in [1,m],\quad a_{n}\notin\{a_1,.., a_{n-1}\}\right)$$

Here for any 2 real numbers $a$ and $b$ we define:

1) $a \mid b$ means $(\exists k \in \mathbb{Z})(b=ka)$.

2) $[1,m] =\{x\in \mathbb{R}|1\leq x \leq m\}$.

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No, that notation is used for integers $a,b$. Otherwise, $a$ can always "divide" $b$ as long as $a\neq 0$. Since $n+1$ is never $0$ any sequence satisfies your "definition" – L. F. Feb 5 '13 at 23:45
Are you sure that your question is right? Note that $(n+1)|\frac{n(n+1)}{2}$ only for even integers.... – N. S. Feb 6 '13 at 0:09
thanks. I edited. – user59671 Feb 6 '13 at 0:11

$$1+2+...+n=\sum_{k=1}^n k=\frac{n(n+1)}{2}=\sum_{k=1}^n a_k$$ $$a_k=k$$

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I edited my post. there was a mistake in it. – user59671 Feb 5 '13 at 23:35

I think it's not hard to see that even with the extension to real numbers you only get half-integers. For the integer $n+1$ to divide the sum, the sum has to be an integer, so each $2a_k$ has to be an integer.

But it looks like you do get some more sequences this way. Unless I miscalculated, there's one beginning $1,2,3,3/2,9/2,11/2,5/2,\dots$ and another starting $1,2,3,3/2,9/2,11/2,13/2,\dots$

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good point. I encountered this problem for integers in high-school (and remembered it hardly after years). I wondered if this can be generalized to real numbers. However I noticed another mistake in the question and I edited it. I think real version may return to integer version somehow. – user59671 Feb 7 '13 at 23:48

The idea of divisibility can be extended to the gaussian integers $a+bi$, where $w | z$ if there exists $x$ Gaussian integer such that $wx = z$.

However, note that $n \mid a+bi$ if and only if $n \mid a$ and $n \mid b$. Hence, all you're saying is that $a_n, b_n$ must be one of your sequences.

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