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

Question

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

How about real sequences?

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

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

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

Answer 1

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

Answer 2

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.

Answer 3

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 $1,2,3,3/2,9/2,11/2,13/2,\dots$