Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove or disprove existence of sequence that satisfies:

  1. $a_i\neq a_j, i\neq j$,
  2. $\{a_i\}_{i=1}^{\infty}=\mathbb{N}$,
  3. $n^2\mid\sum_{i=1}^n a_i$ for all $n$

Odd numbers could satisfy the third condition, but it isn't bijecton onto $\mathbb{N}$, obviously.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Suppose you have started a sequence $a_1, \ldots, a_n$ where $i \neq j \implies a_i \neq a_j$ and $k^2 | \sum_{i=1}^k a_i$.

Pick any $a \in \mathbb{N} - \{a_1 \ldots a_n\}$. Say you want to find some a suitable integer $a_{n+1}$ such that the sequence continued by $a_{n+2} = a$ works. Note $s = \sum_{i=0}^n a_n$.

You need an $a_{n+1}$ satisfying $s+a_{n+1} \equiv 0 \pmod{(n+1)^2}$ and $s + a_{n+1} + a \equiv 0 \pmod{(n+2)^2}$. Write $s+a_{n+1} = k(n+1)^2$. You get the condition $k(n+1)^2+a \equiv 0 \pmod {(n+2)^2}$.

Since $(n+1)$ and $(n+2)$ are coprime, $(n+1)$ is invertible in $\mathbb{Z}/(n+2)^2\mathbb{Z}$, and this condition is equivalent to $k \equiv -a/(n+1)^2 \pmod{(n+2)^2}$.

So there exists infinitely many suitable $k$, and thus infinitely many possibles values for $k(n+1)^2-s = a_{n+1}$. By choosing $k$ large enough, you can choose $a_{n+1}$ larger than $a$ and than any $a_i$ so that the sequence is still injective.

Therefore, you can extend any given sequence $a_1 \ldots a_n$ into a larger sequence $a_1 \ldots a_{n+2}$ where $a_{n+2}$ is any number you want. Repeat the process to fill all the holes (for example by systematically choosing $a_{n+2} = \min (\mathbb{N} - { a_1 \ldots a_n } $) and obtain a complete sequence satisfying your conditions.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.