Prove or disprove that there exists a unique positive integer sequence $\{a_{n}\}$ satisfying a condition Question:

Prove or disprove: there exists a unique positive integer sequence $\{a_{n}\}$ satisfying the following condition:
$\forall m\in N^{+}$, there exists a unique integer sequence $\{b_{i,m}\},i=1,2,\cdots,m$, such that
  $$|b_{i,m}|\le i$$ for $i=1,2,\cdots,m$, and $$m=\sum_{i=1}^{m}\left(a_{i}\cdot b_{i,m}\right)$$
Question: does there exist a sequence $\{a_{n}\}$ that satisfies this condition?

I can't find a solution. Can someone find one? This problem is from Peking University Middle school students' mathematics Olympic problem 3. I found this problem is similar this interesting problem. So I guess this problem was maybe created by Paul Erdös?

A sequence exists.
Define $a_1 = 1$, and $ a_n$ recursively as
$$ a_n = 1 + 2 \sum_{i=1}^{n-1} i a_i. $$
For example, we get
$ a_2 = 1 + 2 \times (1\times 1) = 3$,
$a_3 = 1 + 2 \times ( 1 \times 1 + 2\times 3) = 15$,
$a_4 = 1 + 2 \times ( 1 \times 1 + 2 \times 3 + 3 \times 15 ) = 105$.
For the first few terms of $m$, we have
$\begin{array}{llll} 
1 & = 1  \times 1 & &\\
2 &= (-1) \times 1 &+ 1 \times 3 \\   
3 &= 0 \times 1 &+ 1 \times 3\\   
4 &= 1 \times 1 &+ 1 \times 3\\  
5 &= (-1) \times 1 &+ 2 \times 3\\   
6 &= 0 \times 1 &+ 2 \times 3\\   
7 &= 1 \times 1 &+ 2 \times 3\\
8 &= (-1) \times 1 &+ (-2) \times 3 &+ 1 \times 15\\
9 & = 0 \times 1 & + (-2) \times 3 & + 1 \times 15 \\
 \end{array}$
(Is there a pattern? You bet! We're just building the next "block" of terms from the previous.)
We will show that this sequence satisfies the conditions.
Lemma 1: Fix $m$. If for some $K$, $m + \sum_{i=1}^{k-1} i a_i < a_K$, then $b_{K,m} = 0 $.
It suffices to prove this for the largest value of $K$, and then induct downwards. From the construction of the sequence $a_i$, $b_{K,m}$ must be positive. Then, we cannot subtract off enough in $\sum -i a_i$ to reach $m$. Explicitly, we have
$$ \sum{ b_{i,m} a_i} \geq a_K - \sum_{i=1}^{K-1} i a_i  > m . _\square $$
Lemma 2: Fix $m$. For the largest $K$ such that, $m + \sum_{i=1}^{K-1} i a_i \geq a_K$, then $b_{K,m} = \lfloor  \frac{ m + \sum_{i=1}^{K-1} i a_i}{ a_K }   \rfloor $.
Since it is the largest $K$, Lemma 1 tells us that the coefficients of higher terms are all 0. Hence, this coefficient must be positive. If $b_{K,m}$ is not large enough, then you cannot make the sum large enough. Conversely, if $b_{K,m}$ is not small enough, then you cannot make the sum small enough. Explicitly, we get that 
$$   - \sum_{i=1}^{K-1} i a_i \leq m - b_{K,m} a_K \leq \sum_{i=1}^{K-1} i a_i, $$
$$  m - a_K + \sum_{i=1}^{K-1} i a_i   < m - \sum_{i=1}^{K-1} i a_i  \leq b_{K,m} a_K \leq m + \sum_{i=1}^{K-1} i a_i $$
The result follow by dividing by $a_K$. $_\square$
Proof of existence and existence: Apply Lemma 1 to determine the first non-zero coefficient $b_{K,m}$. Apply Lemma 2 to determine $b_{K,m}$, which is unique. Apply induction to $m' = m - b_{K,m} a_K$.

Note: OEIS informs me that this is the sequence of double factorial of odd numbers. I currently don't see how this fact lends itself to the solution, but that is very suspicious to me, especially when thinking in bases. However, I wasn't able to push through thinking on that front, but I believe that is valuable in understanding this question.
My suspicion is that this sequence $a_i$ is unique. However, it relies on the constructive approach, where I have a ton of $b_{i,m} = 0$. It's not obviously clear to me why we cannot allow for massive cancellations somehow.
Rough working. Suppose $a^* _i$ is another sequence that satisfies the condition.
1) We will show that $a^*_i$ is an increasing sequence.
2) We will show that $ a^*_i \geq a_i$.   
