TITLE: Aternating sum of an increasing sequence of positive integers
QUESTION [9 upvotes]: Suppose $A = (a_n) = (a_1, a_2, a_3, . . .)$ is an positive, increasing sequence of integers.
Define an $A$- expressible number $c$ if $c$ is the alternating sum of a finite subsequence of $A.$ To form such a sum, choose a finite subset of the sequence $A,$ list those numbers in increasing order (no repetitions allowed), and combine them with alternating plus and minus signs. We allow the trivial case of one-element subsequences, so that each an is $A-$expressible.
Definition. Sequence $A = (a_n)$ is an “alt-basis” if every positive integer is uniquely $A-$ expressible. That is, for every integer $m > 0,$ there is exactly one way to express $m$ as an alternating sum of a finite subsequence of $A.$
Examples. Sequence $B = (2^{n−1}) = (1, 2, 4, 8, 16, . . .)$ is not an alt-basis because some numbers are B-expressible in more than one way. For instance $3 = −1 + 4 = 1 − 2 + 4.$
Sequence $C = (3^{n−1}) = (1, 3, 9, 27, 81, . . .)$ is not an alt-basis because some numbers (like 4 and 5) are not C-expressible.
An example of an alt-basis is $\{2^n-1\}=\{1,3,7,15,31,\ldots\}$
Is there a fairly simple test to determine whether a given sequence is an alt basis?
I have attempted to solve this from a limited knowledge in sequences and have found out various kinds of sequences do not work but fail to see what it is that could make it work.

REPLY [2 votes]: I did not collect the set of all alt-bases, but I did find some useful observations, including:

Alt-basis must contain an infinite number of terms of form $a_k=2^{k}-1,k\in N\subseteq\mathbb N$.

The converse does not hold. At the end, I give examples of alt-bases and not-alt-bases in this context.
Do correct me If I missed anything.


Let $A=\{a_1,a_2,\dots\}$ such that $a_1\lt a_2 \lt \dots$ are positive integers. 

Definition. For $A$ to be an "alt-basis", we need to have both the "uniqueness" and "completeness". In other words, every number is expressible in exactly one way via alternating summation of subsets of $A$, which are summed in increasing order.
Definition. A finite (sub)sequence $A|_n=\{a_1,\dots,a_n\}$ is an "alt-prefix" if every integer in $[1,2^{n}-1]$ is uniquely expressible via alternating summation of subsets of $A|_n$ when summed in increasing order. The element $a_n$ is called an "anchor element".
Definition. "Anchor sequence" is a set $\mathcal A(A):=\{a_{n_1},a_{n_2},\dots\}$ of  all "anchor elements" $a_{n_1},a_{n_2},\dots$

Notice that a set has $2^n$ subsets minus the empty set and that every subset can be rearranged in an increasing order. We want to assign a distinct value to each of those subsets via the alternating summation, to have an alt-basis. The alt-prefix is defined to cover exactly those $2^n-1$ subsets. It follows that:

Lemma. $A$ is an alt-basis $\iff$ $A$ is a union of alt-prefixes $A=A|_{n_1}\cup A|_{n_2}\cup \dots$

That is, $A$ is an alt-basis if and only if there exists a corresponding infinite anchor sequence $\mathcal A(A)$.
We add two more definitions to write all of this more easily:

Definition. Let $s(\{b_1,\dots,b_n\})$ be the result of the alternating summation of $b_1\lt b_2\lt \dots\le b_n$. Let $s_+$ and $s_-$ always start the alternating summation with $+,-$ respectively. Then $s_+=-s_-$. If $n$ is odd then $s=s_+$ and if $n$ is even then $s=s_-$. This guarantees $s\gt 0$ because the largest element $b_n$ will have a positive sign.
Definition. Define "$n$-th partial subset-sum set" of a positive increasing integer sequence $A$ as: 
$$\mathcal S_n(A):=\{s(A_i):A_i\in\mathcal P(A|_n)\}$$
Where $\mathcal P(A|_n)$ is the set of all subsets of $A|_n=\{a_1,a_2,\dots,a_n\}$.

The set of all "anchor elements" $\mathcal A(A)=\{a_{n_1},a_{n_2},\dots\}\subseteq A$  satisfies $S_{n_i}=[1,2^{n_i-1}-1]$ for all $n_i$.

Corollary. $A$ is an alt-basis if and only if  it "is covered by the anchor sequence": $\max \mathcal A(A)\to \infty$. 

Notice that $\max S_n = a_n$. If $a_n$ is an anchor element, then $\max S_n = 2^n-1$. This gives:

Proposition. If $a_n$ is an anchor element, then $a_n=2^n-1$.

The converse does not hold. For example, in $\{1,4,7\}$ the $a_3=7=2^3-1$ but $a_3$ is not an anchor element, because $S_3=\{1,3,4,6,7\}\ne[1,7]$. 

Example $1$. It is not hard to see that $\mathcal A(\{2^n-1\})=\{2^n-1\}$. This is because:

$S_1=\{(+1)\}$ $\implies$ $a_1$ is an anchor element.
$S_2=\{(+1),(-1+3),(3)\}$ $\implies$ $a_2$ is an anchor element.
$S_3=\{(+1),(-1+3),(3),(-3+7),(+1-3+7),(-1+7),(7)\}$ $\implies$ $a_3$ is an anchor element.
$\dots$ proceed via induction to show every $a_n$ is an anchor element. 

Since $\mathcal A(\{2^n-1\})$ exists and covers the entire $\{2^n-1\}$, the $\{2^n-1\}$ is an alt-basis.

Example $2$. The $\mathcal A(\{n\})=\{1\}$ does not cover the entire $\{n\}$, hence $\{n\}$ is not an alt-basis. 
It is not hard to see that $\max S_n = n \lt 2^n-1\implies a_n$ is not an anchor element, for every $n\gt 1$. 

Example $3.$ We construct an alt-basis where every $2$nd element is an anchor element.
$$A=\begin{cases}
2^n-1, & n\text{ is even} \\
2^n+2^{n-1}-1, & n\text{ is odd} 
\end{cases}$$
Use an inductive argument. Assume $n=2k$, $a_{n}=2^{n}-1$ is an anchor element, which means we have uniquely constructed all $I_0=[1,2^n-1]=S_{n}$ elements. Now we can subtract numbers in this interval from $a_{n+1}$ to see that:

$a_{n+1}=2^{n+1}+2^{n}-1$ will cover $I_1=[a_{n+1}-a_{n}, a_{n+1}]=[2^{n+1},2^{n+1}+2^{n}-1]$ 

Here we see that $I_0\cup I_1 \ne [1,2^{n+1}-1]$ $\implies$ $a_{n+1}$ is not an anchor elemetn.
To see that $a_{n+2}=2^{n+2}-1$ is an anchor element, lets see what will we cover with it: 

$a_{n+2}$ combined with $I_0$ will cover $I_2=[a_{n+2}-a_{n}, a_{n+2}]=[2^{n+1}+2^{n},2^{n+2}-1]$ 
$a_{n+2}$ combined with $I_1$ will cover $I_3=[a_{n+2}-a_{n+1},a_{n+2}-2^{n+1}]=[2^n,2^{n+1}-1]$ 

Now observe $I=I_0\cup I_3\cup I_1\cup I_2$ is equal to:
$$
I=[1,2^n-1]\cup[2^n,2^{n+1}-1]\cup[2^{n+1},2^{n+1}+2^{n}-1]\cup[2^{n+1}+2^{n},2^{n+2}-1]=[1,2^{n+2}-1]
$$
Implying $a_{n+2}$ covers $I=[1,2^{n+2}-1]=S_{n+2}$, $\implies$ $a_{n+2}$ is an anchor.
It is not hard to check base cases $n=1,2$, and we are done. We have: 
$$\mathcal A\left(\left.\begin{cases}
2^n-1, & n\text{ is even} \\
2^n+2^{n-1}-1, & n\text{ is odd} 
\end{cases}\right\}\right)=\{2^{2n}-1\}$$
So we have an alt-basis $A$.

Example $4$. It is not hard to show that:
$$
A=\{2^k,2^k+1,2^k+3,2^k+7,\dots,2^k+2^{k}-1,2^{k+2}-1,2^{k+3}-1,\dots\}
$$
Is an alt-basis for every $k=0,1,2,\dots$, whose anchors are all elements $a_n,n\gt k$.

Example $5$. The sequence of natural, triangular, tetrahedral,... numbers, or in general, any diagonal of the pascals triangle, is not an alt basis. 
This is because for every fixed $d$, there exists $n_0$, such that for all $n\ge n_0$, we have $\binom{n+d-1}{d}<2^n$ implying that $\max S_n\lt 2^n-1$ for all $n\ge n_0$. This implies the sequence of anchors has at most $n_0$ elements, implying $\max\mathcal A(A)\lt \infty$, hence we do not have an alt-basis becuase of inevetable duplicates.