TITLE: Alternating sum of positive integers
QUESTION [8 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.
Can some sequence $\{E\}$ with first term $1$ and second term $4$ be an alt-basis? What terms would this sequence include?
What about another sequence $\{F\}$ with first term $2$ and second term $3$? What terms would this sequence include?

REPLY [2 votes]: It is not hard to answer the two additional questions OP poses:

Can some sequence $\{E\}$ with first term 1 and second term 4 be an alt-basis? What terms would this sequence include?

This is not possible. The set $\{1,4\}$ generates $1,\_,3,4,\_,\_,\dots=(1),\_,(-1+4),(4),\_,\_,\dots$
Adding $x\gt 4$ as the $3$rd element generates four additional elements: $x,x-1,x-3,x-4$.
The smallest two are consecutive and can never fit inside the singular empty place between $1$ and $3$. 
If you want to fill in $2$ with $(n+1)$th element, because of $1$, the largest two elements already generated will always be $a_n-1,a_n$. A consequence is that, the smallest two elements you will be generating at this step are $a_{n+1}-a_n,a_{n+1}-a_n+1$ which are always consecutive. Since the space for $2$ we are filling is surrounded by two already generated values $1,3$, we always have the same problem.
Hence, if you want to represent $2$ by starting with $\{1,4\}$, you are forced to have at least one duplicate representation, violating the uniqueness requirement. 

What about another sequence $\{F\}$ with first term 2 and second term 3? What terms would this sequence include?

The sequence $a_1=2$ and $a_n=2^n-1,n\ge 2$ is an alt-basis: $F=\{2,3,7,15,31,\dots\}$
We have only changed the way $1,2,3$ are represented from $(1),(−1+3),(3)$ to $(−2+3),(2),(3)$ and preserved all the unique representations given by alt-basis $\{2^n-1\}$ (which is a known alt-basis due to another answer on this question.)
Alternatively,
It is not hard to show by induction, that every $a_n,n\ge 2$ of $\{F\}$ is an "anchor element". This implies $\{F\}$ is an alt-basis. See my answer to Aternating sum of an increasing sequence of positive integers for "anchor element", for more information.