I am pondering a question in number theory that touches upon the representation of natural numbers as sums of consecutive elements from a subset S of $\mathbb{N}$. Specifically, the question is:
Does there exist a subset $S$ of $\mathbb{N}$ such that all but finitely many natural numbers are the sum of multiple consecutive elements (see addendum) in $S$?
Considering the case where $S = \mathbb{N}$, the natural numbers that cannot be represented as a sum of multiple consecutive elements in $S$ are precisely the powers of 2, indicating that there are infinitely many such numbers. However, does this observation extend to every possible subset $S$ of $\mathbb{N}$?
Intuitively, I believe this might be the case for any subset $S$, but I'm at a loss for how to approach proving or disproving this conjecture.
I'm eager to see if anyone can provide insights, suggest methods for proof, or offer counterexamples. Your expertise and thoughts on this matter would be greatly appreciated.
By sum of multiple consecutive elements, I mean that if you write $S = \{s_1, s_2, \dotsc\}$ with $s_1 < s_2 < \dotsb$, then we're considering sums of the form $s_i + \dotsc + s_{i + n}$ for $n \ge 1$.
I conjecture that there is no such subset $S$. It follows from a more concrete
Conjecture. For each subset $S$ of $\mathbb N$ and each $n\in\mathbb N$, the set of natural numbers which are at most $2^n$ and are not sums of multiple consecutive elements of $S$, has size at least $n$.
I wrote a program to check Conjecture for small $n$, and it is already checked for $n=4$ and $n=5$.
Technical remarks. Adding $1$ to $S$, if needed, we can assume that $1\in S$, that is $s_1=1$. Moreover, it is convenient to add to $S$ also an element $s_0=0$. Now for each nonnegative integer $n$ we put $t_n=\sum_{i=0}^n s_i$. Then $t_0=0$, $t_1=1$, and $t_{n+1}-t_n> t_n-t_{n-1}$ for each natural $n$. The sums of multiple consecutive elements of $S$ are exactly the differences $t_n-t_m$ for nonnegative integers $n$ an $m$ such that $n-m\ge 2$.