# Infinite difference length of integer subsets

Let $A$ be a set of integers. In our recent researches, we've faced to the following property and definition:

We say $A$ has infinite difference length, if

(a) For every integer $n$ there exist a number $k=2^q$ (for some positive integer $q$) and $a_1,\cdots,a_k\in A$ such that $$n=a_1+\cdots+a_{\frac{k}{2}}-(a_{\frac{k}{2}+1}+\cdots+a_k).$$

Now, denote by $k(n)$ the least $k$ obtained from (a).

(b) The set of all $k(n)$, where $n$ runs over all integers, is unbounded above.

For example, if $\gcd\{a,b\}=1$ then $A=\{a,b\}$ has infinite difference length, but not $A=\mathbb{Z}^+$ (it does not have the second condition (b)).

Now, my questions are:

(1) Does the set of all Fibonacci numbers have infinite difference length? (see https://math.stackexchange.com/questions/1989375/representation-of-integers-by-fibonacci-numbers)

(2) What about the Euler numbers?

(3) Does anybody know some important well-known integer sequences with infinite difference lengths?

(4) Did anyone see something like the above property (definition) yet?

• In (a), do you assume $a_i\ne a_j$? – Seva Aug 11 '18 at 8:09
• Condition 1 is satisfied if and only if $\mathcal{A}$ contains two coprime integers. Condition 2 is satisfied if and only if $\mathcal{A}$ is not a basis of the integers. In particular all sets containing two coprime integers and having counting function $x^\epsilon$. In particular this holds for the FIbonacci numbers. – Jan-Christoph Schlage-Puchta Aug 11 '18 at 12:50
• $\mathcal A = \{6,10,15\}$ satisfies condition (a) but does not contain two coprime integers. – Greg Martin Aug 11 '18 at 17:02
• @Greg Martin: Oops, right. Since the condition is shift invariant, we may assume $0\in\mathcal{A}$. Then I claim that $\mathcal{A}$ satisfies condition 1 if and only if the greatest common divisor of all elements in $\mathcal{A}$ is 1. Every sufficiently large integer can be written as the sum of elements of $\mathcal{A}$, since we are allowed to take elements repeatedly. Using 0 as padding we can take $k$ as large as we want. In particular if $m, n$ are sufficiently large integers, we can arrange $a_1+\dots+a_{k/2}=n$, $a_{k/2+1}+\dots+a_k=m$. Thus every integer can be written as $n-m$. – Jan-Christoph Schlage-Puchta Aug 11 '18 at 19:26