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Here $\mathbb{N}=\{1,2,3,\dots\}$.

We say that a set $A\subset\mathbb{N}$ is an additive base of natural numbers if there is a positive integer $h$ such that every natural number can be written as $a_1+\cdots+a_h$ for some (not necessarily distinct) $a_i\in A\cup\{0\}$. If there exists we call the smaller $h\in \mathbb{N}$ with this property order of the additive base $A$.

Some famous results about additive bases are:

  1. By Langrange's theorem we know that the set of perfect squares is an additive base of order 4.

  2. More generally Hilbert was the first to prove Waring's problem which says that the sets of $k$-powers is an additive base of natural numbers for every $k\in \mathbb{N}$. (The exact order is another very interesting problem.)

  3. As Schnirelmann first proved, the set of prime numbers along with $1$ is an additive base of natural numbers. The exact order is an open problem which is directly connected with Goldbach's conjecture.

My question formulated when I studied a variation of Hilbert's first proof on Waring's problem. Hilbert first proved the theorem for even powers $2k$. Then in a very beatifull way he managed to "fill" the "gaps" and prove the theorem for the remaining cases of odd powers. Inspired by this technique I started to think about the limitations of this method.

My question has two parts:

Let $A=\{a_n\}_{n=1}^{\infty}\subset \mathbb{N}$ (the terms are in ascending order always) be an additive base of $\mathbb{Ν}$. Observe here that one must have $a_1=1$. Now let $B=\{b_n\}_{n=1}^{\infty}\subset \mathbb{N}$ another set with these properties:

  1. $b_1=1$.
  2. For every $n\geq 2$ one has $a_{n-1}< b_n\leq a_n$.

I was wondering if we can extract that $B$ is also an additive base of $\mathbb{N}$. My intuition says that the answer is no, but I cannot find a counterexamble.

It would be even more surprising to me if I had a counterexamble when the set $A$ above was the set of perfect squares. In that case for the set $B$ we have:

  1. $b_1=1$.
  2. For every $n\geq 2$ one has $(n-1)^2< b_n \leq n^2$.

I would be very grateful if someone interested in these questions could enlightened me in some way. Thanks in advance!

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