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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 $h\in \mathbb{N}$ such that every natural number can be written as $a_1+\dots+a_h$ for some (not necessarily distinct) $a_i\in A\cup\{0\}$.

Some famous examples of such additive bases are the $k$-powers (Waring's problem) and the set of primes including $1$ (a theorem by Schnirelmann).

All the examples I encountered so far had the same property. If $A\subset\mathbb{N}$ was an additive base and $a_n$ was the $n$-th term of the set then $$\lim_{n\to \infty}\frac{a_{n+1}}{a_n}= 1.$$

It wasn't very hard to prove that indeed this is true for every additive base. Then I started to look the reverse direction. I tried to find a set $A=\{a_n\}_{n=1}^{\infty}$ (the terms are in ascending order) with $a_1=1\in A$ and $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}= 1$ which is not an additive base but I couldn't.

I would be glad if someone could enlightened me with such an example. Thanks in advance!

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    $\begingroup$ @lulu Note $1\in A$. But $1$ along with all multiples of $3$ works. $\endgroup$
    – Arthur
    Commented Feb 7, 2022 at 20:09
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    $\begingroup$ @Arthur Ah, thank you. Though the same principle works with a small modification (as you remark). $\endgroup$
    – lulu
    Commented Feb 7, 2022 at 20:09
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    $\begingroup$ @Arthur I think that the multiples of 3 along with 1 is an additive base with h=3. I am trying to find a non additive base with these properties. Yes with repeats. $\endgroup$
    – Iliopoulos Alexandros
    Commented Feb 7, 2022 at 20:20
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    $\begingroup$ @Arthur note that every element of $\mathbb{N}$ is of the form $3n+\varepsilon_1+\varepsilon_2$ for some $\varepsilon_i\in\{0,1\}$, so every element of $\mathbb{N}$ is indeed a sum of at most $3$ elements from $1\cup 3\mathbb{N}$. $\endgroup$
    – Atticus Stonestrom
    Commented Feb 7, 2022 at 20:21
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    $\begingroup$ @AtticusStonestrom Hence the question about repeats. $\endgroup$
    – Arthur
    Commented Feb 7, 2022 at 20:23

1 Answer 1

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Suppose that $A(x)$ denotes the number of elements of $A$ less than $x$. A standard lemma says that an additive basis $A$ of order $h$ one has to have $A(x) \gg x^{1/h}$. In particular, the elements of $A$ can't grow significantly faster than $n^h$. But now simply choose a set where the terms grow faster than $n^h$ for any $h$ but still have the desired limit property $\lim a_{n+1}/a_n = 1$, e.g. take $$a_n = \lfloor n^{\log \log n} \rfloor$$ for $n \ge 2$.

For a reference/proof of the "standard lemma" see Theorem 1 here: http://www.theoryofnumbers.com/melnathanson/pdfs/nath2010-138.pdf

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  • $\begingroup$ thanks a lot for your immediate responce! Very nice counter examble indeed! $\endgroup$
    – Iliopoulos Alexandros
    Commented Feb 8, 2022 at 0:39

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