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Set $U_1 = \{1,2\}$, and for $n\in\mathbb{N}$ with $n\geq 1$ we set $U_{n+1} = U_n\cup A_n$ where $A_n$ is the set of elements of $\mathbb{N}$ that are not in $U_n$ but can be uniquely written as a sum of $2$ elements of $U_n$.

Set $U=\bigcup_{n\in\mathbb{N}} U_n$.

What is the value of $$\lim \inf_{n\to\infty}\frac{|U\cap\{1,\ldots,n\}|}{n}$$?

EDIT: The sequence $\frac{|U\cap \{1,\ldots, n\}|}{n}$ seems to have a trend to grow smaller as $n$ grows large; I conjecture that the value I look for might be zero.

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  • $\begingroup$ Is there some conjecture about the value? Did you try some experiments (by hand or using the computer) to calculate $|U_n|$ at least for some small $n$'s? (The next logical step would be checking whether that sequence appears in OEIS.) $\endgroup$
    – Martin Sleziak
    Commented Jun 2, 2016 at 6:33
  • $\begingroup$ Thanks Martin for your comment; I conjecture the value is $0$ -- I edited the post accordingly. $\endgroup$
    – Dominic van der Zypen
    Commented Jun 2, 2016 at 6:37
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    $\begingroup$ If you read the comments to my answer, you can see that writing out the set $U_n$ at least for a few small $n$'s could help to clarify your question. (Specifically Emil Jeřábek asked whether when you write "sum of two elements", these two elements are supposed to be distinct. In other words, whether $4=2+2$ belongs to $A_2$ or not.) $\endgroup$
    – Martin Sleziak
    Commented Jun 2, 2016 at 11:47
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    $\begingroup$ Several issues worth being mentioned: a) appearance of $13$ in $A_5$, although $\max U_4=16>13$ ($13$ is not representable by $U_3$), as well as of $51$ in $A_7$; b) NOT appearance of $25$ in $U_6$ (it was not representable by $U_4$, but it is multiply representable by $U_5$); and c) NOT appearance of $49$ in $U+U$.(no numbers below $49$ can appear in $A_n$ with $n\geq 7$). This may warn from too simple approaches. $\endgroup$
    – Ilya Bogdanov
    Commented Jun 2, 2016 at 12:56
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    $\begingroup$ After a correction: If my computations are (now) correct, we have $A_2=\{3,4\}$, $A_3=\{7,8\}$, $A_4=\{12,14,15,16\}$, $A_5=\{13,21,26,27,29,31,32\}$, $A_6=\{50,54,55,56,57,59,60,61,62,63,64\}$. $\endgroup$
    – Ilya Bogdanov
    Commented Jun 2, 2016 at 13:30

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