While messing around with the idea of ordinal collapsing functions, I stumbled upon an interesting simple function:

$$C(0)=\{0,1\}\\C(n+1)=C(n)\cup\{\gamma+\delta:\gamma,\delta\in C(n)\}\\\psi(n)=\min\{k\notin C(n),k>0\}$$

The explanation is simple. We start with $\{0,1\}$ and repeatedly add it's elements to themselves:

$C(1)=\{0,1,2\}\\ C(2)=\{0,1,2,3,4\}\\ C(3)=\{0,1,2,3,4,5,6,7,8\}\\\vdots\\C(n)=[0,2^n]$

And $\psi(n)$ is defined as the smallest integer not within $C(n)$, which is $2^n+1$.

I then extended my function. Imagine all the same definition, except that we now have

$$C(n+1)=C(n)\cup\{\gamma+\delta,\color{red}{\gamma\cdot\delta}:\gamma,\delta\in C(n)\}$$

This simple change gives us something a bit more complicated. The first few sets are

$C(1)=\{0,1,2\}\\ C(2)=\{0,1,2,3,4\}\\ C(3)=\{0,1,2,3,4,5,6,7,8,9,12,16\}\\ C(4)=\small\left\{\begin{align}0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 27, 28, 30,\\ 32, 35, 36, 40, 42, 45, 48, 49, 54, 56, 60, 63, 64, 72, 80, 81, 84, 96, 108, 112, 128, 144, 192, 256\end{align}\right\}\\ C(5)=\{0,\dots,177,179,\dots\}\\ \vdots$

If $\psi(n)$ is the smallest natural number not found in $C(n)$, asymptotically, how fast does $\psi$ grow?

The first few values of $\psi$ are


Here's a program that outputs $\psi$ and here's a program that outputs $C$.

I'm looking for better bounds and/or asymptotic formulas in the form of

$$\psi(n)\approx x^{y^n}$$

Simply Beautiful Art
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  • For testing purposes, you could use [this program](https://tio.run/##Vcc7DoMwDADQqzAmBSwiMUFNhxzD8tBKDAjh8pVISM6eirHbe@vxcSlZLDu55GleVBWGG3Wi8ujQkpSGNWyD71tVAcCpYXrPVxjDXyW4HMnTyLkn4eLW4xbH2Do4ZFhg@667jmnOLNWcfg) to output $C(n)$. However, the numbers get fairly large fairly fast... – Simply Beautiful Art Dec 08 '17 at 23:18
  • So $\psi(n)\le4^n+1$? Hm... – Simply Beautiful Art Dec 08 '17 at 23:25
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    It would seem to me that $C(5)$ contains all of the natural numbers from $0$ to $177$, as well as a lot of other numbers. As long as the gap from one consecutive number to the next is less than $178$, we can fill it by adding all of our numbers to the lower number. The smallest gap I can find in $C(5)$ which is larger than $178$ is above $6000$, and $4^6$ is only $4096$. – Simply Beautiful Art Dec 08 '17 at 23:32
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    yes, I see the pattern of the last element of $C(n)$ as $2^1,2^2,2^4,2^8,...$ so a "brutal" upper bound could be $\psi(n)\le 2^{(2^{n-1})}+1$ – Masacroso Dec 08 '17 at 23:41
  • Haha, yes, that is extremely brutal of an upper bound. – Simply Beautiful Art Dec 08 '17 at 23:43
  • 2 times the smallest prime not in the previous set is one number that will never be present in the current set. Maybe that will give a bound. – mathreadler Dec 09 '17 at 09:42
  • @mathreadler The smallest prime not in $C(4)$ is $29$, but $29\times2=58$, which is in $C(5)$. One way to construct $58$ would be to take $3\times16+20$, and $3,16,20\in C(3)$. – Simply Beautiful Art Dec 09 '17 at 12:53
  • what, are you allowed to take any combination of multiplication and addition? i thought you had to choose either multiplication **or** addition of two elements in the previous set. – mathreadler Dec 09 '17 at 12:55
  • @mathreadler Yes, hence why $C(3+2)$, allowing two steps of operations between the elements of $C(3)$. – Simply Beautiful Art Dec 09 '17 at 12:58
  • yes if two steps between then you can do that, since all numbers which once get in will remain in the successors. – mathreadler Dec 09 '17 at 13:00
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    But the point is still there. $29$ is the smallest prime $\notin C(4)$, yet $29\times2\in C(5)$. – Simply Beautiful Art Dec 09 '17 at 13:01
  • @SimplyBeautifulArt 20 isn't in C(3), rather it's a result of an addition of more than two elements in C(3) – Abr001am Dec 09 '17 at 15:04
  • @Abra But $20=4\times5$. – Simply Beautiful Art Dec 09 '17 at 15:05
  • @SimplyBeautifulArt yes it is 16+4 too, kind of a tie that makes me go reserve a third color for numbers considered "bigender". – Abr001am Dec 09 '17 at 15:25
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    @simply look [here](https://jsfiddle.net/ftju3qm1/21/) just get the gaps and remove additions, they are pretty much familiar prime gaps that are filled with supplementary primes not existing in originated set multiplied by numbers from the same set! my conjecture is that $\phi(n)$ is either a prime not included in $C(n-1)$ or multiplied by a number included in it! – Abr001am Dec 10 '17 at 16:47
  • @SimplyBeautifulArt It's worth noting that $C(n)$ is actually *larger* than the set of numbers (individually) obtainable by up to $n$ steps of addition or multiplication starting from 0 or 1. Let's say $x,y \in C(9)$, then the intermediate values used to construct $x$ may be unavoidably different from those used to construct $y$, so it isn't necessarily true that $xy$ can be built in $10$ steps. – Erick Wong Dec 19 '17 at 17:23
  • Just for the reference: my quick-and-dirty program suggests the next two values of $\psi(n)$, following $178$, are $7391$ and $6550891$. For comparison, the cardinality of the $C(n)$ sets would be $2, 3, 5, 12, 53, 679, 67145, 357306081$. – Peter Košinár Dec 20 '17 at 11:27
  • Cheers: https://oeis.org/A296261 – Larry B. Dec 20 '17 at 15:59

2 Answers2


Let's suppose that $\psi(n)=k+1$. It thus follows that

$$\{x:x\in[0,k]\}\subset C(n)$$

By adding these together, we may find that

$$\{k+x:x\in C(n)\}\subset C(n+1)$$

And thus, the simple bound of

$$\psi(n+1)\ge k+k+1=2\psi(n)-1$$



By adding and multiplying, we find that

$$\{k+x,k\cdot x:x\in C(n)\}\subset C(n+1)$$

And by adding these again, we find that

$$\{k\cdot x_0+x_1:x_0,x_1\in C(n)\}\subset C(n+2)$$

Which encompasses all of the numbers from $0$ to $k^2+2k$, hence

$$\psi(n+2)\ge k^2+2k+1=\psi(n)^2$$



By adding and multiplying instead of just adding in the previous step, we find that

$$\{x_0,x_1k^2:x_0,x_1\in C(n+1)\}\subset C(n+2)$$

And by adding these, once again, we find that

$$\{x_1k^2+x_0:x_0,x_1\in C(n+1)\}\subset C(n+3)$$

which will encompass all of the numbers from $0$ to $2k^3+2k$, hence


which is not at all pretty, but since $2\psi(n)\ge1$ for all $n$, we get


And if $\psi(n)\ge\frac1{1-2^{-1/3}}\approx4.8$, then


And thus,


Doing this again gives


And for large enough $k$,




and in general, I'm expecting


For all $x$, large enough $n$, and some $y$.

Simply Beautiful Art
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  • How are you getting $\psi(n_4) \ge 2k^5 + 4k^4 + k^2 + 2k$? – Deedlit Dec 19 '17 at 06:34
  • @Deedlit It's $(k^2+2k)(2k^3+2k)$, that is, we can represent all multiples of $2k^3+2k$ up to that, and the gap may be filled with addition. (IIRC) – Simply Beautiful Art Dec 19 '17 at 15:01
  • I'm not quite seeing it... to fill in the gaps by addition, we need all the multiples of $2k^3 + 2k$ by C(n+3), and to get those multiples, we need $2k^3 + 2k$ by C(n+2). How do you get there? – Deedlit Dec 19 '17 at 20:58
  • @Deedlit Note that $$[0,k^2+2k]\cup\{2k^3\}\subset C(n+2)\\\implies\{2k^3x_0:x_0\in[0,k^2+2k]\}\subset C(n+3)\\ [0,2k^3+2k]\subset C(n+3)\\ \implies\{2k^3x_0+x_1:x_0\in [0,k^2+2k],x_1\in [0,2k^3+2k]\}\subset C(n+4)\\ [0,2k^3(k^2+2k)+2k^3+2k]\subset C(n+4)]$$ – Simply Beautiful Art Apr 14 '18 at 22:49

Building from Simply Beautiful Art's answer, let $A(n)$ be the greatest member of $C(n)$ less than $\psi(n+1)$. Then, $\psi(n+2) \geq \psi(n+1) + A(n)\psi(n)$. If we define $K = \lim \inf_{n->\infty} \frac{A(n)}{\psi(n)}$, then $\psi(n+2) > K\psi(n)\psi(n+1)$, and we should have that $\psi = O(b^{\phi^{n}})$ for some $b$.