A weird but an interesting question I thought of ...

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Consider this random 100-digit integer that I generated using [random.org](http://random.org/)

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> $$9771602964370316251552537368279107346523948777589513994616004391156991741564023185294939440725424639$$
> 
> 
> 

This number will most likely have $1,2,3..$ upto $9$ in them. But when we further look into we can also find $10$, (i.e $1$ then $0$ right after) and also $11$. But there's no $12$ in them (i.e no $1$ then a $2$ right after).

> 
> So what I want to find out is what will be the approximation of $n$, such that there will be $1,2,3,...,(n-2),(n-1),n$ integers that can be seen inside a random x-digit. Also note that $(n+1)$ can never be found in a random x-digit.
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If you say that the 'n' value will change randomly and there cannot be any asymptotic or approximations of that. You're kind of wrong. See, I did this with 9 other random 100-digit numbers and also the 100 decimal places of the famous constants $\pi,e$ and $\phi$. The results are pretty close to $11$. Here are them; (scroll right to see the corresponding $n-values$)

$$\begin{array} {|r|r|}\hline Sl. No. & 100-Digit Integer & n-value \\ \hline 1 & 9771602964370316251552537368279107346523948777589513994616004391156991741564023185294939440725424639 & 11 \\ \hline 2 & 5933798531736924945959336111487355483181228762319970946972332473586986158448822614193445434946714049 & 9 \\ \hline 3 & 0748006010018385701305255708540275105163777493821503992289412752198434315022707697359036044788900096 & 10 \\ \hline 4 & 8895177092192596874320826762636449513223106780359367744719402357085023730983517329137267473940940810 & 10 \\ \hline 5 & 4921918080733246056200145588743524919616898465382373461652822273753544023393089458416653653351525383 & 9 \\ \hline 6 & 6386614054105909456194325500596228770058096441918965347422338768650600045406879715848359336828536144 & 10 \\ \hline 7 & 8599263737601951772329677396723688342052445287633091648167742178784189581850841772769226465303321562 & 9 \\ \hline 8 & 0660623044515625611084768937485195387862394776640249639087054616789402273433078233077669093361145164 & 11 \\ \hline 9 & 8747230800762985911116404157733707704590034002122077023051291424897238915375434573976156208269944527 & 9 \\ \hline 10 & 1177720655605689615530670722522895831819411456667323162088720568188745848931083487687853204730731860 & 11 \\ \hline π & 1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679 & 11 \\ \hline e & 7182818284590452353602874713526624977572470936999595749669676277240766303535475945713821785251664274 & 9 \\ \hline φ & 6180339887498948482045868343656381177203091798057628621354486227052604628189024497072072041893911374 & 9 \\ \hline \end{array}$$

Now take the average of all the n-values and you'll get

$n ≈ 9.8$, for $x=100$

I'm starting to work on $x=1000$ and for a hint $n ≈ 99$ or $100$ and this is just like approximate value of an approximate value, I'll find a good approximate value soon. So what are your thoughts on this? What will be a good expression that could take in the $x$ value and give a good approximate?

I'm got a formula, a crude, anecdotal that could give a value close to what is got. I'll post it soon after a few modifications. 

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**EDIT**

Some observations that I did.

$$n ≈ \frac{x}{10}\tag{1}\label{1}$$

Or I can say that for a $x$, the $n$ value is; (Big thanks to @glowstonetrees for an inspiration.)
$$n≈9\*10^{m−3}\tag{2}\label{2}$$
Where $m$ is the no. of digits of $x$. But we know $x$ is the no. of digits of the given $x$-digit value. So $m$ is like no. of digits of the no. of digits of $x$-digit integer

Now for my own anecdotal estimate of this;

$$n≈\pi^{2\left(\log\left(x\right)-1\right)}\tag{3}\label{3}$$

The reason I chose log is that it is a "slow-moving" function so it doesn't change rapidly. And don't ask me why $\pi$ is there, I don't know I kinda like it. But I also edited it and made it more efficient; so it is;

$$n≈\pi^{2\left(\log\left(x\right)-\sin\left(79.4\right)\right)}\tag{4}\label{4}$$

The reason I chose $\sin(79.4)$ is that $$\sin(79.4)=0.9829353491$$ So I didn't want an exact $1$ as in Eq.3 but a rather a number close to 0.98. And I believe this is an irrational number and $sin(79.8)$ is just an estimate for this. 

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Now to test all these expressions with $x=10000$, from data I got that $n≈999$

$$n≈\frac{10000}{10}=1000\tag{By Eq.1}\label{5}$$
$$n≈9\*10^{5−3}\tag{By Eq.2}\label{6}=900$$
$$n≈\pi^{2\left(\log\left(10000\right)-1\right)}≈961.38\tag{By Eq.3}\label{7}$$
$$n≈\pi^{2\left(\log\left(10000\right)-\sin\left(79.4\right)\right)}≈999.6\tag{By Eq.4}\label{8}$$

So yeah even know my equation, i.e Eq 4 did well; I have to say Eq 1 of $n≈\frac{x}{10}$ is really the best bet for simplicity and accuracy.