From 4f670b750af5c11e1eac16d9cd8556455f89f46a Mon Sep 17 00:00:00 2001
From: Juan Marín Noguera <juan@mnpi.eu>
Date: Fri, 16 May 2025 22:18:44 +0200
Subject: Changed layout for more manageable volumes

---
 3.5.lyx | 1191 ---------------------------------------------------------------
 1 file changed, 1191 deletions(-)
 delete mode 100644 3.5.lyx

(limited to '3.5.lyx')

diff --git a/3.5.lyx b/3.5.lyx
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--- a/3.5.lyx
+++ /dev/null
@@ -1,1191 +0,0 @@
-#LyX 2.4 created this file. For more info see https://www.lyx.org/
-\lyxformat 620
-\begin_document
-\begin_header
-\save_transient_properties true
-\origin unavailable
-\textclass book
-\begin_preamble
-\input defs
-\end_preamble
-\use_default_options true
-\maintain_unincluded_children no
-\language english
-\language_package default
-\inputencoding utf8
-\fontencoding auto
-\font_roman "default" "default"
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-\use_package amsmath 1
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-\cite_engine basic
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-\biblio_style plain
-\use_bibtopic false
-\use_indices false
-\paperorientation portrait
-\suppress_date false
-\justification true
-\use_refstyle 1
-\use_formatted_ref 0
-\use_minted 0
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-\index Index
-\shortcut idx
-\color #008000
-\end_index
-\secnumdepth 3
-\tocdepth 3
-\paragraph_separation indent
-\paragraph_indentation default
-\is_math_indent 0
-\math_numbering_side default
-\quotes_style english
-\dynamic_quotes 0
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-\end_header
-
-\begin_body
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc1[10]
-\end_layout
-
-\end_inset
-
-Can a periodic sequence be equidistributed?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Not if it's a real sequence,
- because the period must be finite and,
- if 
-\begin_inset Formula $0\leq x_{1}<\dots<x_{t}\leq1$
-\end_inset
-
- are the numbers that appear in the period,
- then 
-\begin_inset Formula $\text{Pr}(\frac{1}{3}(2x_{1}+x_{2})\leq x<\frac{1}{3}(x_{1}+2x_{2}))=0\neq\frac{1}{3}(x_{2}-x_{1})$
-\end_inset
-
- (a similar proof can be made for 
-\begin_inset Formula $t=1$
-\end_inset
-
-).
- If it's an integer sequence it can happen;
- for example for the 
-\begin_inset Formula $b$
-\end_inset
-
--ary sequence with period 
-\begin_inset Formula $0,1,2,\dots,b-1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc2[10]
-\end_layout
-
-\end_inset
-
-Consider the periodic binary sequence 0,
- 0,
- 1,
- 1,
- 0,
- 0,
- 1,
- 1,
- 
-\begin_inset Formula $\dots$
-\end_inset
-
-.
- Is it 1-distributed?
- Is it 2-distributed?
- Is it 3-distributed?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-It is clearly 1-distributed and 2-distributed,
- but not 3-distributed because 
-\begin_inset Quotes eld
-\end_inset
-
-111
-\begin_inset Quotes erd
-\end_inset
-
- never appears.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc5[HM22]
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $U_{n}=(2^{\lfloor\lg(n+1)\rfloor}/3)\bmod1$
-\end_inset
-
-.
- What is 
-\begin_inset Formula $\text{Pr}(U_{n}<\frac{1}{2})$
-\end_inset
-
-?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-We have 
-\begin_inset Formula $\lfloor\lg(n+1)\rfloor=0$
-\end_inset
-
- for 
-\begin_inset Formula $n=0$
-\end_inset
-
-;
- 1 for 
-\begin_inset Formula $n=1,2$
-\end_inset
-
-;
- 2 for 
-\begin_inset Formula $n=3,4,5,6$
-\end_inset
-
-,
- 3 for 
-\begin_inset Formula $n=7,\dots,14$
-\end_inset
-
-,
- etc.,
- so the sequence 
-\begin_inset Formula $(2^{\lfloor\lg(n+1)\rfloor})_{n}$
-\end_inset
-
- has 1 1's,
- followed by 2 2's,
- 4 4's,
- 8 8's,
- etc.
- It is easy to prove by induction that,
- when 
-\begin_inset Formula $k\in\mathbb{N}$
-\end_inset
-
- is even,
- 
-\begin_inset Formula $2^{k}\equiv1\bmod3$
-\end_inset
-
-,
- and when it's odd,
- 
-\begin_inset Formula $2^{k}\equiv2\bmod3$
-\end_inset
-
-,
- and so 
-\begin_inset Formula $U_{n}<\frac{1}{2}$
-\end_inset
-
- precisely when 
-\begin_inset Formula $\lfloor\lg(n+1)\rfloor$
-\end_inset
-
- is even,
- which is when 
-\begin_inset Formula $2^{k}\equiv1\bmod3$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-If 
-\begin_inset Formula $\nu(n)=|\{m\leq n\mid U_{n}<\frac{1}{2}\}|$
-\end_inset
-
-,
- then 
-\begin_inset Formula $\nu(n)/n$
-\end_inset
-
- clearly increases when 
-\begin_inset Formula $n$
-\end_inset
-
- is between 
-\begin_inset Formula $2^{2k}-1$
-\end_inset
-
- and 
-\begin_inset Formula $2^{2k+1}-1$
-\end_inset
-
-,
- and it decreases between 
-\begin_inset Formula $2^{2k-1}-1$
-\end_inset
-
- and 
-\begin_inset Formula $2^{2k}-1$
-\end_inset
-
-,
- for 
-\begin_inset Formula $k\in\mathbb{N}^{*}$
-\end_inset
-
-.
- The limit exists if the subsequence made from these infinite local minima and the one made from these infinite local maxima both have a limit and these limits match.
-\end_layout
-
-\begin_layout Standard
-For the maxima,
- 
-\begin_inset Formula $\nu(1)=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $\nu(7)=5$
-\end_inset
-
-,
- 
-\begin_inset Formula $\nu(31)=21$
-\end_inset
-
-,
- etc.
- In general,
- 
-\begin_inset Formula 
-\[
-\nu(2^{2k+1}-1)=\sum_{i=0}^{k}2^{2k}=\frac{1-4^{k+1}}{1-4}=\frac{4^{k+1}-1}{3},
-\]
-
-\end_inset
-
-so
-\begin_inset Formula 
-\[
-\lim_{k}\frac{\nu(2^{2k+1}-1)}{2^{2k+1}-1}=\frac{\frac{4^{k+1}-1}{3}}{2\cdot4^{k}-1}=\frac{1}{3}\frac{4\cdot4^{k}-1}{2\cdot4^{k}-1}=\frac{2}{3}.
-\]
-
-\end_inset
-
-For the minima,
- 
-\begin_inset Formula $\nu(3)=1$
-\end_inset
-
-,
- 
-\begin_inset Formula $\nu(15)=5$
-\end_inset
-
-,
- etc.,
- and in general 
-\begin_inset Formula $\nu(2^{2k}-1)=\nu(2^{2k-1}-1)=\frac{4^{k}-1}{3}$
-\end_inset
-
-,
- so
-\begin_inset Formula 
-\[
-\lim_{k}\frac{\nu(2^{2k}-1)}{2^{2k}-1}=\frac{1}{3}\frac{4^{k}-1}{4^{k}-1}=\frac{1}{3}.
-\]
-
-\end_inset
-
-Since 
-\begin_inset Formula $\frac{1}{3}\neq\frac{2}{3}$
-\end_inset
-
-,
- this probability is undefined.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc10[HM22]
-\end_layout
-
-\end_inset
-
-Where was the fact that 
-\begin_inset Formula $m$
-\end_inset
-
- divides 
-\begin_inset Formula $q$
-\end_inset
-
- used in the proof of Theorem C?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-This is used for the sums in the second page of proof,
- when telling the range of 
-\begin_inset Formula $t$
-\end_inset
-
-.
- In particular,
- it is needed when evaluating the sum over 
-\begin_inset Formula $t$
-\end_inset
-
- in Equation (22).
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-exerc11[M10]
-\end_layout
-
-\end_inset
-
-Use Theorem C to prove that if a sequence 
-\begin_inset Formula $\langle U_{n}\rangle$
-\end_inset
-
- is 
-\begin_inset Formula $\infty$
-\end_inset
-
--distributed,
- so is the subsequence 
-\begin_inset Formula $\langle U_{2n}\rangle$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Since it's 
-\begin_inset Formula $\infty$
-\end_inset
-
--distributed,
- it's also 
-\begin_inset Formula $(2,2k)$
-\end_inset
-
--distributed for all 
-\begin_inset Formula $k\in\mathbb{N}^{*}$
-\end_inset
-
-,
- so 
-\begin_inset Formula $\text{Pr}(u_{1}\leq U_{2n}<v_{1},\dots,u_{2k}\leq U_{2n+2k-1}<v_{2k})=(v_{1}-u_{1})\cdots(v_{k}-u_{k})$
-\end_inset
-
- for any 
-\begin_inset Formula $u_{1},v_{1},\dots,u_{k},v_{k}\in[0,1)$
-\end_inset
-
- with each 
-\begin_inset Formula $u_{i}<v_{i}$
-\end_inset
-
-,
- and in particular,
- if we let 
-\begin_inset Formula $u_{2},u_{4},\dots,u_{2k}=0$
-\end_inset
-
- and 
-\begin_inset Formula $v_{2},v_{4},\dots,v_{2k}=1$
-\end_inset
-
- we get the formula that shows that 
-\begin_inset Formula $\langle U_{2n}\rangle$
-\end_inset
-
- is 
-\begin_inset Formula $k$
-\end_inset
-
--distributed.
- And since this 
-\begin_inset Formula $k$
-\end_inset
-
- is arbitrary,
- 
-\begin_inset Formula $\langle U_{2n}\rangle$
-\end_inset
-
- is 
-\begin_inset Formula $\infty$
-\end_inset
-
--distributed.
- Note that this argument applies to any 
-\begin_inset Formula $\langle U_{mn+j}\rangle$
-\end_inset
-
- with 
-\begin_inset Formula $m\in\mathbb{N}^{*}$
-\end_inset
-
- and 
-\begin_inset Formula $j\in\mathbb{N}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc18[HM22]
-\end_layout
-
-\end_inset
-
-Prove that if 
-\begin_inset Formula $U_{0},U_{1},\dots$
-\end_inset
-
- is 
-\begin_inset Formula $k$
-\end_inset
-
--distributed,
- so is the sequence 
-\begin_inset Formula $V_{0},V_{1},\dots$
-\end_inset
-
- where 
-\begin_inset Formula $V_{n}=\lfloor nU_{n}\rfloor/n$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Take any 
-\begin_inset Formula $u_{1},v_{1},\dots,u_{k},v_{k}\in[0,1)$
-\end_inset
-
- such that each 
-\begin_inset Formula $u_{i}<v_{i}$
-\end_inset
-
-.
- If 
-\begin_inset Formula $u_{i}\leq V_{n}<v_{i}$
-\end_inset
-
-,
- then 
-\begin_inset Formula $u_{i}-\frac{1}{n}\leq U_{n}<v_{i}+\frac{1}{n}$
-\end_inset
-
-,
- so if 
-\begin_inset Formula $S(n)\coloneqq\{\forall i,u_{i}\leq V_{n+i}<v_{i}\}$
-\end_inset
-
-,
- then 
-\begin_inset Formula 
-\begin{align*}
-\overline{\text{Pr}}(S(n)) & \leq\text{Pr}\left(\forall i,u_{i}-\frac{1}{n+i}\leq U_{n+i}<v_{i}+\frac{1}{n+i}\right)\\
- & \leq\text{Pr}\left(\forall i,u_{i}-\frac{1}{n}\leq U_{n+i}<v_{i}+\frac{1}{n}\right)\\
- & \leq\text{Pr}\left(\forall i,u_{i}-\frac{1}{n_{0}}\leq U_{n+i}<v_{i}+\frac{1}{n_{0}}\right)
-\end{align*}
-
-\end_inset
-
-for any 
-\begin_inset Formula $n_{0}\in\mathbb{N}$
-\end_inset
-
-,
- since the first finitely many terms of the sequence 
-\begin_inset Quotes eld
-\end_inset
-
-don't matter,
-\begin_inset Quotes erd
-\end_inset
-
- and since 
-\begin_inset Formula $n_{0}$
-\end_inset
-
- is arbitrary,
- taking limits on it we see that 
-\begin_inset Formula $\overline{\text{Pr}}(S(n))\leq\text{Pr}(\forall i,u_{i}\leq U_{n+i}<v_{i})=\prod_{i}(v_{i}-u_{i})$
-\end_inset
-
-.
- Similarly,
- if 
-\begin_inset Formula $u_{i}+\frac{1}{n}\leq U_{n}<v_{i}-\frac{1}{n}$
-\end_inset
-
-,
- then 
-\begin_inset Formula $u_{i}\leq V_{n}<v_{i}$
-\end_inset
-
-,
- so
-\begin_inset Formula 
-\[
-\underline{\text{Pr}}(S(n))\geq\text{Pr}\left(\forall i,u_{i}+\frac{1}{n}\leq U_{n+i}<v_{i}+\frac{1}{n}\right)\geq\text{Pr}\left(\forall i,u_{i}+\frac{1}{n_{0}}\leq U_{n+i}<v_{i}-\frac{1}{n_{0}}\right),
-\]
-
-\end_inset
-
-this time taking 
-\begin_inset Formula $n_{0}$
-\end_inset
-
- such that 
-\begin_inset Formula $\frac{1}{n_{0}}\leq v_{i}-u_{i}$
-\end_inset
-
- for every 
-\begin_inset Formula $i$
-\end_inset
-
-.
- Again we reach the conclusion that 
-\begin_inset Formula $\underline{\text{Pr}}(S(n))\geq\prod_{i}(v_{i}-u_{i})$
-\end_inset
-
-.
- We get the result by the same argument used at the end of Theorem A.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc28[HM21]
-\end_layout
-
-\end_inset
-
-Use the sequence (11) to construct a 
-\begin_inset Formula $[0..1)$
-\end_inset
-
- sequence that is 3-distributed,
- for which 
-\begin_inset Formula $\text{Pr}(U_{2n}\geq\frac{1}{2})=\frac{3}{4}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Let 
-\begin_inset Formula $(W_{n})_{n}$
-\end_inset
-
- be an 
-\begin_inset Formula $\infty$
-\end_inset
-
--distributed real-valued sequence,
- and let 
-\begin_inset Formula $(X_{n})_{n}$
-\end_inset
-
- be the 3-distributed binary sequence from (11),
- then 
-\begin_inset Formula $(U_{n}\coloneqq\frac{1}{2}(W_{n}+1-X_{n}))_{n}$
-\end_inset
-
- satisfies the properties.
- For if 
-\begin_inset Formula $0\leq u_{i}<v_{i}<1$
-\end_inset
-
-,
- 
-\begin_inset Formula $i\in\{1,2,3\}$
-\end_inset
-
-,
- and if we assume that,
- for each 
-\begin_inset Formula $i$
-\end_inset
-
-,
- 
-\begin_inset Formula $v_{i}\geq\frac{1}{2}$
-\end_inset
-
- implies 
-\begin_inset Formula $u_{i}\geq\frac{1}{2}$
-\end_inset
-
- (so the 
-\begin_inset Quotes eld
-\end_inset
-
-rectangle
-\begin_inset Quotes erd
-\end_inset
-
- is contained in one quadrant),
- then 
-\begin_inset Formula $u_{i}\leq U_{n}<v_{i}$
-\end_inset
-
- if,
- and only if,
- 
-\begin_inset Formula $\lfloor2u_{i}\rfloor=\lfloor2U_{i}\rfloor=1-X_{n}$
-\end_inset
-
- and 
-\begin_inset Formula $2u_{i}\bmod1\leq2U_{n}\bmod1=W_{n}\leq2v_{i}$
-\end_inset
-
-.
- Since 
-\begin_inset Formula $W_{n}$
-\end_inset
-
- is 
-\begin_inset Formula $(16,3)$
-\end_inset
-
--distributed,
- the triplets 
-\begin_inset Formula $(W_{n},W_{n+1},W_{n+2})$
-\end_inset
-
- starting at positions where 
-\begin_inset Formula $(X_{n},X_{n+1},X_{n+2})$
-\end_inset
-
- has a given value have the same density as those starting at positions where it has any other value,
- so 
-\begin_inset Formula 
-\begin{multline*}
-\text{Pr}(\forall i,u_{i}\leq U_{n}<v_{i})=\text{Pr}(\forall i,\lfloor2u_{i}\rfloor=1-X_{n})\text{Pr}(\forall i,2u_{i}\bmod1\leq W_{n}\leq2v_{i}\bmod1)=\\
-=\frac{1}{8}\prod_{i}(2v_{i}-2u_{i})=\prod_{i}(v_{i}-u_{i})
-\end{multline*}
-
-\end_inset
-
-and the sequence is 3-distributed (the cases where some 
-\begin_inset Formula $\lfloor2u_{i}\rfloor\neq\lfloor2v_{i}\rfloor$
-\end_inset
-
- can be split into cases where this is not the case).
- In addition,
- 
-\begin_inset Formula $\text{Pr}(U_{2n}\geq\frac{1}{2})=\text{Pr}(X_{2n}=0)=\frac{3}{4}$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc34[M25]
-\end_layout
-
-\end_inset
-
-Define subsequence rules 
-\begin_inset Formula ${\cal R}_{1}$
-\end_inset
-
-,
- 
-\begin_inset Formula ${\cal R}_{2}$
-\end_inset
-
-,
- 
-\begin_inset Formula ${\cal R}_{3}$
-\end_inset
-
-,
- ...
- such that Algorithm W can be used with these rules to give an effective algorithm to construct a 
-\begin_inset Formula $[0..1)$
-\end_inset
-
- sequence satisfying Definition R1.
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-
-\begin_inset Note Greyedout
-status open
-
-\begin_layout Plain Layout
-(I had to look up the solution.)
-\end_layout
-
-\end_inset
-
-The 
-\begin_inset Quotes eld
-\end_inset
-
-algorithm
-\begin_inset Quotes erd
-\end_inset
-
- gives us a potentially infinite amount of sequences 
-\begin_inset Formula $\langle U_{n}\rangle{\cal R}_{k}$
-\end_inset
-
- that are 1-distributed,
- so we may encode the properties that we want to check for in the value 
-\begin_inset Formula $k$
-\end_inset
-
-.
- Specifically,
- we want to check that,
- for an increasing sequence of bases 
-\begin_inset Formula $(b_{n})_{n}$
-\end_inset
-
-,
- 
-\begin_inset Formula $k\in\mathbb{N}^{*}$
-\end_inset
-
-,
- and 
-\begin_inset Formula $a_{1},\dots,a_{k}\in\{0,\dots,b-1\}$
-\end_inset
-
-,
- 
-\begin_inset Formula $U_{n-k}=a_{k},\dots,U_{n-1}=a_{1}$
-\end_inset
-
-,
- so if,
- for example,
- 
-\begin_inset Formula $k=10^{b}10^{a_{1}}10^{a_{2}}1\cdots10a^{j}$
-\end_inset
-
- with each 
-\begin_inset Formula $a_{i}<b$
-\end_inset
-
-,
- we may set 
-\begin_inset Formula ${\cal R}_{k}(x_{0},\dots,x_{n-1})=1$
-\end_inset
-
- if,
- and only if,
- 
-\begin_inset Formula $\lfloor bU_{n-1}\rfloor=a_{1}\land\dots\land\lfloor bU_{n-k}\rfloor=a_{k}$
-\end_inset
-
-.
- For every other value of 
-\begin_inset Formula $k$
-\end_inset
-
-,
- we may as well set 
-\begin_inset Formula ${\cal R}_{k}\equiv1$
-\end_inset
-
-.
-\end_layout
-
-\begin_layout Standard
-\begin_inset Note Note
-status open
-
-\begin_layout Plain Layout
-TODO 44 (
-\begin_inset Formula $<$
-\end_inset
-
-4pp.,
- 
-\begin_inset Formula $<$
-\end_inset
-
-1:53)
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-rexerc44[16]
-\end_layout
-
-\end_inset
-
-(I.
- J.
- Good.) Can a valid table of random digits contain just one misprint?
-\end_layout
-
-\begin_layout Standard
-\begin_inset ERT
-status open
-
-\begin_layout Plain Layout
-
-
-\backslash
-answer 
-\end_layout
-
-\end_inset
-
-Yes;
- for example,
- both 23456782019372837458 and 23456782019372837459 are random.
- Of course,
- this is not true for any misprint,
- as then all numbers would be random.
- For example,
- 23456782019372828221 is random but 23456782019372828222 isn't,
- as it contains too many 2's.
- This has been calculated with the following (terrible) code:
-\end_layout
-
-\begin_layout Standard
-\begin_inset listings
-lstparams "language=Python,numbers=left,basicstyle={\footnotesize\sffamily},breaklines=true"
-inline false
-status open
-
-\begin_layout Plain Layout
-
-import math
-\end_layout
-
-\begin_layout Plain Layout
-
-\end_layout
-
-\begin_layout Plain Layout
-
-def israndom(digs):
-\end_layout
-
-\begin_layout Plain Layout
-
-    N = len(digs)
-\end_layout
-
-\begin_layout Plain Layout
-
-    dev = math.sqrt(N)
-\end_layout
-
-\begin_layout Plain Layout
-
-    for k in range(0,
- math.floor(math.log10(N)) + 1):
-\end_layout
-
-\begin_layout Plain Layout
-
-        pos = 10**k
-\end_layout
-
-\begin_layout Plain Layout
-
-        expect = N/pos
-\end_layout
-
-\begin_layout Plain Layout
-
-        for ss in range(pos,
- 2*pos):
-\end_layout
-
-\begin_layout Plain Layout
-
-          sb = str(ss)[1:]
-\end_layout
-
-\begin_layout Plain Layout
-
-          amt = len([n for n in range(len(digs)) if digs[n:n+k] == sb])
-\end_layout
-
-\begin_layout Plain Layout
-
-          if abs(amt-expect) > dev:
-\end_layout
-
-\begin_layout Plain Layout
-
-              print(
-\begin_inset Quotes eld
-\end_inset
-
-FAIL
-\begin_inset Quotes erd
-\end_inset
-
-,
- digs,
- sb)
-\end_layout
-
-\begin_layout Plain Layout
-
-              return False
-\end_layout
-
-\begin_layout Plain Layout
-
-    return True
-\end_layout
-
-\begin_layout Plain Layout
-
-\end_layout
-
-\begin_layout Plain Layout
-
-for x in range(123456782019372800000,
- 123456790000000000000):
-\end_layout
-
-\begin_layout Plain Layout
-
-    if israndom(str(x)[1:]):
-\end_layout
-
-\begin_layout Plain Layout
-
-        print(str(x)[1:])
-\end_layout
-
-\end_inset
-
-
-\end_layout
-
-\end_body
-\end_document
-- 
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