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| author | Juan Marín Noguera <juan@mnpi.eu> | 2025-05-16 22:18:44 +0200 |
|---|---|---|
| committer | Juan Marín Noguera <juan@mnpi.eu> | 2025-05-16 22:18:44 +0200 |
| commit | 4f670b750af5c11e1eac16d9cd8556455f89f46a (patch) | |
| tree | e0f8d7b33df2727d89150f799ee8628821fda80a /3.5.lyx | |
| parent | 16ccda6c459c0fd7ca2081e9d541124c28b0c556 (diff) | |
Changed layout for more manageable volumes
Diffstat (limited to '3.5.lyx')
| -rw-r--r-- | 3.5.lyx | 1191 |
1 files changed, 0 insertions, 1191 deletions
diff --git a/3.5.lyx b/3.5.lyx deleted file mode 100644 index 0268000..0000000 --- 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" -\font_sans "default" "default" -\font_typewriter "default" "default" -\font_math "auto" "auto" -\font_default_family default -\use_non_tex_fonts false -\font_sc false -\font_roman_osf false -\font_sans_osf false -\font_typewriter_osf false -\font_sf_scale 100 100 -\font_tt_scale 100 100 -\use_microtype false -\use_dash_ligatures true -\graphics default -\default_output_format default -\output_sync 0 -\bibtex_command default -\index_command default -\float_placement class -\float_alignment class -\paperfontsize default -\spacing single -\use_hyperref false -\papersize default -\use_geometry false -\use_package amsmath 1 -\use_package amssymb 1 -\use_package cancel 1 -\use_package esint 1 -\use_package mathdots 1 -\use_package mathtools 1 -\use_package mhchem 1 -\use_package stackrel 1 -\use_package stmaryrd 1 -\use_package undertilde 1 -\cite_engine basic -\cite_engine_type default -\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 -\use_lineno 0 -\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 -\papercolumns 1 -\papersides 1 -\paperpagestyle default -\tablestyle default -\tracking_changes false -\output_changes false -\change_bars false -\postpone_fragile_content false -\html_math_output 0 -\html_css_as_file 0 -\html_be_strict false -\docbook_table_output 0 -\docbook_mathml_prefix 1 -\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 |
