$\pi$ is irrational, and of course its digits are not a repeating sequence. However I cannot wrap my head around the following...  Is there a digit $x$ in $\pi$ such that the digits before are repeated in reverse order after that digit? $$\pi = 3.14159\ldots zyxxyz \ldots 951413 [\ldots]$$ In other words, is there a positive integer $n$ such that $\lfloor 10^n\pi\rfloor$ is ?  $\pi$ has infinite digits, but it is unknown whether any sequence of digits will appear at some point. Moreover, this isn't just a requirement for $n$ digits appearing at some position, but for a sequence of $n$ digits has to appear at position $n$.
284
I came across the :    Which states:     $\pi$ Pi      Pi is an infinite, nonrepeating $($sic$)$ decimal - meaning that   every possible number combination exists somewhere in pi.  Converted   into ASCII text, somewhere in that infinite string of digits is the   name of every person you will ever love, the date, time and manner    of your death, and the answers to all the great questions of    the universe.   Is this true? Does it make any sense ?
60971
$\pi\approx 3.141592654$   Why is it so close to $3$?   I find this intriguing, this cannot be a coincidence.
68700
I came across this when learning about sequences and series. No one proved it to me tho! Is there a link anywhere? Or would it be beyond what I know? (First term of analysis)
64298
i.e. can we write  $\pi = 3.14159\dots X\dots$  where $X$ consists of (say) $10^{100}$ consecutive zeroes?  [Originally asked on reddit without response :-( ]
80866
$\text{What are some interesting cases of $\pi$ appearing in situations that do not seem geometric?}$  Ever since I saw the identity $$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$  and the generalization of $\zeta (2k)$, my perception of $\pi$ has changed. I used to think of it as rather obscure and purely geometric (applying to circles and such), but it seems that is not the case since it pops up in things like this which have no known geometric connection as far as I know. What are some other cases of $\pi$ popping up in unexpected places, and is there an underlying geometric explanation for its appearance?  In other words, what are some examples of $\pi$ popping up in places we wouldn't expect?
61995
What's the $n$-th digit in the sequence $S$ of numbers formed by the natural numbers, i.e.,  $n$-th digit in the sequence 1 2 3 4 5 6 7 8 9 10 11 12... ? For example 11th digit in the sequence is 0.
91620
It is easy to argue that the  converges:  $$  \sum\limits_{\substack{n \text{ s.t. 9 is}\\\text{ not a digit} \\\text{ of } n}}  \frac{1}{n} &lt; \infty$$  Let $E \subset \Bbb N_{&gt;0}$  the subset of the positive integers  such that $9$ is not a digit of the decimal expansion of $1/n$ (the decimal expansion is not allowed to have a trailing infinite sequence of "$9$"s. For instance $0.24999...$ is not allowed).  Here are the  that don't belong to $E$ : $11,13,17,19,21,23,29,31,34,38,41,…$ (not known by the OEIS, by the way).  My question is:     Does the series    $$ \sum\limits_{n \in E} \frac{1}{n} \tag 1$$ converge?   My attempt is :   Let $1/n = 0,a_1 a_2 \dots a_k \overline{b_1 b_2 \dots b_m}$ with $n \in E$. Since $1/n$ has no digit "9", we have at most $9^{k+m}$ possibilities for the $a_i$'s and $b_j$'s. Moreover, $1/n ≥ 0,00...0\overline{00...01}≥1/10^{k+m}$. But then I can only bound my series $(1)$ from below, by some real number. So, this is not a clue for the divergence of the series. Apparently, the numbers of the form $n=10k+1$ don't belong to $E$. Maybe we can find sufficiently many numbers that have $9$ in the decimal representation of their reciprocals, so that $(1)$ could converge...   Any comment will be appreciated !
105372
If you consider the (decimal) numbers 10, 101, 1010, 10101... where the 1 and 0 alternate, what is the largest prime number in the sequence?  Thanks in advance for your help, I appreciate it.
102472
Question     Numbers $n$ of the form $10^{m}(2^{k}−1)+2^{k-1}−1$, where $m$ is the number of decimal digits of $ 2^{k-1}$. For example:         $k=1$ then $n=10$.   $k=2$ then $n=31$.    $k=3$ then  $n=73$.   $k=4$ then $n=157.$      Conjecture:   the number $(2^k-1)\cdot 10^m+2^{k-1}-1$ where $m$ is the number of decimal digits of $2^{k-1}$ is never prime when it is of the form $7s+6$, that is when it is congruent to $6$ $\pmod 7$. Examples: $n=1023511$ ($k=10$)$\equiv 6 \pmod 7$ and thus it is composite $(1023511=19\times103\times523)$, $n=20471023$ ($k=11$) $\equiv 6 \pmod 7$ and thus it is composite ($20471023=479\times42737)$. With PFGW we arrived to $k=565000$ and all the $n's$ congruent to $6 \pmod 7$ are composite. According to Giovanni Resta's calculations in a post which has been canceled, there should be no probable prime congruent to 6 $\pmod 7$ upto k=800.000. The residue $6$ $\pmod 7$ occurs when either $m=6t+3$ and $k=3l+1$ or $m=6t+4$ and $k=3l+2$ with $k$ and $l$ some non-negative integers, but amazingly when it occurs the number is not prime. Can you find a counter-example or give a proof for the conjecture? Here a link to other interesting questions:  and  For primes of this form see:  The following vector contains all the exponents k&lt;=366800 leading to a prime  $[2, 3, 4, 7, 8, 12, 19, 22, 36, 46, 51, 67, 79, 215, 359, 394, 451, 1323, 2131, 3336, 3371, 6231, 19179, 39699, 51456, 56238, 69660, 75894, 79798, 92020, 174968, 176006, 181015, 285019, 331259, 360787, 366770]$  Exponent $541456$ leads to another probable prime with residue 5 mod 7 and 325990 digits, but it need not be the next in increasing order.  Remark: we found five-in-a-row probable primes with res 5 mod 7. Probable primes with residue 5 are now twice frequent than expected. Exponents of these primes seem to be NOT random at all. Another thing I noticed, i don't know if it has some importance: the exponents leading to a probable prime $215, 69660, 92020, 541456$ are multiples of $43$. I noticed that $\frac{215}{41}, \frac{69660}{41}, \frac{92020}{41}, \frac{541456}{41}$ all have a periodic decimal expansion equal to $\overline{24390}=29^3+1$. This is equivalent to say that when k is a multiple of 43 and the number $10^{m}(2^{k}−1)+2^{k-1}−1$ is prime, then k is of the form $41s+r$ where r is a number in the set (1,10,16,18,37). Is there some mathematical reason for that?
17281
Does there exist a positive irrational number $\alpha $, such that for any positive integer $n$ the number $\lfloor n\alpha \rfloor$ is not a prime? My try if $\alpha=\sqrt{17}$ then $\lfloor n\alpha \rfloor=4n$
48976
I'm assuming this theorem was found by someone else before, but I found this relationship between square numbers of 3 digits or less. The theorem is this: If you reverse the digits in a square number, then the result will also be a square number. Take the square 961. It is 31 squared, and if you reverse the digits you will get 169, which is also a square number. Plus, 31 and 13(the roots of these reversed squares) are also reverses of eachother. The problem is this breaks with 4 or more digits. If I take the square 1024 and reverse the digits, I get 4201, which is not a square. How can I expand this theorem to fit 4 or more digits?
72019
Show that the sequence $\{a_n\}_{n\in \mathbb{N}} = \{x: x=$ the nth decimal digit of Champernowne's constant$\}$ is not periodic.  For those who don't know what Champernowne's constant is, it's the following number:  $C_{10} = 0.123456789101112131415161718192021...$  Well, before someone says that $\{a_n\}_{n\in \mathbb{N}}$ is definitely aperiodic because Champernowne's constant is irrational(and even transcendental), I have to say that I'm well aware of the fact that a number is rational if and only if the sequence of its digits is periodic, but I'm looking for another proof that the sequence of the digits of $C_{10}$ is aperiodic that doesn't use anything but a smart way of showing that $a_{n+k} = a_n$ can't hold for any $k\in \mathbb{N}$.  I've observed that by grouping the terms of a sequence in a clever way we can show that no such $k$ can exist for some sequences. For example I created the following sequence:  $1,0,0,1,1,1,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,1,1,1,1,1,\cdots$  The sequence is created this way: First you put one $1$, then two $0$'s, and in the nth step you add either $1$'s or $0$'s n times depending on whether n is odd or even respectively.   It's easy to show this sequence is not periodic, because no matter how large $k$ is you can always find $n$ such that $a_n = 1$ but $a_{n+k}=0$ or $a_n=0$ but $a_{n+k}=1$. That proves the sequence is not periodic and as a corollary it' proved that the real number $0.1001110000111110000001111111\cdots$ is irrational.  After showing that I've been trying to apply this technique of grouping to the sequence in problem without much success so far.  I think if the periodicity of a sequence $\{a_n\}$ is denoted by $\Omega(\{a_n\})$ then if we define a new sequence $\bar{a}_m(n) := \bar{a}_{m,n} = [a_n]_m$ then $\Omega(\{\bar{a}_{m,n}\}) \mid \Omega({a_n})$. This shouldn't be very hard to prove  I think.  Now, to show that   $1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,2,0,2,1,\cdots$  is not periodic it's sufficient to prove that the sequence   $1,0,1,0,1,0,1,0,1,1,0,1,1,1,0,1,0,1,0,1,\cdots$  is not periodic. But I haven't progressed much from here  :/
124256
Why are the ten digits in base 10 noted $0, 1, 2 ,3 ,4 ,5 ,6 ,7 ,8 ,9$ ? I heard it has to do with angles, but since I can't see how $9$ has $9$ angles, this may not be the historical reason ?
115219
Here is the sequence of irrational number that converges to $\alpha\in \mathbb{Q}$. Take $x_n=\alpha -\frac{\sqrt{2}}{n}$. Clearly $\{x_n\}\rightarrow\alpha .$  But I'm trying to find a sequence of rationals that converges to arbitrary irrational number $\beta$. Can you give me such a sequence.
13134
There's a problem statement:     For each $m \in \mathbb{N}$, we construct a sequence $m_0$, $m_1$, $m_2,\dots$ denoted $S_m$, recursively via $m_0=m$ and      $$m_{i+1} = m_i + \left\lfloor \sqrt{m_i} \right\rfloor$$      for all $i \ge 0$.   Here, $\lfloor x \rfloor$ is the floor of $x$, the greatest integer less than or equal to $x$.   Hence, we have $\left\lfloor \sqrt{10} \right\rfloor=3$ and $\left\lfloor \sqrt{29} \right\rfloor=5$.      Show that for each positive integer $m$, the sequence $S_m$ contains the square of some integer.   I'm pretty certain that this can be proved with induction. I am just not quite sure what to induct on. Examining examples shows that $S_m$ always results in a perfect square eventually, though I'm not sure how to prove it.
59256
Let $n$ be a positive integer. I know the traditional proof that $e$ is irrational. How do we show that $e^n$ is irrational in some sort of similar line? I am of course assuming it is but I would be astounded if not. It occurs to me that since $e$ is transcendental, of course $e^n$ is irrational, but I don't want to use that fact.  Googling gives me something for $e^2$, but I could not easily find anything for $e^3$.
89025
Take a look at this symbol:   $$ \pi=3 + \underset{k=1}{\overset{\infty}{\large{\mathrm K}}} \frac{(2k-1)^2} 6 $$   Does it look familiar to you? If so please help me!