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Number of ways a natural number can be written as sum of smaller natural number It is easy to realize that given a natural number N the number of doublets that sum N are $\frac{N+(-1)(N \pmod 2)}{2}$ , so I thought I could reach some recursive formula in the sense that found the number of doublets I could find the number of triplets and so on ...,  example: N=3 the only doublet is 2+1=3 -not said yet but 2+1, and 1+2 count as one- then I could count the number of way the number 2 can be expressed as the indicated sum and got the total number of ways 3 can be written as a sum. But this seems not so efficient, so I was wondering if there is other way to attack the problem and if there is some reference to this problem such as if it is well known its used, once I read that this have a chaotic behavior, and also read It was used in probability but don't remember where I got that information. So if you know something I would be grateful to be notice, thanks in advance. How to remove wine completely How can I remove the wine folder and all the applications in that folder?  I've tried via synaptic but it keeps showing up in the application menu. Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$ Prove that $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$     It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain:  $\begin{matrix}  \tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}  &amp;  \tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13} &amp; \tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13} &amp; \tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13} &amp; \tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}  \\  \tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}  &amp;  \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13} \\ \tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13} &amp; \tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13} \\ \tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13} &amp; \tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13} \\ \tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13} &amp; \tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13} \\ \tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13} &amp; \tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13} &amp; \tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13} &amp; \tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}  \end{matrix}$  one can notice then, that the first column vanish the fourth one :  $\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$  $\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$  and the third column vanish the fifth one :  $\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$  $\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$  $\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$  $\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$  while the second column is self-vanishing:  $\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$  $\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ .  So the equality occurs. But how to generalize the proof? Points of a Measure Zero Sets Covered by Intervals Infinitely Many Times Given a measure zero set $E$, by definition we have forall $\varepsilon &gt; 0$, a covering of $E$ by intervals whose lengths sum to $&lt; \varepsilon$.  I want to prove that we can cover $E$ in intervals such that the sum of the lengths the intervals is $&lt; 1$ and each point in $E$ is contained in infinitely many of the intervals.  Do you know how to prove this? Thank you Probability of a 2 pair in poker Configuring dpkg stops processing when calling an HTTP request to Adobe Flash Player I recently moved to Linux, and I am in love with the experience. Opinions apart, I am having problems when running sudo dpkg --configure -a  My computer stops processing the command here:  You may know Adobe Flash Player is no longer supported, and I believe that could be why the command is unable to load pepperflashplugin-nonfree. I have tried fixing the problem by checking older posts here, but none of them was able to fix this problem. I keep seeing custom flags in the queue from question askers who are incredibly irritated because there is a big, prominent box at the top of their question. It's just a list of other questions which may have the answer they desire, and is only even displayed to them because the question is not closed yet.  This question may already have an answer here: [list]  But to an ordinary user, it is not at all clear that the box is only shown to them, or that it will go away on its own if their question never actually gets closed. Could the text for this notice be altered to make it more clear that:  They are just suggestions for the person to view. Potentially if they agree with one, they should vote to close their own question as a duplicate of it.  Their question is not closed yet. If they disagree with the duplicates suggested, they should clarify their question and explain why it's not a duplicate.   I was thinking of an altered wording such as:  You might be able to find an answer to your question here: [list] If those answers do not fully address your question, please [edit your question] to clarify your exact problem.  I went with a more direct wording for the bolded title. Since the notice is only displayed to the asker, it makes more sense to talk directly to them in the message rather than making it feel like it's supposed to be addressing the entire audience of the site. Emphasize that only that person is seeing it by talking to them. It would also be helpful to append a &quot;More info&quot; link at the end of the message that links them to some further documentation about duplicates, and explains how the notice works - particularly that it will go away eventually or change to a proper duplicate notice if the question does get closed. How to see this improper integral diverges? $$\int^\infty_1\frac{1}{x^{1+1/x}}dx$$  I'm preparing for exams. I would also like to know what are some commonly used methods to show an improper integral diverges? Difference between if () { } and if () : endif; Are there any differences between...  if ($value) {  }   ...and...  if ($value):  endif;   ? Evaluation of $ \sum_{k=0}^n \cos k\theta $ I just wanted to evaluate  $$ \sum_{k=0}^n \cos k\theta $$  and I know that it should give  $$  \cos\left(\frac{n\theta}{2}\right)\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin(\theta / 2)}   $$  I tried to start by writing the sum as  $$ 1 + \cos\theta + \cos 2\theta + \cdots  + \cos n\theta $$  and expand each cosine by its series representation. But this soon looked not very helpful so I need some clue about how this partial sum is calculated more efficiently ... Phone : nexus 5 version 2013 Android version : 6.0 Kernel version : 3.4.0-g2aa165e Build number : MRA58N   For the last few days, 10118 has been draining my phone's battery. It drains battery by using GPS. I typed 'ps' in a terminal emulator, and there was no application with PID 10118. I don't know what exactly 10118 is. My guess is that it's one of recently installed weather apps.  How can I fix this?    Screenshots (click images for larger variants) What constitutes a legal block in volleyball in regard to it not counting as the first hit? Today in a match the setter set the ball too low for me to spike it, so I just dinked it over.  The bottom of the ball was lower than the top of the net at the point where I made contact.  When the defender made contact, again, the bottom of the ball was lower than the top of the net (the top of the ball was over the top of the net though).  I felt that this wasn't technically a block and that he shouldn't have been allowed to hit it a second time, but I wasn't sure so I just let it go.  Was this a legal block or should it have been our point when he hit it a second time in a row?  Edit:  I don't think I was clear about the part where, after I hit the ball, the ball travelled over the net onto the other team's side of the court.  The other player did not reach over the net so there is no issue there.  It's just an issue of whether or not their first contact was a block or a hit.  I believe it was not a block but I can't find a specific ruling to address it. why __getitem__ cannot be classmethod? On several occasions, I was reading a question thinking "I know a guy that knows the answer to this" and that guy (or gal) also happens to be a registered user. How can I bring this to the attention of this user? SCP fails without error What special characters must be escaped in regular expressions? Hello [Comma?] John, I'm trying to show that for $b&gt;1$, $x&gt;0$ and $x$ irrational, that $$\sup \{b^p:p\in Q,\;0&lt;p&lt;x\} = \inf\{b^q:q\in Q,\;x&lt;q\}$$ I know this follows immediately if we define $b^x=\exp(x\log(b))$ and use the fact that the exponential function is strictly increasing and continuous over $\mathbb R$, but I'm not allowed to do that. In fact, I only have that $b^{m/n}=(b^{1/n})^m$, where $b^{1/n}$ means the positive $n^{th}$ root of $b$; $b^x$ is not yet defined for irrational $x$.  I know that the supremum and infimum both exist because if $b&gt;1$ and $p&lt;q$, then $b^p&lt;b^q$. Also, by assuming that the inf $&lt;$ sup, I got a contradiction. The part I'm stuck on is deriving a contradiction from assuming that sup $&lt;$ inf. Does anybody have any ideas?  Thank you. is there any way to decode string? How to set spawnpoint?