I'm an Indian passport holder, travelling to Buenos Aires via London Heathrow (Flight departs same day within 24 hours). I am also a valid B1/B2 US Visa Holder. Would I still require a DATV (Direct Airside Transit Visa) despite that?
Is there a website or some other way to find out if I need a transit visa for a short stop in the UK on my way somewhere else? Ideally this should give me the answer for all nationalities, and take into account any other visas I may have.
What free/open source image design programs can you recommend? Are there any free/ open source image designing programs that you can recommend? Derivative of $f(x)=|x|$ Is it possible to shutdown remote Computers without inputing the password? I used the following bash to shutdown remote computers,but for each computer,I must input the password for it.Is it possible to input the password Automatically?   #!/bin/bash  ssh -t root@192.168.1.11 "sudo poweroff" ssh -t root@192.168.1.12 "sudo poweroff" ssh -t root@192.168.1.13 "sudo poweroff" ssh -t root@192.168.1.14 "sudo poweroff" ssh -t root@192.168.1.15 "sudo poweroff" ssh -t root@192.168.1.16 "sudo poweroff" ssh -t root@192.168.1.17 "sudo poweroff" ssh -t root@192.168.1.18 "sudo poweroff" ssh -t root@192.168.1.19 "sudo poweroff" ssh -t root@192.168.1.20 "sudo poweroff" ssh -t root@192.168.1.21 "sudo poweroff" ssh -t root@192.168.1.22 "sudo poweroff" ssh -t root@192.168.1.23 "sudo poweroff" ssh -t root@192.168.1.24 "sudo poweroff" ssh -t root@192.168.1.25 "sudo poweroff" ssh -t root@192.168.1.26 "sudo poweroff" I've read about Fermat's little theorem and generally how congruence works. But I can't figure out how to work out these two:   $13^{100} \bmod 7$ $7^{100} \bmod 13$   I've also heard of this formula:  $$a \equiv b\pmod n \Rightarrow a^k \equiv b^k \pmod n $$  But I don't see how exactly to use that here, because from $13^1 \bmod 7$ I get 6, and $13^2 \bmod 7$ is 1. I'm unclear as to which one to raise to the kth power here (I'm assuming k = 100?)  Any hints or pointers in the right direction would be great. In  I prove, using generating functions, a statement equivalent to  $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}$$ when $n$ is even.  (Clearly the sum is $0$ when $n$ is odd.)  The nice expression on the right-hand side indicates that there should be a pretty combinatorial proof of this statement.  The proof should start by associating objects with even parity and objects with odd parity counted by the left-hand side.  The number of leftover (unassociated) objects should have even parity and should "obviously" be $2^n \binom{n}{n/2}$.  I'm having trouble finding such a proof, though.  So, my question is     Can someone produce a combinatorial proof that, for even $n$, $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} (-1)^k = 2^n \binom{n}{n/2}?$$   Some thoughts so far:   Combinatorial proofs for $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$ are given by  and by .  The proofs are basically equivalent.  In Phira's argument, both sides count the number of paths of length $2n$ starting from $(0,0)$ using steps of $(1,1)$ and $(1,-1)$.  By conditioning on the largest value of $2k$ for which a particular path returns to the horizontal axis at $(2k,0)$ and using the facts that there are $\binom{2k}{k}$ paths from $(0,0)$ to $(2k,0)$ and $\binom{2n-2k}{n-k}$ paths of length $2n-2k$ that start at the horizontal axis but never return to the axis we obtain the left-hand side.  With these interpretations of the central binomial coefficients $2^n \binom{n}{n/2}$ could count (1) paths that do not return to the horizontal axis by the path's halfway point of $(n,0)$, or (2) paths that touch the point $(n,0)$.  But I haven't been able to construct the association that makes these the leftover paths (nor do all of these paths have even parity anyway).  So perhaps there's some other interpretation of $2^n \binom{n}{n/2}$ as the number of leftover paths.    Update. Some more thoughts:  There's another way to view the identity $\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k}  = 4^n$.  Both sides count the number of lattice paths of length $n$ when north, south, east, and west steps are allowed.  The right side is obvious. The left side has a similar interpretation as before: $\binom{2k}{k}$ counts the number of NSEW lattice paths of length $k$ that end on the line $y=0$, and $\binom{2n-2k}{n-k}$ counts the number of NSEW lattice paths of length $n-k$ that never return to the line $y =0$.  So far, this isn't much different as before.  However, $2^n \binom{n}{n/2}$ has an intriguing interpretation: It counts the number of NSEW lattice paths that end on the diagonal $y = x$ (or, equivalently, $y = -x$).  So maybe there's an involution that leaves these as the leftover paths.  (Proofs of all of these claims can be found on , for those who are interested.) $Path hijacked by PacletManager? In a ring homomorphism we always have $f(1)=1$? SEO regarding using separate  tags for the logo and website name Is is true &lt;h1&gt; tag should not have more than 1 on every page for SEO?   Every pages on the header, it includes and &lt;h1&gt; tag. Website Name which show an image from logo class, like this:  &lt;div id="header"&gt;   &lt;h1 class="logo"&gt;      &lt;a href="#"&gt;Website Name&lt;/a&gt;   &lt;/h1&gt; &lt;/div&gt;   and I also include &lt;h1&gt; on the contact page, help, etc and article page for example:  &lt;h1&gt;Contact Us&lt;/h1&gt; &lt;h1&gt;Name of the Article Title here&lt;/h1&gt;   On the homepage, it just has one &lt;h1&gt; for the logo Site Name... other pages have 2 &lt;h1&gt; tags phpmyadmin doesn't connection with true data Prove that $f$ has a fixed point . For $f:[a,b]\rightarrow [a,b]$ is a continuous . Prove that $f$ has a fixed point . Is that true if we change $[a,b]$ by $[a,b)$ or $(a,b)$. Is there a "best" way to automatically restart SQL Server on a regular basis? I read that I could create a batch file to net start, net stop the agent and service in a batch file and schedule that, but I was curious if there was a better way of doing this.  Please nevermind how this is indicative of a larger problem of operation, I've been arguing this for a month and we've ended up here.  Thanks :). What is the proper way not to scare civilians? We know the escape velocity from the Earth is 11.2km/s. Isn't it the velocity required to escape from earth if we go normal to the surface of earth? i.e while we derive the formula for the escape velocity from earth we never consider the slanted motion of the object. So when we launch a rocket we need to use very less value of velocity compare to escape velocity to escape from earth because rocket follows a slanted path so curvature of earth has an effect?  If what I said above is right, then how we can say escape velocity is 11.2 km/s? Running OS X Apps On Ubuntu 14.04 Is there any way that still works to run OS X apps on Ubuntu 14.04? If Darling still works, please tell me how to install it.  Thanks. :) "In a party people shake hands with one another" Should there be a place to ask about programming libraries and technologies? How to make dried desert/cracked ground I am trying to create a cracked ground appearance for an animation that takes place in the desert. something like the image here:I would like to do this all in blender, and have blender generate the texture so I will not have to worry about unwanted ridges like those you would get in an image from the internet: I am wondering if there is a way to procedurally generate these cracks and use them with microdisplacements. Thank you! If $H\leq G$ and $[G:H]=2$, then $H$ is normal in $G$. Do votes get canceled if voting privilege is lost? If I'm below 15 reputation, I can't alter the public score of an answer/question.  Suppose I gain the privilege by getting the required rep and vote some posts up. Then, for any reason, I fall below 15 and lose the ability to vote. What about the votes I already did cast?  Do post scores get altered by ignoring my vote, or will they stay the same? In other words, is losing a privilege retroactive?  If yes, what about when I gain my privilege back? Are my votes automatically restored, or do I have to cast them again from scratch?  I made the easiest example, but this applies to any higher privilege How do tag merge requests work? I had raised a  to merge pre-build-event, and pre-buildevent; and postbuild-event and post-build-event. That was about 11 months ago, and I still see these two tags even now.  How are tag merge requests handled? Is there a better way for tag merge requests other than posting on Meta?