How do I get rid of them? A: Log in to the profile you’re having issues with, then find the settings button in the top right corner of the drop down tab and select the “Sign out” option. This will reset your email address and everything else to a default state. If you can’t log in to your account for some reason then try setting a password for that account from the settings window, then log in again, then reset your password and hit the account settings button. A: Go to : you will get an URL like: for login password reset Then Enter your email and send a password reset code. Q: Complexity of the Collatz conjecture? The Collatz conjecture is: If $k \geq 1$ and $n \geq 0$, then $c_k(n)$ is the first integer $m \geq 0$ with $m = k \cdot 2^n$. Is this conjecture open? What is its complexity? A: The Collatz conjecture is $\mathbf{Z}$-hard. In fact, $\mathbf{Z}$-hardness implies that (the equivalence of) the problem of whether a function is $\mathbf{Z}$-recursive is undecidable. To see that the Collatz conjecture is $\mathbf{Z}$-hard, let $p(n) = n$ and $k=1$, and consider the $n^{th}$ iterate $f_n(x) = p(x)^n$. Then for every $n \in \mathbf{Z}$, $f_n$ is total and so $f_n \in \mathbf{Z}$. However, it’s easy to see that $f_n(x) = kx$ for exactly the $n$ values $x$ that solve the Collatz conjecture for $k=1$. So the Collatz conjecture is $\mathbf{Z}$-hard. The equivalence of $\mathbf{Z}$-recursion and the Collatz conjecture follows from the fact that any Collatz-like