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 Dec 2 comment Is Lehmer's polynomial solvable? @IgorRivin I agree, it's a bit curious. I could presumably construct a clumsy proof that it's not the full hyperoctahedral, but I wonder if there's a better reason. (I am inclined to trust Magma.) Dec 1 comment Is Lehmer's polynomial solvable? Magma quickly computes that it's a nonsolvable group of order 1920. Feb 5 comment Is there a Poisson Summation formula for imprimitive Dirichlet characters? You might be interested in a preprint of Daileda and Jones, where they show that by modifying the way in which one extends primitive characters to imprivitive characters (in particular, by making a choice other than $\chi(n)=0$ for $n$ not coprime to $q$ -- and, iirc, by choosing it so that the Gauss sum is well-behaved), these new imprimitive characters behave nicely analytically. It's available here: olemiss.edu/working/ncjones/primitivity9.pdf . May 3 comment A square-squareroot integer race sequence involving primes Legendre conjectured that there is always a prime between $n^2$ and $(n+1)^2$, and this is generally believed to be true. Assuming this, for any $n$, $g(f(n))=n$, so you're just looking at a random walk. May 3 comment A question on primitive roots There's surely a dependence on the choice of the primitive roots (assuming my interpretation of the question is correct). Look at the primes $q$ for which $x_p \equiv a\pmod{n}$, and then choose a different primitive root for each such $q$. Why should the congruence $x_p\equiv a\pmod{n}$ be preserved for almost all of these $q$? It might also be possible to hack together a sequence of $x_p's$ which are not equidistributed modulo any $n$, but I'm too tired to think through the details of this right now. May 3 comment A question on primitive roots Seva: Despite the notation, it seems that $q$ varying is more interesting. If we fix $q$ and vary $p$, Dirichlet's theorem shows that each possible value of $x_p$ occurs equally often. May 2 comment Is the alternating sum of primes $2 - 3 + 5 + . . P_N$ asymptotic to $- N ln N/2$ for even $N$? And, I guess, regarding Ruiz's purported proof: I haven't read it, but the discussion suggests that it's replacing $p_n$ by the asymptotic $p_n\sim n\log n$. This cannot work: there are many sequences which satisfy all the properties we know primes to have (e.g., the prime number theorem), and yet are such that, using the notation from my answer, $S(N;a,q)/p_N$ can tend to any value in $[0,1]$, or for which the limit doesn't exist, etc. To prove that $S(N;1,2) \sim p_N/2$, you have to rule out the possibility that all small prime gaps occur with odd index. But I have no idea how to do that! May 2 comment Is the alternating sum of primes $2 - 3 + 5 + . . P_N$ asymptotic to $- N ln N/2$ for even $N$? Regarding the generalization with $k\geq 2$: Using essentially the same techniques and that $p_{n+1}^{k+1}-p_n^{k+1}=(p_{n+1}-p_n)(p_{n+1}^k+\dots+p_n^k)=(p_{n+1}-p_n)((k+1‌​)p_n^k + O_k(p_n^{k-1}))$, it would again be possible to prove a lower bound of the right order of magnitude for the alternating sum of $k$-th powers of primes. May 1 comment An introduction to sieve method and their application, Cojocaru & Murty Actually, wait. Unless I'm missing something, their claimed error term is wrong, and it should be $O(1/\log x)$ rather than $O(1/x^{1/4})$ (this is fine for their purposes, as the error on the next line is $O(1/\log x)$ anyway). To see it has to be at least this big, just look at the integral of the error from 3 to 4. To see that it's no bigger, split the integral into two parts, integrating from 3 to $x^{1/2}$, say, and from $x^{1/2}$ to $x$. The latter decays like $x^{-1/8}$, while the former can be bounded by $O(1/\log x)$. May 1 comment An introduction to sieve method and their application, Cojocaru & Murty Plugging in the lemma, you'd get an $A_1 \log t$ term. This is the same as $-A_1(1+\log(x/t))+A_1\log x + A_1$. Is that your question? Mar 17 comment Question about a certain class of primes Greg: Good point. I was pretty sure there was an easy check, but I didn't bother to think about it. Mar 7 comment The behavior of a certain greedy algorithm for Erdős Discrepancy Problem Fixed! Sorry about that. Apparently all the files I'd hosted at Emory before moving have gone away. Nov 18 comment Long gaps between primes Following up on @GerhardPaseman's comment, the expected order of magnitude is $c \log^2 n$ (though proving this is hard!). See, for example, this paper of Granville: www.dms.umontreal.ca/~andrew/PDF/cramer.pdf. It discusses Cramer's conjecture, why it's probably wrong (at least in a precise form), and what the correct modification should be. Nov 18 comment Long gaps between primes The Wikipedia article en.wikipedia.org/wiki/Prime_gap on prime gaps covers this topic. In particular, C can be taken to be arbitrarily large. Nov 3 comment What is known about the sum x^{n^2}/n? I'm not an expert, but the Eichler integral can be defined, purely formally, as the operator that sends, with your notation, $x^n \mapsto x^n/n$ (recall that for modular forms $x=exp(2\pi i z)$, so this is integration $dz$). Your series is then a formal half-integral'' of the standard theta function. I've seen such things arise in talks, though I don't know anything about them myself (and I trust your ability to google as much as my own). I'd recommend searching for "half-derivative" rather than "half-integral", since the latter appears quite frequently with another meaning. Oct 23 comment On extended Riemann Hypothesis and coefficients of Selberg Class L-functions Even with an infinite sum, I think it's impossible if $d>1$. Let $G(s)=L(s,f)H(s)$. We need to balance three things: 1) The sum of $f*h$ being bounded, 2) $H(s)$ analytic in $\Re(s)>0$, maybe by asking for $S_h(X) \ll X^\epsilon$, and 3) The non-existence of zeros of $H(s)$ in $\Re(s)>0$. It's not even obvious to me that 1 and 2 are compatible once $d>1$: each is equivalent to a system of inequalities on $h(n)$. 1 has coefficients $S_f(X/n)$, whereas 2 has coefficients all one. Once $S_f(X)$ has size (as in $d>1$), these systems are on different scales, so a common solution is not obvious. Apr 17 comment Off critical line zeros for half integer weight $L$-functions Have you done any computations yourself? While I'm dubious that this should be true for almost any form, it's worth noting that $L(s,\theta_\chi)=L(2s-1/2,\chi)$ for a non-trivial Dirichlet character $\chi$, so RH presumably holds in this case. In general, though, the multiplicative structure of half-integral weight eigenforms is more complex, and I'd be very surprised if it were to hold if the form is orthogonal to the space of unary theta functions. Nov 30 comment Is this extension of the Selberg class trivial? You're absolutely right that there are issues with $L(s,f)^{1/2}$, which is why it's not actually something I want to consider. I brought it up mostly to clarify points 1 and 2. Maybe a prototype question would be this: Can $L(s,f)^{1/2}L(s,g)^{1/2}$ ever be sensibly continued to an entire function, where $L(s,f)$ and $L(s,g)$ are primitive elements of the Selberg class? It's known that each has zeros disjoint from the other, but maybe all the simple zeros coincide, or are there are no simple zeros, or... Conjecturally, this can't happen, but that's the sort of thing I'm imagining. Nov 30 comment Is this extension of the Selberg class trivial? Right, this absolutely falls under the purview of pretentiousness; in fact, my question about more than square root cancellation can be viewed as a counterpart to Halasz's theorem which says that anything with large sums must come from (pretend to be) one of a natural set of examples. There, though, the natural examples are not Dirichlet characters. Instead, they are the additive characters, $n^{it}$. Indeed, non-principal Dirichlet characters don't have large partial sums - the partial sums are bounded! Thus, they are the natural examples for exceptional (more than square root) cancellation. Aug 25 comment The behavior of a certain greedy algorithm for Erdős Discrepancy Problem I will add data for the squarefree problem to my answer in a second, but let me just quickly give you the gist -- it's the same basic behavior, with $1/3$ appearing just as clearly.