bio | website | stanford.edu/~rjlo |
---|---|---|

location | Palo Alto, CA | |

age | ||

visits | member for | 5 years |

seen | 14 mins ago | |

stats | profile views | 400 |

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. |

Aug 25 |
comment |
The behavior of a certain greedy algorithm for Erdős Discrepancy Problem
His. And the ratio of $\log D/\log N$ I alluded to, not in graph form: 0.3010299957, 0.345687124, 0.3788254007, 0.3972102138, 0.3260186782, 0.3302216042, 0.3378116896, 0.3352640758, 0.3368385727, 0.3590944053, 0.336264762, 0.3602395359, 0.3305424391, 0.3367194215. Close enough to 1/3 that I suspect something is happening. |

Aug 24 |
comment |
The behavior of a certain greedy algorithm for Erdős Discrepancy Problem
Are we requiring $f(n)$ to be $\pm 1$ only? |

Jul 19 |
comment |
Axioms for Riemann $\zeta$ function
As I stated it, yes, that is necessary. However, Kaczorowski and Perelli define what they call the extended Selberg class, where there is no Euler product, and they provide a classification of the elements of degree up to 1. Degree 0 elements are Dirichlet polynomials satisfying a certain symmetry condition, and degree 1 elements are linear combinations of the product of a degree 0 element with a (potentially shifted) Dirichlet L-function. Sound's proof also works for this class, and shows that the Dirichlet coefficients are periodic, so that multiplicativity implies Dirichlet character. |