Impact
~15k
people reached
- 0 posts edited
- 0 helpful flags
- 146 votes cast
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. |
Nov
24 |
awarded | Nice Answer |
Nov
24 |
awarded | Yearling |
Nov
22 |
awarded | Critic |
Nov
22 |
answered | BSD and congruent numbers |
Nov
18 |
awarded | Nice Answer |
Nov
18 |
answered | Primes $P_{2n-1}$ that are $2$ mod $3$ |
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 . |
Dec
16 |
answered | Least supersingular prime |
Nov
24 |
awarded | Yearling |
May
15 |
answered | S integral points of an elliptic curve, with S of positive density |
May
4 |
answered | A square-squareroot integer race sequence involving primes |
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
2 |
awarded | Nice Answer |
May
1 |
answered | Is the alternating sum of primes $2 - 3 + 5 + . . P_N$ asymptotic to $- N ln N/2$ for even $N$? |