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palindromicprime

 Hilbert's hotel

A room for posing problems and for solving problems and also f...
palindromicprime
Apr 11, 2020 13:58
do let me know if you figure it out
palindromicprime
Apr 11, 2020 13:57
because that’s the freshest info for me
palindromicprime
Apr 11, 2020 13:57
yeah hopefully. I’m just applying what I’ve been applying to my own research
palindromicprime
Apr 11, 2020 13:55
problem is that the error terms don’t let me analyze divisibility by 6. Something to consider is using the explicit formulas for these in terms of the zeros of the zeta function. But that would take me a good while to do
palindromicprime
Apr 11, 2020 13:53
some known floor expressions for the arithmetic functions involved
palindromicprime
Apr 11, 2020 13:53
Yeah it didn’t work. I tried using Perron’s Formula and
palindromicprime
Apr 11, 2020 13:28
Interesting I’ve got something that’s looking nice but we’ll see probably doesn’t work
palindromicprime
Apr 11, 2020 13:05
Pretty much all that my number theory research is based on
palindromicprime
Apr 6, 2020 00:35
Sentences loop around and I can’t even see what I’m typing
palindromicprime
Apr 6, 2020 00:34
Ugh StackExchange chat on iPad is worthless
palindromicprime
Apr 6, 2020 00:33
That to a side for now. Interesting to checkout nonethelessas a side project I’m also working on Wolfram’s rule 30 problems. But I’ve taken
palindromicprime
Apr 6, 2020 00:31
Still working on the integral I mentioned yesterday
palindromicprime
Apr 5, 2020 23:31
Hey
palindromicprime
Apr 5, 2020 23:26
About that yesterday*
palindromicprime
Apr 5, 2020 23:25
@Knight I have already provided you a proof about yesterday at Mathematics chat
palindromicprime
Apr 5, 2020 14:51
In particular, if the proof is done with an abstract point of view of L-functions, then wouldn't skip my imagination that one could generalize this to other L-functions
palindromicprime
Apr 5, 2020 14:51
Well depends how the proof looks like. But mostly likely I think not.
palindromicprime
Apr 5, 2020 14:44
oh well, thanks for the upvote anyway it helps
palindromicprime
Apr 5, 2020 14:43
you can imagine the frustration of having missed it during preliminary research on the topic
palindromicprime
Apr 5, 2020 14:43
took me years to find that integral expression for D(x), only to find it has already been found and its even generalized for any arithmetic function
palindromicprime
Apr 5, 2020 14:42
just need to take the line integral over over the critical strip and that just break my head every time I try
palindromicprime
Apr 5, 2020 14:42
yeah me neither, but it has been an interesting path to take. the zeta function has a really beautiful laurent series expansion, so I worked with that and got most of it
palindromicprime
Apr 5, 2020 14:38
3
Q: Contour Integral involving Zeta function

palindromicprimeI'm trying to compute the contour integral $$\frac{1}{2 \pi i} \int_{c - i \infty}^{c + i \infty} \zeta^2(\omega) \frac{8^\omega}{\omega} \ d \omega$$ where $c > 1$, $\zeta(s)$ is the Riemann zeta function. Using Perron's Formula and defining $D(x) = \sum_{k \leq x} \sigma_0(n)$, where $\sigma_0...

number-theory contour-integration analytic-number-theory complex-integration riemann-zeta
palindromicprime
Apr 5, 2020 14:38
by the way, I have a problem I'm trying to solve. is it okay if i share it here?
palindromicprime
Apr 5, 2020 14:37
yes of course, it flew straight over my head, haven't slept at all
palindromicprime
Apr 5, 2020 14:35
@Peter oh i see so you exploit the form of the number in the modulo, makes sense
palindromicprime
Apr 5, 2020 14:33
@Ante oh nice! would be happy to check it out if/when you are ready to share
palindromicprime
Apr 5, 2020 14:33
@Peter then my question becomes, how can we tell a number is a factor of a really huge number?
palindromicprime
Apr 5, 2020 14:31
@Ante haha yeah, I guess it must be a common pitfall one falls into after one learns that 2^p - 1 prime implies p prime
palindromicprime
Apr 5, 2020 14:31
@Peter Do you know, how is it partially verified? Wouldn't a single digit, change the whole result completely?
palindromicprime
Apr 5, 2020 14:24
what I was thinking maybe is to feedback those large mersenne primes back into the mersenne expression. So taking a mersenne prime 2^p - 1 larger than mentioned N, 2^(2^p - 1) is prime as well. But yeah, I think im dead wrong here, since I think the observations above still apply
palindromicprime
Apr 5, 2020 14:20
from that (and here is the part I might be wrong) this impacts the asymptotic behaviour of primes
palindromicprime
Apr 5, 2020 14:18
If composite Mersenne numbers of prime exponent are finite then there exists a (large) N such that for all Mersenne numbers 2^p - 1 larger than N, 2^p-1 is prime
palindromicprime
Apr 5, 2020 14:16
yeah, I'm not talking about proof. What I mean by this is as follows:
palindromicprime
Apr 5, 2020 14:15
Sorry about that my ipad had a stroke while writing that sentence. I'm on my laptop now. I meant to write "I don't know if my heuristic is correct. But if composite Mersenne numbers with prime exponent are finite, wouldn't this imply (or at least) suggest a prime number density, probably different from the prime number theorem?"
palindromicprime
Apr 5, 2020 14:08
If* mersenne numbers....
palindromicprime
Apr 5, 2020 14:08
Mersenne numbers with prime exponent are finite, wouldn’t this imply a different prime density than the one implied in the prime number theorem?The one I don’t know if my heuristic here is correct. But if composite
 

 Mathematics

Associated with Math.SE; for both general discussion & math qu...
7
1
palindromicprime
Apr 5, 2020 15:22
I was wondering if anyone here could shed a light on this?
palindromicprime
Apr 5, 2020 15:22
4
Q: Contour Integral involving Zeta function

palindromicprimeI'm trying to compute the contour integral $$\frac{1}{2 \pi i} \int_{c - i \infty}^{c + i \infty} \zeta^2(\omega) \frac{8^\omega}{\omega} \ d \omega$$ where $c > 1$, $\zeta(s)$ is the Riemann zeta function. Using Perron's Formula and defining $D(x) = \sum_{k \leq x} \sigma_0(n)$, where $\sigma_0...

number-theory contour-integration analytic-number-theory complex-integration riemann-zeta
palindromicprime
Apr 5, 2020 15:22
:53996159 He was trying to prove a lemma
palindromicprime
Apr 5, 2020 15:19
Ask yourself, for what choice of $f(x)$ is $f(x) = f'(x)$ true for all $x \in (0,1)$. Then a simple modification of that yields the answer naturally
palindromicprime
Apr 5, 2020 15:12
Let $g(x) = e^{-x} f(x)$ which yields $g(0) = e^{-0} f(0) = 0 = e^{-1}f(1)$. $g$ is also continuous on $[0,1]$ and differentiable on $(0,1)$, so Rolle's Theorem applies. Hence there exists an $x \in (0,1)$ such that $g'(0) = 0$. This implies that $-e^{-x} f(x) + e^{-x} f'(x) = 0 \implies e^{-x} (f'(x) - f(x)) = 0$ and since $e^{-x} \neq 0$ for all $x \in \mathbb{R}$ then $f'(x) - f(x) = 0 \implies f'(x) = f(x)$ for that choice of $x$.
palindromicprime
Apr 5, 2020 14:50
@Knight Are you looking to solve this problem without the MVT then?