# Agno

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## Registered User

 Name Agno Member for 2 years Seen 2 days ago Website Location Netherlands Age
 2d comment Values where infinite products of primes and composites are equal@unFortunately. Went through all my old notes, pdf's and links, but have so far not managed to retrieve my source. I do recall that the formula was only mentioned as a premise (in an on-line book or arxiv). It was used for a different topic and I am certain there was no proof of this formula provided. In those days I googled on "infinite product", but also recall I searched on specific values of this particular product at the integers > 1. Will keep digging a bit further and post it when I find more. Apr5 answered Does there exist a closed form for the factors of this infinite product ? Apr2 comment Alternating sums of the non-trivial zeros of $\zeta(s)$.@Joro, I don't believe these sums are the same. The secondzeta-function in mpmath sums $\displaystyle \sum_{n=1}^\infty \frac{1}{\gamma_n^s}$ and can indeed be analytically continued (via four components). However, $Z_1(s)$ sums over all (paired) $\rho_n$'s. The secondzeta-function for instance has a pole at $s=1$ whereas $Z_1(1)$ doesn't. Apr1 comment Alternating sums of the non-trivial zeros of $\zeta(s)$.Hi Joro. Complex $s$ do not seem to converge in my Maple program. What does work though is to make $\beta \in \mathbb{C}$ as long as $\beta \pm \gamma_n \ne 0$ since that would induce a pole. Apr1 asked Alternating sums of the non-trivial zeros of $\zeta(s)$. Mar29 revised Does there exist a closed form for the factors of this infinite product ?Deleted my own answer with the follow up question Mar28 revised Does there exist a closed form for the factors of this infinite product ?added the zeta function tag for easier reference Mar28 comment Does there exist a closed form for the factors of this infinite product ?Brilliant. Many thanks, Carlo! I have followed up with a (very) provocative second question, that might not have an answer like this, but maybe it does... Mar28 asked Does there exist a closed form for the factors of this infinite product ? Mar26 comment What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are “scaled” linearly?@joro. Interesting thought for which I don't have an immediate answer and need to explore further. P.S. In the mean time I have found the closed form (and a proof assuming RH) for $Had(s,\sigma,x)$ and posted it here: mathoverflow.net/questions/117874/… with an additional observation here: mathoverflow.net/questions/122582/… Mar26 comment What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are “scaled” linearly?In the mean time I have found the closed form for the above and posted it here: mathoverflow.net/questions/117874/… with an additional observation here: mathoverflow.net/questions/122582/… Mar12 answered Is there a connection between the closed forms of these two infinite products? Feb22 revised Is there a connection between the closed forms of these two infinite products?Added similarities 4) and 5) Feb22 revised Is there a connection between the closed forms of these two infinite products?Included a few clarifications in the last 2 formulae. Feb21 asked Is there a connection between the closed forms of these two infinite products? Feb16 comment The distribution of balls in a Bean Machine that omits all the “prime pegs”?@Aaron. Many thanks, this is really great. Nice approach with the Pascal Triangle. Is there anything to say about the ratio between the total number of permanently empty bins and the total number of bins $r+1$ (as function of $-y$). Will that ratio converge? Feb14 asked The distribution of balls in a Bean Machine that omits all the “prime pegs”? Jan30 awarded ● Popular Question Jan25 awarded ● Yearling Jan14 comment A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? Have now added the proof that when assuming RH, the formula can be logically derived. Jan14 comment A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? To be complete I should have added the constraint: $a \pm ix\gamma_n \ne 0$ Jan10 revised A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? added 21 characters in body Jan10 revised A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? fixed the link and included reference to my answer. Jan10 answered A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? Jan6 revised A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? Fixed a couple of small errors and added the XI(1/2) formula. Jan4 revised A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? Fixed the subscripts for the mu's. Jan4 answered A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? Jan3 comment A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? @Juan. Many thanks! The outcome is also in line with my expectations :-) Now that there is evidence that "The proposed formula is not true if RH is not true", a logical follow up question would be: Can the formula actually be derived assuming RH (i.e. similar to what you did prove for $x$ in the previous question)? (\sloppy math mode on) Have again tested the formula a lot today against the brute force calculations with 2mln $\rho$s and I believe it just works too well to not be correct.(\sloppy math mode off). Jan3 comment A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? @Juan. I believe each $H(s,a,x)$ generates a unique zero when $(s-a)=\frac12+i\gamma$ via the $\zeta$s in either $\xi(\frac12 - \frac{a}{x} + \frac{s}{x})$ or $\xi(\frac12 - \frac{a}{x} + \frac{1}{x} - \frac{s}{x})$. This also implies that when $H(s,a,x)=0$, then also $H(s,1-a,x)=H(1-s,a,x)=H(1-s,1-a,x)=0$. The values correspond to $s=\mu, s=\overline{1-\mu},1-s=1-\mu, 1-s=\overline{\mu}$ respectively. Also when $a=\frac12$ there will be 4 zeros associated with it, of which two pairs are equal. Jan2 revised A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? Turned the title into a question (it was not comprehensive) and added a P.S. Jan2 asked A closed form of infinite products of complex zeros involving $\Im(\rho_n)$. Does a proof of this closed form imply RH? Dec16 answered What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are “scaled” linearly? Dec2 comment What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are “scaled” linearly?Juan, you're right. I was way too quick. This is not the correct formula $Had(s,x,\sigma)$. Dec2 comment What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are “scaled” linearly?@Juan. It is indeed easy to expand and: $$Had(s,x, \sigma)=\frac{x^2-4(s-\sigma)^2}{x^2-1}\pi^{-s/2x} \frac{\Gamma(\sigma^2-\frac{1}{4x}+\frac{s}{2x})} {\Gamma(\sigma^2-\frac{1}{4x})}\frac{\zeta(\sigma+\frac{s}{x}-\frac{1}{2x})} {\zeta(\sigma-\frac{1}{2x})}$$ For $x \rightarrow 1$ it reduces to the known form for $\sigma=\frac12$. What does this mean? Dec2 comment What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are “scaled” linearly?Many thank Juan. Checked your formula against the numbers I have computed with 'brute force' and it all looks correct! The structure of your closed form also looks very similar to the ones for the infinite products with $n$ and therefore might be easily expandable to $Had(s,x, \sigma)$. Could the RH just be that $Had(s,1, \frac12) = \dfrac{2(s-1)\Gamma(1+\frac{s}{2}){\zeta(s)}}{ \pi^{\frac{s}{2}}}$ (i.e. reduced form of a more general product formula)? Dec2 revised What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are “scaled” linearly?Fixed an error and added a follow up question Dec2 asked What happens to $\zeta(s)$ when all its $\Im(\rho_n)$ are “scaled” linearly? Nov28 comment Question about the function $\zeta(s) \pm \dfrac{1}{\zeta(1-s)}$@Thanks Joro. The master of root finding at work! Think these are indeed exceptions for the first part of the conjecture, but what about the second part? Could there be a zero on $\Re(s)=\frac12$ for $|\zeta(s) + \dfrac{1}{\zeta(1-s)}|-2$ ? Nov27 asked Question about the function $\zeta(s) \pm \dfrac{1}{\zeta(1-s)}$