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This forum brings together a broad enough base of mathematicians to collect a "big list" of equivalent forms of the Riemann Hypothesis...just for fun. Also, perhaps, this collection could include statements that imply RH or its negation.

Here is what I am suggesting we do:

Construct a more or less complete list of sufficiently diverse known reformulations of the Riemann Hypothesis and of statements that would resolve the Riemann Hypothesis.

Since it is in bad taste to directly attack RH, let me provide some rationale for suggesting this:

1) The resolution of RH is most likely to require a new point of view or a powerful new approach. It would serve us to collect existing attempts/perspectives in a single place in order to reveal new perspectives.

2) Perhaps the resolution of RH will need ideas from many areas of mathematics. One hopes that the solution of this problem will exemplify the unity of mathematics, and so it is of interest to see very diverse statements of RH in one place. Even in the event where no solution is near after this effort, the resulting compilation would itself help illustrate the depth of RH.

3) It would take very little effort for an expert in a given area to post a favorite known reformulation of RH whose statement is in the language of his area. Therefore, with very little effort, we could have access to many different points of view. This would be a case of many hands making light work. (OK, I guess not such light work!)

Anyhow, in case this indeed turns out to be an appropriate forum for such a collection, you should try to include proper references for any reformulation you include.

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    $\begingroup$ You can find a few at: en.wikipedia.org/wiki/Riemann_hypothesis $\endgroup$
    – S. Carnahan
    Sep 25, 2010 at 12:41
  • $\begingroup$ There was an AIM Problem List which included a bunch of examples in connection with last year's "RH Day." Although the link: aimpl.org/pl seems not to be working anymore, at least there is this Archive.Org link: web.archive.org/web/20120731034246/http://aimath.org/pl/… (hat tip to John Washburn). $\endgroup$ Sep 25, 2010 at 15:15
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    $\begingroup$ I think this is a great question and have more than once wished for such a list in the past. $\endgroup$ Sep 25, 2010 at 20:04
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    $\begingroup$ Maybe people should vote for what they think is currently the most promising approach, based on an equivalent reformulation $\endgroup$
    – Alex R.
    Sep 25, 2010 at 23:47
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    $\begingroup$ $$\text{Riemann Hypothesis} \iff \sum\limits_{\rho}\frac{1}{|\rho|^2}=2+\gamma-\log4\pi;$$ where $\gamma$ is Euler's constant and $\rho$ are the non-trivial zeros of the Riemann zeta function. The paper Transcendental Sums Related to The Zeros of The Riemann Zeta Function by Gun for proof and context. $\endgroup$
    – Anthony
    Dec 20, 2019 at 16:00

23 Answers 23

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In the article Seized opportunities (Notices of the AMS, April 2010), Victor Moll gives the following, which he credits to V.V.Volchkov. Establishing the exact value $$\int_{0}^{\infty}\frac{(1-12t^2)}{(1+4t^2)^3}\int_{1/2}^{\infty}\log|\zeta(\sigma+it)|~d\sigma ~dt=\frac{\pi(3-\gamma)}{32}$$ is equivalent to the Riemann Hypothesis. Moll cheekily adds that evaluating that integral might be hard.

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    $\begingroup$ Oh, and $\gamma$ is Euler-Mascheroni. $\endgroup$ Mar 9, 2011 at 13:07
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    $\begingroup$ Someone get Cleo! $\endgroup$ Nov 25, 2022 at 20:20
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I like Lagarias "elementary" reformulation of Robin's theorem: that RH is true iff

$\sigma(n)\leq H_n+e^{H_n}\log(H_n)$

holds for every $n\geq 1$, where $\sigma(n)$ is the sum of divisors function and $H_n$ is the nth harmonic number.

Its major appeal is that anyone with rudimentary exposure to number theory can play with it. Having spent the better part of my youth fiddling with this reformulation really brought out the enormous difficulty of proving RH. In a way I think this reformulation is evil, because it looks tractable, but is ultimately useless and perhaps even harder to work with than other more complex reformulations. On the other hand I hope a future proof of RH will involve this reformulation because then I might have a chance of understanding the proof!

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    $\begingroup$ I used to teach a computer algebra class, and at the exam I asked the student to write a program checking this inequality on random examples. The possibility of winning a million dollars if they find a counter example (minus my share, of course) is a nice incentive :). $\endgroup$
    – Adrien
    Aug 3, 2018 at 12:16
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    $\begingroup$ Where can I find the proof that it is equivalent to the originial RH statement? $\endgroup$ Apr 17, 2019 at 3:54
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    $\begingroup$ @NewBornMATH: math.lsa.umich.edu/~lagarias/doc/elementaryrh.pdf $\endgroup$
    – Alex R.
    Apr 17, 2019 at 7:06
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The following is given without source here:

RH is equivalent to the assertion that for all $n\ge3$ $$| \log \operatorname{lcm}(1,2,\dots, n) - n | < \sqrt{n}\log^2(n)$$ where $\operatorname{lcm}$ denotes the least common multiple.

More details about this function (also called second Chebyshev function) can be found in this Wikipedia entry (thank you Wojowu!).

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  • $\begingroup$ Very interesting. Is it anyhow related to the conjectural upper bound for the quantity $\alpha_{n}$ defined in mathoverflow.net/questions/61842/about-goldbachs-conjecture ? $\endgroup$ Feb 29, 2016 at 13:56
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    $\begingroup$ For the record, the function $\log lcm(1,2,\dots,n)$ is equal to second Chebyshev function $\psi(n)$. $\endgroup$
    – Wojowu
    Feb 29, 2016 at 17:25
  • $\begingroup$ Hello, could you give a reference for where the implication of the RH from the statement above can be found. The direction RH=>statement above is found in many places, but the opposite direction I cannot find. Thanks $\endgroup$
    – EGME
    May 11, 2022 at 9:36
  • $\begingroup$ Good point... maybe the user Lekraj Beedassy who posted this on oeis just made up the opposite direction. But I know that in oeis (unlike e.g. Wikipedia) info added by users is usually first checked by them, so it should in theory be trustworthy. Possibly not here, though :( $\endgroup$
    – Wolfgang
    May 11, 2022 at 11:42
  • $\begingroup$ @EGME Granville and Martin mention this equivalence on pg. 9 of their article "Prime Number Races" dms.umontreal.ca/~andrew/PDF/PrimeRace.pdf They do not provide a reference, but at least this is a strong vote in favor of the result being true. $\endgroup$
    – D.R.
    Dec 13, 2022 at 22:55
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This one is not too bad though not big:

Equivalences to the Riemann Hypothesis, ed. J. Brian Conrey and David W. Farmer

Yet there are many (above a hundred at least) and it depends on the type you are looking for. Analytic elementary number theory ....

There is the two volume work "Equivalences of the Riemann hypothesis: vol I Arithmetic Equivalences, vol II Analytic Equivalences" Cambridge, 2017, by Kevin Broughan. Information is linked to the site https://web.archive.org/web/20120731034246/http://aimath.org/pl/rhequivalences . Volume III is currently being drafted and should contain some recent equivalences, such as those relating to the number of divisors function, the de Bruijn-Newman constant and Jensen polynomials.

ADDED LATER : My favorite is very elementary:
Among the square free integers below $N$:
Let $D(N)$ denote the absolute value of the difference between the number of those divisible by an even number of primes and the number of those divisible by an odd number of primes .

R.H. says that $D(N)$ comes close to the square root of $N$.

More precisely: for any $\epsilon > 0 $ there is $N_0$ such that any $N > N_0$ verifies $ {D(N)} \leq N^{1/2+\epsilon}$.

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Robin's criterion has been written in various places in MO: define Gronwall's function $$ G(n) = \frac{\sigma(n)}{n \log \log n}. $$ In 1984, Robin showed that RH is equivalent to $$ G(n) < e^\gamma, \; \; \forall n \geq 5041.$$

Robin's adviser was Jean-Louis Nicolas. There is a new equivalence due to Nicolas, G. Caveney, and J. Sondow. Define a positive integer $N$ to be $GA1$ if $N$ is composite and $G(N) \geq G(N/p)$ for all primes $p |N.$ Let $N$ be called $GA2$ if $G(N) \geq G(aN)$ for all positive integers $a,$ where in this case we allow $N$ to be prime or composite. Then RH is equivalent to the assertion that the only number that is both $GA1$ and $GA2$ is 4. See arXiv and arXiv

I learned of this because Sondow wrote to me asking for a pdf of Robin 1984. And I wrote back. Which people ought to do.

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  • $\begingroup$ Since I am still worried you never saw my reply, I will seize this opportunity to thank you again for your kind offer to send me this paper when we both answered a question related to Robin's criterion this January. (I only replied via a comment with a delay of a couple days as I was off-line; and only via a comment as I did not need the paper sended.) In any case, thanks again for the offer! $\endgroup$
    – user9072
    Aug 22, 2012 at 12:46
  • $\begingroup$ @quid, I suspect I never saw that comment. I have noticed that there are some bugs in the comment notification system. Even though people correctly write after an answer of mine, or add @Will when the comments are at some other location, i do not always get notified (the little envelope at top turning orange). I generally check my own activity from the day before to see for changes or comments, but if an extra day passed I might not have thought to check any more. $\endgroup$
    – Will Jagy
    Aug 22, 2012 at 18:19
  • $\begingroup$ Then good I wrote this, like, better late than never. Just one tangential point: the @name notification does not work here 'by design'. This feature was only added in newer version of SE than currently in use here (but after the move it should work); and the thing was a reply to your comment on my answer, so there was no notification by design. $\endgroup$
    – user9072
    Aug 22, 2012 at 18:45
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    $\begingroup$ @quid, found it, mathoverflow.net/questions/84266/on-robins-criterion-for-rh $\endgroup$
    – Will Jagy
    Aug 22, 2012 at 19:37
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    $\begingroup$ @Mr. It depends on the number $n$. When you take $n$ to be prime for example, $G(n)$ can get arbitrarily small. On the other hand, if you take $N_k$ to be the least common multiple of the first $k$ positive integers, then $N_k/G(N_k)$ does converge to $e^{\gamma}$ $\endgroup$
    – Woett
    Nov 26, 2021 at 11:40
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Lapidus and Maier show that “One can hear the shape of a fractal string of dimension $D \neq \frac12$” if and only if the Riemann hypothesis is true.

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Found in this question

DMR 1974: http://books.google.ca/books?id=4lT3M6F745sC&pg=PA335

$$\eta(j)=p \text{ if } j=p^k, \; p \text{ is prime}$$ $$\eta(j)=1 \text{ otherwise}$$ $$ \delta(x)=\prod_{n < x }\prod_{ j \le n} \eta(j) $$

RH is equivalent to the assertion that $$\left( \sum_{k \le \delta(n)}\frac1k - \frac{n^2}{2}\right)^2 < 36n^3$$

for $n \ge 1$.

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    $\begingroup$ Since it may be hard to look inside that Google Book, let me mention that DMR 1974 stands for Martin Davis, Yuri Matiyasevic, and Julia Robinson (1974), Hilbert's tenth problem. Diophantine equations: positive aspects of a negative solution. I like joro's statement for its being so nakedly expressible in Peano arithmetic (there are others of course, but this one is most obviously so expressible). $\endgroup$
    – Todd Trimble
    Mar 25, 2018 at 16:31
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Using Corollary 1 in Schoenfeld's 1976 paper "Sharper bounds for the Chebyshev functions $\theta(x)$ and $\psi(x)$. II", we see with a bit of numeric work that the Riemann Hypothesis is equivalent to the following inequality: $$|\pi(x)-\mathrm{li}(x)|<\sqrt{x}\log x,\qquad x\geq 2.$$

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  • $\begingroup$ Good morning! Do you know of a textbook wherein this equivalence is established in detail? Thanks in advance for your reply. $\endgroup$ Sep 6, 2021 at 16:51
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    $\begingroup$ @JoséHdz.Stgo. The RH implies $|\pi(x)-\mathrm{li}(x)|<\frac{1}{8\pi}\sqrt{x}\log x$ for all $x\geq 2657$. This is Corollary 1 in Schoenfeld: Sharper bounds for the Chebyshev functions $\theta(x)$ and $\psi(x)$. II, Math. Comp. 134 (1976), 337-360. Checking the finite range $2657>x\geq 2$ by a computer program, we find that the weaker inequality $|\pi(x)-\mathrm{li}(x)|<\sqrt{x}\log x$ holds for all $x\geq 2$. Conversely, if we assume this inequality for all $x\geq 2$, then the RH is true by Theorem 15.2 in Montgomery-Vaughan: Multiplicative number theory I. $\endgroup$
    – GH from MO
    Sep 6, 2021 at 20:46
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Li's criterion ?

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    $\begingroup$ For the interested reader, there is a variant of this criterion by Voros (arXiv:2204.01036) which is much better behaved for computational investigations. $\endgroup$ Jun 10, 2022 at 9:44
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A good resource is The Riemann Hypothesis by Borwein, Choi, Rooney, and Weirathmueller, CMS, 2008. It has equivalences plus much more.

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Two statements I find quite interesting involve Farey sequences, and, very roughly, they can be interpreted as saying that elements of Farey sequence are "not far" from being evenly distributed in the unit interval.

To be precise, suppose $n$-th Farey sequence is $0=a_{0,n}<a_{1,n}<\dots<a_{m_n,n}=1$ and let $d_{k,n}=a_{k,n}-\frac{k}{m_n}$. Then RH is equivalent to any of the two statements below:

  • $\sum_{k=0}^{m_n} d_{k,n}^2=O(n^{-1+\varepsilon})$ for any $\varepsilon>0$ as $n\rightarrow\infty$.
  • $\sum_{k=0}^{m_n} |d_{k,n}|=O(n^{1/2+\varepsilon})$ for any $\varepsilon>0$ as $n\rightarrow\infty$.
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There are many matrix formulations of the RH. This one is my favorite, due to Jean-Paul Cardinal: A sequence of symmetric matrices related to the Mertens function. (There is also an English translation.)

One considers the set $$S_n = \left\{ s \in \mathbb{N}\ \left| \ \ s = \left\lfloor \frac{n}{k}\right\rfloor , \ k \in \mathbb{N} \right.\right\}$$ and the $|S_n| \times |S_n|$ matrix $A^{(n)}_{ \ \ ij} = \text{Mertens}\left(\left\lfloor \frac{n}{s_i s_j} \right\rfloor\right)$.

Given that $\text{Mertens}(n) \le \|A^{(n)}\| = \max_{\|u\|=1} \|A^{(n)} u\|$ , the Riemann hypothesis would be implied by $$ \|A^{(n)}\| = \mathcal{O}(n^{1/2+\epsilon}) \tag{$\star$} .$$

What is amazing is that the sequence $\|A^{(n)}\|$ is one of the smoothest you'll ever see being related to the Riemann hypothesis, and the fact that $\text{Mertens}(n)$ only depends on the $|S_n|{\scriptstyle-}1 < 2\sqrt{n}$ previous values of $\text{Mertens}\left(s_i\right)$ which is perfectly encoded in that matrix sequence.

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    $\begingroup$ A preprint of Lagarias and Montague, Notes on Cardinal's matrices, proves that RH is equivalent to the bound ($\star$) on the norms of Cardinal's matrices, if the norm is taken to mean the Frobenius norm rather than the $\ell_2$ norm. $\endgroup$ Apr 29, 2020 at 14:21
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I met a guy today who I convinced to sign up on MathOverflow. His name is Kevin Broughan and he has a couple of volumes on this subject, divided into arithmetic and analytic equivalents. You might check it out.

Gerhard "Oh, The People You Meet...." Paseman, 2018.08.02.

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By Value Distribution Theory Related to Number Theory, Riemann's Hypothesis is true if and only if $$\frac{1}{\pi}\int_0^{\infty} \log\left|\frac{\zeta(\frac{1}{2}+it)}{\zeta(\frac{1}{2})}\right|\ \frac{dt}{t^2}=\frac{\pi}{8}+\frac{\gamma}{4}+\frac{\log 8\pi}{4}-2$$

And a more general theorem has be proved in this book:

Take $a\in R$ with $\frac{1}{2}\leq a<1$. Riemann's $\zeta$-function has no zeros in $\Re(s)>a$ if and only if $$\frac{1}{\pi}\int_0^{\infty} \log\left|\frac{\zeta(a+it)}{\zeta(a)}\right|\ \frac{dt}{t^2}=\frac{\zeta'(a)}{2\zeta(a)}-\frac{1}{1-a}$$

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Not especially sophisticated, but there is the note by Tuck: When does the first derivative exceed the geometric mean of a function and its second derivative?. This is studied some more in

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  • $\begingroup$ Which of these papers contains an equivalent form of RH? Brief look at the papers didn't reveal any such statement. $\endgroup$
    – Wojowu
    Feb 12, 2023 at 2:36
  • $\begingroup$ @wojowu I saw a talk by Ernie Tuck 15 or so years ago where he spoke about the results in the notes, and at the very least the condition, involving the particular combinations of the first few deriviatives of a function cooked up out of zeta, is sufficient. I thought at the time I wrote this it was an equivalent condition. $\endgroup$
    – David Roberts
    Feb 12, 2023 at 6:02
  • $\begingroup$ But I might be wrong! $\endgroup$
    – David Roberts
    Feb 12, 2023 at 6:05
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I can't seem to find it in any of the answers here, so:

In three papers, Roesler presents an $n\times n$ matrix $\mathbf A_n$ whose determinant is $n! \sum_k \frac{\mu(k)}{k}$, (from which the statement that is equivalent to the hypothesis is $\det \mathbf A_n=O(n! n^{\varepsilon-\frac12})$ for all positive $\varepsilon$). He then proceeds to study the eigenvalues of this matrix to glean insights into the hypothesis from this viewpoint.

Relatedly, this paper by Barrett and Jarvis deals with a matrix originally studied by Redheffer, whose determinant is the Mertens (summatory Möbius) function $\sum_k\mu(k)$. They also then study the eigenvalues of this matrix.

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On the arXiv this morning, The Landau function and the Riemann Hypothesis by Marc Deleglise and Jean-Louis Nicolas:

The Landau function $g(n)$ is the maximal order of an element of the symmetric group of degree $n$; it is also the largest product of powers of primes whose sum is $\le n$. The main result of this article is that the property

For all $n > 0$ , $\log g(n) < li^{-1} (n))$

(where $li^{-1}(n)$ denotes the inverse function of the logarithmic integral) is equivalent to the Riemann hypothesis.

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  • $\begingroup$ Something very like this was posted (but then deleted) by user68208 in March 2016. $\endgroup$ Jul 18, 2019 at 23:22
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Two other articles, FWIW:

http://arxiv.org/pdf/0808.0640.pdf (mainly a criteria of Riesz and Baez-Duarte)
http://arxiv.org/pdf/1003.3392.pdf (introducing the "Acid zeta function")

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Let $C_{c}^{r}(\mathbb{R}^{\ast})$ denote the set of all functions $f \colon (0,\infty) \to \mathbb{C}$ that are $r$ times differentiable and have compact support. For any $y \in (0,\infty)$ and any $f \in C_{c}^{2}(\mathbb{R}^{\ast})$, let $m_{y}(f) := \sum_{n \in \mathbb{N}} y\phi(n) f(y^{\frac{1}{2}}n)$ where $\phi$ is the Euler totient function; in addition, let $m_{0}(y) := \int_{0}^{\infty} \left(\frac{6}{\pi^{2}}\right)u f(u)\, du$.

In the paper Discrete measures and the Riemann hypothesis (Kodai Math. J. 17 (1994), no. 3, 596–608.), Prof. Alberto Verjovsky proved the folllowing:

The Riemann Hypothesis is true if and only if $m_{y}(f) = m_{0}(f) + o(y^{\frac{3}{4}-\epsilon})$ as $y \to 0$ for every $f \in C_{c}^{r}(\mathbb{R}^{\ast})$ with $r \in [2,\infty]$ and every $\epsilon > 0$.

You can find the aforementioned paper here: http://projecteuclid.org/euclid.kmj/1138040054

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    $\begingroup$ This was proven much earlier by Peter Sarnak ("Asymptotic behavior of periodic orbits of the horocycle flow and eisenstein series") based on prior work of Zagier (or at-least so Peter said to me once). $\endgroup$
    – Asaf
    Mar 7, 2016 at 5:55
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    $\begingroup$ Didn't know that... In any case, it seems to me that neither of the said papers is well-known. $\endgroup$ Mar 8, 2016 at 21:26
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    $\begingroup$ Sarnak's paper is very well-known in the field (and its "younger cousin" - the Sarnak-Ubis paper which is relatively recent, and related works by Strombergsson). The reason why it cited so few times is that qualitatively the result (without the RH connection) can be found from the Margulis' mixing argument (from his thesis), which is extremely famous and useful and its a bit more general, the problem with Margulis' approach is that at-best-case you'll hit the Selberg-Ramanujan conjecture (if you generalize the result in the obvious manner for general congruence surface), see Burger's article. $\endgroup$
    – Asaf
    Mar 8, 2016 at 21:59
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    $\begingroup$ The beauty of Sarnak's paper is that it goes deeper from Ramanujan, just because of explicitly analyzing the location of the poles of the Eisenstein series.Those quantities (and in general,the(constant term of the-)Eisenstein series for congruence surfaces) will include the Zeta function and then is when you get RH, so the "purely dynamical approach" won't give you RH in any case, but it might recover Ramanujan (related work is Eisendler-Margulis-Venkatesh where they recovered most of Clozel property(Tau) result). anyhow,the reference should be Sarnak's or Zagier's work,and not Verjovsky's $\endgroup$
    – Asaf
    Mar 8, 2016 at 22:04
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Polya-Hurwitz program.

This may become more interesting in light of the recent progress in the Polya-Jensen program by Griffin, Ono, Rolen, Zagier.

We will first provide definitions of some functions involved.

The Riemann Xi-function $\Xi(z)$ is related to the Riemann zeta-function $\zeta(s)$ via ([A], [B]): $\Xi(z)=\xi(\tfrac{1}{2}+iz)$, $\xi(s)=\tfrac{1}{2}s(s-1)\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s)$.

Riemann $\Xi(z)$ can be expressed as a Fourier transform of a positive, fast decaying, and even kernel $\Phi(t)$ ([A], [B]):

\begin{equation} \Xi(z)=2\int_0^{\infty}\Phi(t)\cos(zt)\mathrm{d}t,\tag{1} \end{equation}

where

\begin{equation} \Phi(t)=\sum_{k\geqslant 1}\phi_k(t)=\Phi(-t),\tag{2}\\ \end{equation}

\begin{equation} \phi_k(t)=\left(4\pi^2 k^4 e^{9t/2}-6\pi k^2e^{5t/2}\right)\exp\left(-\pi k^2 e^{2t}\right)\tag{3}. \end{equation}

The Polya aspect of this approach is the following:

Truncate the Kernel $\Phi(t)$ of (2) and/or the integration range in (1) such that the resulting Fourier transform leads to a family of entire functions which only have real roots.

One such candidate is given in [C]: \begin{equation} \Phi_{\color{red}n}(t)=(1/2)\sum_{1\leqslant k\leqslant {\color{red}n}}\left(\phi_k(t)+\phi_k(-t)\right)=\Phi_{\color{red}n}(-t)\tag{4} \end{equation}

\begin{equation} \Xi_{\color{red}n}(z)=2\int_0^{(1/2)\log {\color{red}n}}\Phi_{\color{red}n}(t)\cos(zt)\mathrm{d}t=\Xi_{\color{red}n}(-z),\tag{5} \end{equation}

We refer to [D] and [E] for a near complete review on the zeros of entire functions as Fourier transforms.

The Hurwitz aspect of this approach is the following:

Corollary of Hurwitz's theorem in complex analysis (applied to our case) [F]:

If $\Xi(z)$ and $\{\Xi_n(z)\}$ are analytic functions on a domain $S_{1/2}(z)=\{z: 0<Im(z)<1/2\}$, $\{\Xi_n(z)\}$ converges to $\Xi(z)$ uniformly on compact subsets of $S_{1/2}(z)$, and all but finitely many $\Xi_n(z)$ have no zeros in $S_{1/2}(z)$, then either $\Xi(z)$ is identically zero or $\Xi(z)$ has no zeros in $S_{1/2}(z)$.

The functional equation for $\zeta(s)$ becomes $\Xi(-z)=\Xi(z)$. The candidate of $\Xi_n(z)$ in (5) automatically satisfies this functional equation.

Another benefit of Polya-Hurwitz approach is that the entrance barrier is relatively low (comparing to other approaches that usually require the advance knowledge of analytical number theory).

To get started, one only needs to know Fourier transform, basic complex analysis, some knowledge of entire functions, polynomials etc. So anyone who has math training with the college undergraduate math major may start to work on the Polya-Hurwitz approach and learn other necessary new math as he/she goes.

The most difficult part of Polya-Hurwitz approach seems to be the following:(for example,) proving that all the zeros of $\Xi_n(z)$ in (5) are real in $S_{1/2}(z)$.

One may need to have several iterations: guess one form of the Kernel like $\Phi_n(1,t)$ and complete the integration to get explicit expression for $\Xi_n(1,z)$. If all the zeros of $\Xi_n(1,z)$ are found not to be all real in $S_{1/2}(z)$, then move on to $\Phi_n(2,t)$ and $\Xi_n(2,z)$...

References:

[A] Titchmarsh,"The Theory of the Riemann Zeta-Function",
(1986).

[B] Edwards, "Riemann's Zeta Function", (1974).

[C] Shi, "On the zeros of Riemann Xi-function", (2017) arXiv:1706.08868.

[D] Dimitrov and Rusev, “ZEROS OF ENTIRE FOURIER TRANSFORMS” (2001), 108 page review paper.

[E] Hallum, “ZEROS OF ENTIRE FUNCTIONS REPRESENTED BY FOURIER TRANSFORMS” (2014), Master thesis.

[F] Conway, "Functions of One Complex Variable",(1978)

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First formulation, the Möbius function:
From Wikipedia:

The statement that the equation:

$$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$$

is valid for every $s$ with real part greater than 1/2, with the sum on the right hand side converging, is equivalent to the Riemann hypothesis.

This in turn can be rewritten as:

$$\displaystyle \frac{1}{\lim\limits_{k\to \infty } \, \left(\sum\limits_{a=1}^{k} \frac{1}{a^s}+\frac{1}{(s-1) k^{s-1}}\right)} =$$ $$\displaystyle \lim_{k \rightarrow \infty} \left( \underbrace{1 - \sum_{2 \leq a \leq k} \frac{1}{a^{s}} + \underset{ab \leq k}{\sum_{a \geq 2} \sum_{b \geq 2}} \frac{1}{(ab)^{s}} - \underset{abc \leq k}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} \frac{1}{(abc)^{s}} + \underset{abcd \leq k}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} \frac{1}{(abcd)^{s}} - \cdots}_{\text{number of alternating sums} > \frac{\log(k)}{\log(2)}} \right)$$

where:

$\Re(s)>0:\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\zeta(s)=\lim\limits_{k\to \infty } \, \left(\sum\limits_{n=1}^{k} \frac{1}{n^s}+\frac{1}{(s-1) k^{s-1}}\right)$


Second formulation, the von Mangoldt function:
Reuns answer:
The Riemann hypothesis is equivalent to that the following sum converges:
$$\displaystyle \sum_{n=2}^{\infty} \left( \underbrace{-\frac{1}{\sqrt{n} \log^{3+\epsilon}(n)}+\underset{a = n}{\sum_{a \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} - \underset{ab = n}{\sum_{a \geq 2} \sum_{b \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} + \underset{abc = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} - \underset{abcd = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} \frac{\log(a)}{\sqrt{n}\log^{3+\epsilon}(n)} + \cdots}_{\text{number of alternating sums} > \frac{\log(n)}{\log(2)}} \right)$$

Mathematica: https://pastebin.com/gxAE6ZgY


Third formulation, Liouville Lambda function:
The Riemann hypothesis is equivalent to: $$\lim_{n\to \infty } \, \frac{\sum\limits_{k=1}^n \lambda (k)}{n^{\frac{1}{2}+\epsilon}}=0$$ where $\lambda(k)$ is the Liouville Lambda function.

This in turn can be rewritten as:

$$\lim\limits_{n \rightarrow \infty}\frac{1}{n^{\frac{1}{2}+\epsilon}}\left(\underbrace{\underset {a^2 \leq n} {\sum_ {a\geq 1}} 1 - \underset {a^2 b \leq n} {\sum_ {a\geq 1}\sum_{b\geq 2}} 1 + \underset {a^2 bc \leq n} {\sum_ {a\geq 1}\sum_ {b\geq 2}\sum_ {c\geq 2}} 1 - \underset {a^2 bcd \leq n} {\sum_ {a\geq 1}\sum_ {b\geq 2}\sum_ {c\geq 2}\sum_ {d\geq 2}} 1 + \cdots}_{\text{number of alternating sums} > \frac{\log(n)}{\log(2)}}\right)=0$$

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  • $\begingroup$ @Granvik do you know of a way to proove in an elementary way (without complex analysis) the RH-bouding between summing moebius (mu) and summing Liouville (lambda). $\endgroup$ Sep 6, 2023 at 15:24
  • $\begingroup$ @JérômeJEAN-CHARLES The multiple sums come from matrix power series. Here is a picture of the matrices: i.stack.imgur.com/C47sM.png OEIS comment: oeis.org/A008683 Calculate matrix powers: A175992^0 - A175992^1 + A175992^2 - A175992^3 + A175992^4 - ... Then the Mobius function is found in the first column. Compare this to the binomial series for (1+x)^-1 = 1 - x + x^2 - x^3 + x^4 - ... For the Möbius function, see link to Mathematica program here: math.stackexchange.com/a/3316344/8530 For Liouville see link here: math.stackexchange.com/a/4190159/8530 $\endgroup$ Sep 6, 2023 at 17:10
  • $\begingroup$ Better picture: i.stack.imgur.com/RqSsN.png With row sums I mean from the second column onwards in each previous matrix power. $\endgroup$ Sep 6, 2023 at 17:36
  • $\begingroup$ I see now what you are asking. No I don't know, other than that the möbius function sum converges to the right of the critical line. I would expect an alternating sum to converge when $\Re(s) > 0$. Not only when $\Re(s) > 1/2$. $\endgroup$ Sep 6, 2023 at 17:59
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Matiyasevich has reformulated RH as a computer science problem : "a particular explicitly presented register machine with 29 registers and 130 instructions never halts", see this reference .

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  • $\begingroup$ I have a Mathematica version of the first program in the article. I you want to post it in your answer, I am happy to provide it. One of the paraters (d) grows very large very quickly, so the program gets slow $\endgroup$
    – EGME
    Apr 28, 2022 at 14:47

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