ABSTRACT I consider a sequence of rational numbers defined by a set which is taken to be the Engel Expansion of the rational number. MAIN The Engel Expansion of a number x ∈ ℝ is the unique set of integers a₁, a₂, a₃, a₄, ⋯ such that x = {a_1}+{a_1a_2}+{a_1a_2a_3} + \cdots if the set of integers all become 0 or repeat indefinately in a regular pattern, after some fixed point, then the number x is likely to be rational, if the set of integers a is infinite, then x is irrational. There are many famous expansions, have a look. Let us define sets of numbers S_1=\{1,1,1,1,1,1,1,\cdots\}\\ S_2=\{1,2,2,2,2,2,2,\cdots\}\\ S_3=\{1,2,3,3,3,3,3,\cdots\}\\ S_4=\{1,2,3,4,4,4,4,\cdots\} and so on. We can treat these numbers as the terms in the Engel expansion of x, then we may write x=(S) we find that (S_1)=\infty\\ (S_2)=2\\ (S_3)={4}\\ (S_4)={18}\\ (S_5)={32}\\ (S_6)={600}\\ \cdots\\ (S_\infty)=e-1 They converge to e − 1. We might also consider the sets P_1=\{1,1,1,1,1,1,1,\cdots\}\\ P_2=\{1,2,2,2,2,2,2,\cdots\}\\ P_3=\{1,2,3,3,3,3,3,\cdots\}\\ P_4=\{1,2,3,5,5,5,5,\cdots\}\\ P_5=\{1,2,3,5,7,7,7,\cdots\} where we increment over the prime numbers. Then (P_1)=\infty\\ (P_2)={1}\\ (P_3)={4}\\ (P_4)={24}\\ (P_5)={180}\\ (P_5)={2100}\\ (P_6)={27720} the denominators seem to be taking A061720. The numerators are all prime up to P₆, however beyond this it does not seem to be the case. ENGEL EXPANSION SUM EQUIVALENT What about a set of integers mapping to a constant as x={a_1}+{a_1+a_2}+{a_1+a_2+a_3}+\cdots this doesn’t converge as nicely, we have some results ^\infty {^m k} =2 \\ ^\infty {^m k+1} ={9} \\ ^\infty {^m k+2} ={150} \\ ^\infty {^m {2}} =1 \\ ^\infty {^m 2k} =4 \\ ^\infty {^m k^2} =18-24\log(2) \\ ^\infty {^m k^3} ={3}-12 = 8\zeta(2)-12 It seems the number whose engel expansion is i, 2i, 3i, 4i, 5i, 6i, 7i, 8i, 9i, ... is (e−i − 1). Can we find a number whose engel expansion is the non-trivial zeta zeroes? What is the equivalent integer based expansion? It seems to arrive at the set 15i,16+4i,13-18i,-39-30i,-57+36i,-11+95i,45+121i,143+122i,151-131i,-61-421i,\cdots

Research is really f**king important.  This statement is almost self-evident by the fact that you're reading this online.  From research has come the web, life-saving vaccines, pasteurization, and countless other advancements. In other words, you can look at cat gifs all day because of research, you're alive because of research, and you can safely add milk to your coffee or tea without contracting some disease, because of research. But how research is done today is being stymied by how it is being communicated.  Most research is locked behind expensive paywalls \cite{Bj_rk_2010}, is not communicated to the public or scientific community until months or years after the experiments are done \cite{trickydoi}, is biased in how it is reported - only "positive" results are typically published \cite{Ahmed_2012}, does not supply the underlying data to major studies \cite{Alsheikh_Ali_2011}, and has been found to be irreproducible at alarming rates \cite{Begley_2012}.Why is science communication so broken?Many would blame the fault of old profit-hungry publishers, like Elsevier, and in many respects, that blame is deserved. However, here's a different hypothesis: what is holding us back from a real shift in the research communication industry is not Elsevier, it's Microsoft Word. Yes, Word, the same application that introduced us to Clippy is the real impediment to effective communication in research.Today, researchers are judged by their publications, both in terms of quantity and prestige.  Accordingly, researchers write up their documents and send them to the most prestigious journals they think they can publish in.  The journals, owned by large multinational corporations, charge researchers to publish their work and then again charge institutions to subscribe to the content. Such subscriptions can run into the many millions of dollars per year per institution \cite{Lawson_2015} with individual access costing $30-60 per article.The system and process for publishing and disseminating research is inimical to scientific advancement and accordingly Open Access and Open Science movements have made big steps towards improving how research is disseminated. Recently, Germany, Peru, and Taiwan have boycotted subscriptions to Elsevier \cite{Schiermeier_2016} and an ongoing boycott to publish or review for certain publishers has accumulated the signatures of 16,493 researchers and counting.  New developments such as Sci-hub, have helped to make research accessible, albeit illegally.  While regarded as a victory by many, the Sci-hub approach is not the solution that researchers are hoping for as it is built on an illegal system of exchanging copyrighted content and bypassing publisher paywalls \cite{Priego}.  The most interesting technologist view of the matter is that the real culprit for keeping science closed isn't actually the oligopoly of publishers \cite{Larivi_re_2015}-- after all, they're for-profit companies trying to run businesses and they're entitled to do any legal thing that helps them deliver value to shareholders. We suggest that a concrete solution for true open access is already out there and it's 100% legal.What is the best solution to truly and legally open access to research?The solution is publishing preprints -- the last version of a paper that belongs to an author before it is submitted to a journal for peer review. Unlike other industries (e.g. literature, music, film, etc.), in research, the preprint version copyright is legally held by the author, even after publication of the work in a journal.Pre-prints are rapidly gaining adoption in the scientific community, with a couple of preprint servers (e.g. arXiv which is run by Cornell University and is primarily for physics papers, and bioRxiv which is similarly for biology papers) receiving thousands of preprints per month.Some of the multinationals are responding with threats against authors not to publish (or post) preprints. However they are being met with fierce opposition from the scientific community, and the tide seems to be turning. Multinationals are now under immense pressure not just from authors in the scientific community, but increasingly from the sources of public and private funding for the actual research. Some organizations are even mandating preprints as a condition of funding. But what is holding back preprints and in general a better way for Authors to have more control of their research?We think the inability for scientists to independently produce and disseminate their work is a major impediment and at the heart of that of that problem is how scientists write. How can Microsoft Word harm scientific communication?Whereas other industries, like the music industry, have been radically transformed and accelerated by providing creators with powerful tools like Youtube, there is no parallel in research.  Researchers are reliant upon publishers to get their ideas out and because of this, they are forced into an antiquated system that has remained largely stagnant since it's inception over 350 years ago.Whereas a minority of researchers in math-heavy disciplines write using typesetting formats like LaTeX, the large majority of researchers (~82%) write their documents in Microsoft Word \cite{brischoux2009don}. Word is easy to use for basic editing but is essentially incompatible with online publishing. Word was created for the personal computer: offline, single-author use. Also, it was not built with scientific research in mind - as such, it lacks support for complex objects like tables and math, data, and code. All in all, Word is extraordinarily feature-poor compared to what we can accomplish today with an online collaborative platform. Because publishers have traditionally accepted manuscripts formatted in Word, and because they consistently fail to truly innovate from a technological standpoint, millions of researchers find themselves using Word. In turn, the research they publish is non-discoverable on the web, data-less, non-actionable, not reusable and, most likely, behind a paywall.  What does the scientific communication ecosystem of the future look like?What is needed is a web-first solution. Research articles should be available on distinct web pages, Wikipedia style. Real data should live underneath the tables and figures. Research needs to finally be machine readable (instead of just tagged with keywords) so that it may be found and processed by search engines and machines. Modern research also deserves to have rich media enhancement -- visualizations, videos, and other forms of rich data in the document itself.All told, researchers need to be able to disseminate their ideas in a web first world, while playing the "Journal game" as long as it exists. Our particular dream (www.authorea.com) is to construct a democratic platform for scientific research -- a vast organizational space for scientists to read and contribute cutting edge science. There is a new class of startups out there doing similar things with the research cycle, and we feel like there is a real and urgent demand for solutions right now in research.

ABBSTRACT I consider operations of the ’sets’ that make up the continued fractions of special constants. This leads to some concepts. MAIN We have a continued fraction representation for constants, for example \zeta(2)={6}=[1;1,1,1,4,2,4,7,1,4,\cdots] if we treat this string of numbers like a vector and multiply by a constant, what constant is required to say turn ζ(2) into ζ(3), c_1[1;1,1,1,4,2,4,7,1,4,\cdots]=[c_1;c_1,c_1,c_1,4c_1,2c_1,4c_1,7c_1,c_1,4c_1,\cdots]=\zeta(3)\\ c_1\otimes \zeta(2) = \zeta(3) it would appear that the constant is c₁ ≈ 0.13341786712221396. Exactly how to treat the first term is not clear. For example if we have the golden ratio \varphi=[1;1,1,1,1,1,1,1,\cdots] then 2\otimes \varphi = [2;2,2,2,2,2,2,2,\cdots] = 1+ which is the silver ratio. Here we are using ⊗ to denote the multiplication of the continued fraction bracket. However, if we didn’t scale the first element we would have 2\otimes \varphi = [1;2,2,2,2,2,2,2,\cdots] = I think the second definition is better. Otherwise we are just adding extra integers. Then we can revisit the first definition c_1\otimes \zeta(2) = \zeta(3) and find that c₁ ≈ 4.747311360. Which is a drastic change. Using this we find a potential result. It would appear that \varphi \otimes \varphi = {4}\left(3-+)} \right) SEQUENCES Does there exist a constant that changes one sequence to another in this manner. For example, converting the sequence 1, 1, 1, 1, 1, 1, ⋯ to the sequence 1, 2, 3, 4, 5, 6, ... would mean finding c such that c\otimes \varphi = {I_0(2)} see A052119. Is there a constant that turns [0; 1, 2, 3, 4, 5, 6, 7, 8, ⋯] into [0, 2, 3, 5, 7, 11, 13, ⋯], i.e. c_p \otimes [0;2,3,5,7,11,13,\cdots] = [0;1,2,3,4,5,6,7,8,\cdots] = {I_0(2)} we find cp ≈ 0.416327135295592, also c_q \otimes [0;1,2,3,4,5,6,7,8,\cdots] = [0;2,3,5,7,11,13,\cdots] we find cq ≈ 2.08157073000677. Interestingly {c_p} \approx 5 ADDITION We can also define the addition of the brackets, such that \varphi \oplus \varphi = or [1;1,1,1,1,\cdots] \oplus [1;1,1,1,1,\cdots] = [1;2,2,2,2,\cdots] this would also give us [1;1,2,3,4,5,6,\cdots] \oplus [1;1,2,3,4,5,6,\cdots] = [1;2,4,6,8,10,12,\cdots] where explicitly [1;1,2,3,4,5,6,\cdots] = 1 + {I_0(2)} [1;2,4,6,8,10,12,\cdots] = 1 + {I_0(1)} with I Bessel functions.

MAIN Define the nested zeta function, \nu_k(s)=_{}(s) where ⊙ is a function composition. So then \nu_1(s)=\zeta(s)\\ \nu_2(s)=\zeta(\zeta(s))\\ \nu_3(s)=\zeta(\zeta(\zeta(s))) and so on. Things to note Let ρ be a non trivial zero, ν₁(ρ)=0 by definition, $\nu_2(\rho)={2}$. The imaginary part of zeta(s), is zero from then on, It appears \Im[\nu_k(-1/2)]=-0.2959050055752138.... = C_\zeta a real constant, A069857. So ℑ[nuk(ρ)] = 0 ∀k, which is apparently the only attractive fixed point on the real line for the ζ → ζ map. Then any zeta zero ρ, subject to ν∞ will go to this constant. So we can say \nu_k(\rho) = C_\zeta, \; \forall \; \rho : \zeta(\rho)=0

MAIN If \zeta(s) = {s-1}-s\int_1^\infty {x^{s+1}}\;dx, \; s>0 then let us consider P_{n,m}(s)=-s\int_n^m {x^{s+1}}\;dx it would appear that P_{n,n+1}(s)=-s\int_n^{n+1} {x^{s+1}}\;dx = -}{s-1} + }{s-1} + }{s-1}, \; n \in taking the infinite sum over these then gives that \zeta(s) = {s-1} +{1-s} + ^\infty }{s-1}\\ \zeta(s) = {s-1} + ^\infty }{s-1} and therefore ^\infty }{s-1} = {s-1} or ^\infty n(n+1)^{-s} = \zeta(s-1)-\zeta(s) then ζ(s)=0 for points that ^\infty n(n+1)^{-s_k} = \zeta(s_k-1)

I was accepted into the cell biology program at Virginia Tech under conditional terms due to a mediocre undergraduate GPA. This was the deal: maintain good grades and I’d get to continue, or slip up and I was out. As an undergraduate, I spent a lot of time surfing and very little time  cramming for tests -- what can I say? I wasn’t exactly a traditional grad student applicant.Despite my shortcomings on paper, I was ambitious. Before grad school, I contacted a researcher from Harvard who’d proposed through mathematical models that we could kill cancer cells with cancer cells \cite{Deisboeck_2008}.  I told him I wanted to test his proposal experimentally. When he wrote back and I brought the proposal to my potential PI, I quickly realized that incoming grad students don’t actually do this. You’re supposed to go through rotations first, and then select a lab, pick a project that falls within the scope of your PI’s research, and so on. This wasn’t exactly my style.The deeper I got into my PhD, the more I realized the game you have to play in order to be successful: publish in certain journals, publish with the best coauthors you can manage, publish as much as you can. I played the game and published as much as possible within the scope of cell biology and cancer, but also papers within the scope of the scientific communication process itself -- papers on funding, peer review in high-impact journals, and peer review at the NIH. I wrote about cancer but I also wrote about all the problems I was seeing around me in the process itself.I never thought about actually doing anything about these systemic problems until I read The Trouble with Medical Journals \cite{Smith_2006}. The key tenet  -- that peer review misses most major errors -- is the idea that sent me down the path of building a publishing company to take the whole publishing process and flip it in favor of openness. Instead of filtering results and then publishing, I wanted scientists first to publish and then to filter -- to publish and then winnow, so to speak. That’s why the Winnower was born.From Scientist to EntrepreneurI didn’t know anything about starting a business but I knew I needed some money to do it.  I wrote up some ideas for a new publication, entered a business contest on campus, and lost. It was harder than I thought.  But then I sent that proposal to some people I knew from undergrad and through a lot of luck managed to get 50k from a private investor.The Winnower launched in May 2014 and over the course or two years we shifted away from publishing traditional papers to publishing so-called grey literature -- informal documents that traditional publishers ignore.  We published scholarly reddit AMAs, foldscope images, responses to NIH RFIs, journal clubs, and some of the coolest essays I’ve ever read. We formalized blogs and journal clubs so that they could act as reviewers themselves. People liked what we were doing, as judged by the growth of publications and readership. Why shouldn’t reddit AMA’s have DOIs and be given real scientific consideration? I gave talks around the world, raised more money, and met other academics doing similar things with their own companies. I felt lucky and privileged to be doing what I loved, despite the fact I was making less now running a company than I was as a grad student. The End.Okay, the story doesn’t have an ending yet because the story is still ongoing.  Very recently The Winnower was acquired by Authorea, another early company working on the same problem but from a different direction. Authorea, which is also founded by former academics, is fixing how researchers write, collaborate, and share online.  Together we’re working to become the place where researchers can write and publish whatever they want collaboratively and online. It’s an ambitious goal but so too was cancer research.I can’t say if we’ll achieve our goals and I know the road ahead is still daunting but I think the problems we’re working to solve are as hard as some of the most complex problems in science. What is certainly true is that we must work collaboratively to solve them.  I hope this essay inspires more academics to follow their own “crazy” ideas and I hope you’ll stand with our mission to build a more transparent system of research communication.  Let’s get it right.

MAIN Define the function T(x,t)={t}\arg\left[{} + {x}}\right] then if we plot as a function of x, F(x) = ^N T(x,t_k) for large enough N, where tk is the imaginary part of the kth zeta zero, we see the primes forming as spikes in x, with much smaller spikes for the squares and cubes and so on. This is fairly standard behaviour from a spectral point of view. However, if we plot a function G(x)=^N T(x,{2}) we only really see very pronounced spikes in x for the square numbers. We may also sum H(x)=^N T(x,p_k) where pk is the kth prime, and we see spikes that appear to be at values eπ and e2π/3, relating to Gelfond’s constant. We needn’t necessarily include the factor of 1/t in T(x, t), in fact it supresses the peaks somewhat. We can also just plot, with T(x,t)=-\cos(t \log(x)) and get similar results.

Corresponding author: leouieda@gmail.com This is a part of The Leading Edge “Geophysical Tutorials” series. You can read more about it in . Open any textbook about seismic data processing and you will inevitably find a section about the normal moveout (NMO) correction. There you’ll see that we can correct the measured travel-time of a reflected wave t at a given offset x to obtain the travel-time at normal incidence t₀ by applying the following equation t_0^2=t^2-{v_^2} in which vNMO is the NMO velocity. There are variants of this equation with different degrees of accuracy, but we’ll use this one for simplicity. When applied to a common midpoint (CMP) section, the equation above is supposed to turn the hyperbola associated with a reflection into a straight horizontal line. What most textbooks won’t tell you is _how, exactly, do you apply this equation to the data_? Read on and I’ll explain step-by-step how the algorithm for NMO correction from works and how to implement it in Python. The accompanying Jupyter notebook contains the full source code, with documentation and tests for each function. You can download the notebook at github.com/seg or github.com/pinga-lab/nmo-tutorial.

MAIN I am exploring a unique mapping of the positive integers to polynomials. The mapping expressed the divisibility in a nice way. Define, K_n(x)=^{2n} x^k \\ Q_n(x)=^{2n} (1+x^k) Then we may define the polynomial that represents n a positive integer as P_n(x) = ( K_n(x), Q_n(x) ) where pgcd is the polynomial gcd function, that is, greatest common divisor function. RESULTS Below is a table of the results. Notice how all numbers have a factor of 1, but just once. Then 4 contains 1 and 2, but also something special which is not just 2 squared, but is actually (1 + ix²)(1 − ix²). See more on this below the table. P_1(x) = (1+x) \\ P_2(x) = (1+x)(1+x^2) \\ P_3(x) = (1+x)(1-x+x^2) \\ P_4(x) = (1+x)(1+x^2)(1+x^4) \\ P_5(x) = (1+x)(1-x+x^2-x^3+x^4) \\ P_6(x) = (1+x)(1+x^2)(1-x+x^2)(1-x^2+x^4) \\ P_7(x) = (1+x)(1-x+x^2-x^3+x^4-x^5+x^6) \\ P_8(x) = (1+x)(1+x^2)(1+x^4)(1+x^8) \\ P_9(x) = (1+x)(1-x+x^2)(1-x^3+x^6) \\ P_{10}(x) = (1+x)(1+x^2)(1-x+x^2-x^3+x^4)(1-x^2+x^4-x^6+x^8) Let us explore the hierarchy of these factors. We can say the Fn(x), are the parts of the polynomial that do not appear in any of the polynomials representing n. For example the part of P₆(x) which is not a part of P₁(x),P₂(x) or P₃(x) is F_6(x) = (1-x^2+x^4) where the prescence of 3 terms indicates relation to the divisor 3, but the powers increasing by steps of 2 indicates relation to the divisor 2. Listing these Fn(x) factors gives F_1(x) = (1+x) \\ F_2(x) = (1+x^2) \\ F_3(x) = (1-x+x^2) \\ F_4(x) = (1+x^4) \\ F_5(x) = (1-x+x^2-x^3+x^4) \\ F_6(x) = (1-x^2+x^4) \\ F_7(x) = (1-x+x^2-x^3+x^4-x^5+x^6)\\ F_8(x) = (1+x^8) \\ F_9(x) = (1-x^3+x^6) \\ F_{10}(x) = (1-x^2+x^4-x^6+x^8) we can see that if n is prime, then Fn(x) seems to be given by F_p(x) = ^{p-1} (-x)^k however, much like the cyclotomic polynomials, at n = 105, we see this rule broken, and a coefficient of 2 enter on the power x⁴¹. we see for powers of 2, F_{2^n}(x) = (1+x^{2^n}) SPECIAL PROPERTIES These polynomials appear to be such that Pn(−1)=0 ∀ n. These polynomials appear to be such that Pn(−1)/(x + 1)=n ∀ n. These polynomials appear to be such that Pn(0)=1 ∀ n. These polynomials appear to be such that Pn(1)=2 ⋅ 2k ∀ n. Where 2k is the largest power of 2 that divides n. We see that the factor (1 + x) on each Pn(x) will add this additional factor of 2. If m divides n, then it seems Pm(x) divides Pn(x), leaving some finite remainder polynomial. These polynomials evaluated at Pn ∈ ℙ(−2) appear to generate many primes, semi-primes and other numbers with large prime factors of multiplicity 1. We may conjecture, that the only n such that Pn(−2)∈ℙ are prime n. These numbers begin 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, ..., 521, ..., these appear to be the Mersenne Exponents. It appears the polynomials It seems that $P_n({3})$ tends to  1 for large n, apart for powers of 2, where it tends to 3. (Looking closer, there may be other patterns as well) For $P_n({2})$, it seems to tend towards 2 for powers of 2, 1 in general, and to ${7}$ for the numbers , 6, 12, 24, 48, 96, which are 6 times the powers of 2. Divisibility These polynomials appear to have the property that if x is replaced with the correct complex exponential expression, divisibility can be tested. For a given Pn(x), if we replace x with exp(iπ/2), then if n is divisible by 2, Pn(exp(iπ/2)=0, otherwise, it is not 0. In general P_n\left(\exp({m})\right)=0,\; \; {m}\in This means we can test primality of a number n, by the following \; \min\left[ P_n\left(\exp({m})\right) \right]>0, \; \forall \; 1<m<n\; n \in SOME UNPROVEN IDENTITIES ^\infty {P_n(x)} = {1-x^2} conjecture: the sum \eta(x)=^\infty {P_n(x)} is the alternating sign generating function for A086374. Then if this is true, the coefficient of a power of x in the expansion is −3, then that power is an odd prime! Comparing this to (x)=^\infty }{P_{2n}(x)} where every coefficient of a power that is −1, denotes that power is an odd prime. This sequence seems to be a signed version of A091954, the number of odd proper divisors of n. MORE It seems that {dx^n}P_n(x)\Bigg|_{x=0} = n! \\\\ {dx}P_n(x)\Bigg|_{x=0} = (n) =\bigg\{ 1 & n=2^m\;m\in \\ 0 \\ {dx^3}P_n(x)\Bigg|_{x=0} = (n) =\bigg\{ 3! & n=2^m||3\cdot2^m\;m\in \\ 0 \\ {dx}P_n(x)\Bigg|_{x=1} =\bigg\{ 2n-1 & \;n \\ 6+12m & n = 2+4m \\ ?? & \\ {dx}P_n(x)\Bigg|_{x=-1} =n F_N(X) It appears that Fn(−1) make A014963, the exponentiaation of the von Mangoldt function. When evalauted as Fn(2), they appear to equal A066845.

ABSTRACT Something I noticed. We can not down the binomial expansion in a matrix form and see if there are any nice patterns. MAIN We have the binomial expansion (a+b)^n \\ (a+b)^2= a^2+2ab+b^2 \\ (a+b)^3 = a^3+3a^2b+3ab^2+b^3\\ (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \\ \cdots for some integer power of n, where the general form of the expanded version of this has been known for a long time leading to Pascal’s triangle and so on. I just wanted to write down in matrix form a few steps. Here we have two variables a and b. So let’s apply some matrix to the vector (a, b) and recreated the various results. 1 & 1 a & a \\ b & b a \\ b = a^2 + 2ab+b^2 = (a+b)^2 1 & 1 a & a \\ b & b ^2 a \\ b = a^3+3a^2b+3ab^2+b^3 1 & 1 a & a \\ b & b ^3 a \\ b = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 So there appears to be a general form 1 & 1 a & a \\ b & b ^n a \\ b =(a+b)^{n+1}, \; n \in for positive fractional numbers n → (n/m) the following seems to hold 1 & 1 a & a \\ b & b ^{n/m} a \\ b =(a+b)^{(n+m)/m}, \; n,m \in in fact it even seems to hold for irrational numbers like π. The matrix power automatically generates the binomial expansion. So this may not be very helpful to calculate such a thing, as one still needs to perform many matrix operations, however, we might as what happens if we manipulate the central matrix. What other transforms can be described by this? To make things shorter define = 1 & 1 \\ = a \\ b \\ _1= a & a \\ b & b Then we can write our expression _1^n=(a+b)^{n+1} Now what if we use the transpose of the matrix? Wrting out a few terms (^{T}_1)^1=2(a^2+b^2)\\ (^{T}_1)^2=2(a+b)(a^2+b^2)\\ (^{T}_1)^3=2(a+b)^2(a^2+b^2) then (^{T}_1)^n=2(a+b)^{n-1}(a^2+b^2) We only did a 2 × 2 matrix, but it seems the multinomial expansion is given by a similar expression, given k variables a₁, a₂, ⋯, ak, let $$ be the column vector of those variables, and M be the k × k matrix whose columns are all $$, then let $$ be the length k row vector whose elements are all 1, then we have ^n=\left(^k a_m\right)^{n+1} explicitly for k = 5 1 & 1 & 1 & 1 & 1 a_1 & a_1 & a_1 & a_1 & a_1 \\ a_2 & a_2 & a_2 & a_2 & a_2 \\ a_3 & a_3 & a_3 & a_3 & a_3 \\ a_4 & a_4 & a_4 & a_4 & a_4 \\ a_5 & a_5 & a_5 & a_5 & a_5 \\ ^n a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 =(a_1+a_2+a_3+a_4+a_5)^{n+1}, \; n \in the vector full of ones exists to sum over the resulting vector from the matrix vector multiplication. TABULATE RESULTS a & a \\ b & a^n=-)^2(a-)^n}{2} + +)^2(a+)^n}{2} If we set a = 1 and b = 5 for example, then running n = −1, 0, 1, 2, 3, ⋯ gives an integer sequence 1, 6, 16, 56, 176, 576, 1856, ⋯ which is A084057. In a previous article I was trying to find a nice expression for the coefficients of (x + 1)x + m, for integers x and m. We can now write this in matrix form ...

MAIN Following the notes at the bottom of http://mathworld.wolfram.com/RiemannZetaFunctionZeros.html, if the RH is true, then we can say, for imaginary parts of the zeta zeroes tk \pi={4}\exp\left(2+\gamma-^\infty {1+4t_k^2} \right) this appears to converge quite slowly. Of course we can then write this as \pi={4}\exp(2+\gamma)\exp\left(-^\infty {1+4t_k^2} \right) and then \pi={4}^\infty\exp\left({1+4t_k^2} \right) we can write {1+4t^2} =^\infty (-1)^n 2^{2 n + 1} t^{2n - 2} but this doesn’t converge for all t. so then \pi={4}^\infty\exp\left(^\infty (-1)^n 2^{2 n + 1} t_k^{2n - 2}\right) leading to \pi={4}^\infty^\infty\exp\left( (-1)^n 2^{2 n + 1} t_k^{2n - 2}\right) but is this helpful?

MAIN We write out the terms of the series for the Rieman zeta function into a grid of width k, wrapping down onto the next line. By drawing k + 1 lines through the terms, which also wrap, all of the terms are accounted for. Using this relationships can be drawn upon by using known values of the ζ(s). I.e. for integer s. These results are for example {12^2}+\psi ^{(1)}\left({5}\right)+\psi ^{(1)}\left({5}\right)+\psi ^{(1)}\left({5}\right)+\psi ^{(1)}\left({5}\right)=4 \pi ^2 with ψn the polygamma function. GRID First we write the terms of the zeta function in a width k grid, for example for k = 4 {1^2} & {2^s} & {3^s} & {4^s} \\ {5^2} & {6^s} & {7^s} & {8^s} \\ {9^2} & {10^s} & {11^s} & {12^s} \\ {13^2} & {14^s} & {15^s} & {16^s} we may draw a line through the 1, 6, 11, 16, ... terms, and also lines through the 2, 7, 12, ..., the 3, 8, 13, ..., the 4, 9, 14, ... and the 5, 10, 15, ... terms. Then these five sequences describe all of the terms. Then the sum of the terms (which is the zeta function itself), is equal to the five sequences summed. This is obvious and arbitrary. However, we can also define the next terms to sum in the grid based on the old terms. We may write {1^s} + {6^s} + {11^s} + \cdots = {1^s} + {2^s} + \left( {6^s} - {2^s} \right) + {3^s} + \left( {11^s} - {3^s} \right) + \cdots which is actually equal to ζ(s) plus an additional sum, we can call the additional sum A. If we do this for the other 4 sequences, we get more sums B, C, D, E, and find them to all be polylog expressions. Equationg the whole lot gives the expressions \zeta(s) = 5\zeta(s)-4-{2^s} - {3^s} - {4^s} + A + B + C + D + E where explicitly A=^\infty (nk+n+1)^{-s} - (n+1)^{-s}\\ B=^\infty (nk+n+2)^{-s} - (n+2)^{-s}\\ C=^\infty (nk+n+3)^{-s} - (n+3)^{-s}\\ D=^\infty (nk+n+4)^{-s} - (n+4)^{-s}\\ E=^\infty (nk+n+5)^{-s} - (n+5)^{-s} with k = 4. However, the selection of k was arbitrary, we can do this for any k and write \zeta(s) = (k+1)\zeta(s) -^{k-1} {(m+1)^s} + ^{k+1} S_n where then S_n=^\infty (mk+m+n)^{-s} - (m+n)^{-s}\\ IDENTITIES {54}+\psi ^{(2)}\left({5}\right)+\psi ^{(2)}\left({5}\right)+\psi ^{(2)}\left({5}\right)+\psi ^{(2)}\left({5}\right) -125 \left( \psi ^{(2)}(2)+ \psi ^{(2)}(3)+ \psi ^{(2)}(4)+ \psi ^{(2)}(5) \right)=752 \zeta (3) \psi ^{(1)}\left({3}\right)+\psi ^{(1)}\left({3}\right)={3}-{4} -\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)-\psi ^{(2)}\left({7}\right)+343 \psi ^{(2)}(2)+343 \psi ^{(2)}(3)+343 \psi ^{(2)}(4)+343 \psi ^{(2)}(5)+343 \psi ^{(2)}(6)+343 \psi ^{(2)}(7)=686 \left({36000}-{343}\right) which may be written 7^3^7\psi ^{(2)}(n) - ^6\psi ^{(2)}\left({7}\right) ={2^4 3^2 5^3} - 3432 \zeta(3) also have ^1\psi ^{(2)}(n)= {1} - 2\zeta(3) \\ ^2\psi ^{(2)}(n)= {1} - 4\zeta(3) \\ ^3\psi ^{(2)}(n)= {2^2} - 6\zeta(3) \\ ^4\psi ^{(2)}(n)= {2\cdot3^3}-8 \zeta (3) \\ ^5\psi ^{(2)}(n)= {2^5\cdot3^3}-10 \zeta (3) \\ ^6\psi ^{(2)}(n)= {2^4\cdot3^2\cdot5^3} -12 \zeta(3) \\ ^7\psi ^{(2)}(n)= {2^5\cdot3^2\cdot5^3} - 14 \zeta(3) \\ ^8\psi ^{(2)}(n)= {2^3\cdot3^2\cdot5^3\cdot7^3}-16 \zeta(3) \\ ^9\psi ^{(2)}(n)= {2^8\cdot3^2\cdot5^3\cdot7^3}-18 \zeta(3) \\ ^{10}\psi ^{(2)}(n)={800150400}-20 \zeta (3)\\ ^{11}\psi ^{(2)}(n)={8001504000}-22 \zeta (3)\\ ^{12}\psi ^{(2)}(n)={2^6×3^5×5^3×7^3×11^3}-24 \zeta (3) \\ It would appear that there is a product of powers of primes as the denominator for this constant! This creates a very convoluted prime test. We have the famous \psi^{(n)}(z+1)=\psi^{(n)}(z) + (-1)^nn!z^{-n-1} and therefore \psi^{(2)}(z+1)=\psi^{(2)}(z) + {z^3} using that \psi^{(2)}(1)=-2\zeta(3) we can then write ^{z+1}\psi^{(2)}(n)=^z\psi^{(2)}(1) + {z^3} we need some way of separating the numerator and denominator. For example we may use , {l, 1, 50}]]} and we see the higher prime factors growing (with the exception of glitches for low values). The benefit of this is that if we know a number p is prime. Then we could work out the next prime immediately by looking at (sensible) differences in the value of the polylog until the denominator becomes zero mod p, then we know the new prime and we can repeat the process. Example. Say I know 389 is prime, and no other information. If we work out the denominator D₃₈₉, it is a very large number. (This is the main problem, we need a modular extension to all of this). But it is quick to verify D₃₈₉ mod 389 = 0, we then calculate the next D389 + k’s for k for candidate primes (digit sum rules, odd numbers etc.), checking for each whether D389 + k mod k = 0. ... (unfortunately this is not strong enough to test primality).

MAIN By writing the natural numbers in a special pattern we may find strange patterns in the prime numbers. Make the grid 1 & 3 & 6 & 10 & 15 & 21 & 28 & 36 & 45 \\ 2 & 4 & 7 & 11 & 16 & 22 & 29 & 37 & 46 \\ 5 & 8 & 12 & 17 & 23 & 30 & 38 & 47 & 57 the top row is the triangular numbers, the following rows are m + (n+m-1)(n+m)/2 with m indexed at 0 and n indexed at 1. We can draw a line to trace the numbers from 1, 2, 3, 4, 5, ... and the pattern is obvious. Creating the grid is easy with a turtle like code. When we highlight the prime numbers on this grid we see dense streaking patterns and broken sections of regular structure. For example, it is easy to see that there are many primes of the form n+58 + (2n+58-1)(2n+58)/2 for n < 100000 there are ≈20000, so they are very common. This pattern repeats with even more primes for those of the form n+253 + (2n+253-1)(2n+253)/2 and many of the primes are the same! If we pick n and m such that n + 3m = 59 or n + 3m = 38, we see that every positive combination results in Q_{mn}=m + (n+m-1)(n+m)/2 being prime. That is let m = 1, 2, 3, 4, 5, ⋯, ans then have that n=59-3m\\ n=38-3m\\ n=23-3m\\ n=14-3m\\ n=11-3m\\ n=7-3m finding that Qmn is prime for all of these for suitable m.

MAIN We can package a sum over n into a set of sums with a triangular number like stride. We can see this by taking the integers, removing the triangular numbers, then shifting by one, removing the triangular numbers again 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,\cdots\\ .,2,.,4,5,.,7,8,9,.,11,12,13,14,.,16,17,18,19,20,.,22,23,24,25,26,27,.,29,30,\cdots\\ .,.,.,.,5,.,.,8,9,.,.,12,13,14,.,.,17,18,19,20,.,.,23,24,25,26,27,.,.,30,\cdots\\ .,.,.,.,.,.,.,.,9,.,.,.,13,14,.,.,.,18,19,20,.,.,.,24,25,26,27,.,.,.,\cdots\\ and so on ^\infty f(n) = }f(n) + }f(n) + }f(n) + \cdots the first coefficients of the latter sums take the formula c_{1,0}=1\\ c_{1,n}={2}(3n+n^2) where we are calling the first sum 0 and the later sums 1, 2, 3, 4, ⋯, n, ⋯. Then we have that c_{n,0}={2}\\ c_{n,1}=1+{2} = 1+c_{n,0} \\ c_{n,2}=2+{2} = 1+c_{n+1,1} = 2+c_{n+1,0} \\ c_{n,3}=3+c_{n+2,0} \\ c_{n,4}=4+c_{n+3,0} \\ c_{n,m}=m+c_{n+m-1,0} = m + {2}, \;m>0 so we can rewrite ^\infty f(n) = ^\infty f(c_{n,0}) + ^\infty f(c_{n,1}) + ^\infty f(c_{n,1}) + \cdots\\ ^\infty f(n) = ^\infty ^\infty f(c_{n,m}) \\ ^\infty f(n) = ^\infty f\left({2} \right)+ ^\infty ^\infty f\left(m + {2}\right) Is this useful? It seems to complicate things alot, for example the normal example \zeta(2)=^\infty {n^2}={6} The first term is easy S_0=^\infty \left({2}\right)^{-2} = {3}(\pi^2-9) the rest is then S_1=^\infty {(1+8n)^{3/2}} \left(2(\psi_0(t_-)-\psi_0(t_+))+(\psi_1(t_-)+\psi_1(t_+)) \right) with ψn the polygamma function, and t_+= {2} + n +{2} \\ t_-= {2} + n -{2} we then must have that S₀ + S₁ = π²/6. So S_1={6}(72-7\pi^2) But we can also write a chain of expressions for the mth sum, {6} = {3}(4\pi^2-36) + {27}(4\pi^2-31) + S_2 + \cdots but some of these sums are hard to write here. We find the value indeed converges towards the required value of S₁ ≈ 0.485462. The sum is not particularly efficient. ZETA(3) We can work towards a sum for ζ(3) though! We find from similar treatment we have \zeta(3)= 8(10-\pi^2) - ^\infty {(1+8n)^{5/2}}\Bigg( 12(\psi_0(t_-)-\psi_0(t_+)) + 6( \psi_1(t_-) + \psi_1(t_+)) + (1+8n)(\psi_2(t_-)-\psi_2(t_+)) \Bigg) and the remaining terms converge on the correct number 0.15889211187406554. For triangular numbers, the summand simplifies to a nicer form. SUM ^\infty {2^n} = 1 so then when we perform ^\infty 2^{-(n(n+1)/2)} = {2^{7/8}}\theta_2(0,{})-1 ^\infty ^\infty 2^{-\left(m+{2}\right)} for m = 1 we get {2\cdot2^{7/8}}\theta_2(0,{})-{2} GENERATING FUNCTIONS We may write the set of powers of x as generating functions as follows T={2}\theta_2(0,) for elliptic theta, then -1+x^{-1/8}T = x + x^3 + x^6 + x^{10} + x^{15} + x^{21} + x^{28} + \cdots \\ -x + x^{7/8}T = x^2+x^4+x^7+x^{11}+x^{16}+x^{22} + \cdots\\ -x^2(1+x) +x^{15/8}T = x^5 + x^8 + x^{12} + x^{17} + x^{23} + x^{30} + \cdots \\ and the general term becomes -1+x^{-1/8}T, \;n=0 \\ -x^n\left(^{n-1} x^{k(k+1)/2}\right) + x^{(8n-1)/8}T = ^\infty x^{n+(k+n-1)(k+n)/2}, \;n>0 Thus {1-x} = -1+x^{-1/8}T + ^\infty -x^n\left(^{n-1} x^{k(k+1)/2}\right) + x^{(8n-1)/8}T \\ {1-x} = A + ^\infty -x^n\left(^{n-1} x^{k(k+1)/2}\right) where A= -1+x^{-1/8}T + ^\infty x^{(8n-1)/8}T\\ A = )}{(2-2x)x^{1/8}} -1 where A + 1 is the beautiful series 1 + 2x + 2x^2 + 3x^3 + 3x^4 + 3x^5 + 4x^6 + 4x^7 + 4x^8 + 4x^9 + \cdots with n of each coefficient! This then gives B = ^\infty -x^n\left(^{n-1} x^{k(k+1)/2}\right) = -x-x^2-2x^3-2x^4-2x^5-3x^6-3x^7-3x^8-3x^9-\cdots

MAIN \zeta(s)=^\infty f_k(s)\\ f_k(s)=\exp\left(} {kp^{ks}} \right) but P(s), still seems to exhibit non trivial zeroes in e^{P(s)} which is sufficient for the whole function going to zero due to the product. We see the next term e^{P(2s)/2} seems even to have a local maximum at the first zero. also investigate e^{iP(s)} and see discontinuities in the absolute value across critical line for this one. FURTHER So if then P(s)= \mu(n){n} we have \exp(P(s)) = \exp\left(\mu(n){n}\right) = \zeta(s)}\zeta(ms)^{1/m}}{}\zeta(ns)^{1/n}} so if η(s) has a zero, then it is likely exp(P(s)) has a zero, unless the ratio of products diverges somehow. Then } (1-p^{-s})\exp(p^{-s})=}\zeta(ms)^{1/m}}{}\zeta(ns)^{1/n}} } (1-p^{-s})\exp(p^{-s})=} \left(\prod_p 1-p^{-ns}\right)^{1/n}}{}\left(\prod_p 1-p^{-ms}\right)^{1/m}} } (1-p^{-s})\exp(p^{-s})=}{m:}}\prod_p \right)^{1/n}}{\left(1-p^{-ms}\right)^{1/m}} for every pair, m takes 6, 10, 14, 15, ..., n takes 2, 3, 5, 7, 11, 13, .... } (1-p^{-s})\exp(p^{-s})=}{m:}}\prod_p \right)^{m/nm}}{\left(1-p^{-ms}\right)^{n/nm}} } (1-p^{-s})\exp(p^{-s})=}{m:}}\prod_p \left(\right)^{m}}{\left(1-p^{-ms}\right)^{n}}\right)^{1/nm} } (1-p^{-s})\exp(p^{-s})=}{m:}}\prod_p \left(^m {k}(-1)^kp^{-nsk}}{^n {l}(-1)^lp^{-msl}}\right)^{1/nm} So define a special set of functions \kappa_p^{(n,m)}(s)=^m {k}(-1)^kp^{-nsk}}{^n {l}(-1)^lp^{-msl}} = \right)^{m}}{\left(1-p^{-ms}\right)^{n}} for example for the first pair \kappa_p^{(2,6)}(1)={p^{12}}-{p^{10}}+{p^8}-{p^6}+{p^4}-{p^2}}{1+{p^12}-{p^6}}= {(1+p^2+p^4)^2} if we take this to be an inverted generating function in p, i.e G(1/p) for some sequence, we can perform the inverse Z-transform to get and expression for the sequence, this turns out to be amazingly an aerated sequence of integers 0,-6,0,15,0,-18,0,3,0,24,0,-36,0,12,0,33,\cdots actually, the function is invarient to the transform p → 1/p. When we add the power of 1/nm which is 1/12, we get a function which is the double factorial generating function for the sequence 1, −1, −1, 5, −47, −185, 3055, −45955, ⋯ LIST OF THE SPECIAL FUNCTIONS FOR RELEVANT PAIRS \kappa_p^{(2,6)}(1)={(1+p^2+p^4)^2}\\ \kappa_p^{(3,10)}(1)=}{(p^{10}-1)^3} the 3, 10 function changes sign with p → 1/p

"Like every great river and every great sea, the moon belongs to none and belongs to all. It still holds the key to madness, still controls the tides that lap on shores everywhere, still guards the lovers who kiss in every land under no banner but the sky". E.B. White The New Yorker, July 26, 1969Where does the Moon come from?Scientist believe that our Moon formed out of a ‘giant impact’ that occurred between a Mars-sized planet and the early Earth, some 4.5 billion years ago. The Moon was then formed from the coalescence of the orbiting debris scattered during the impact.Recent results seem to confirm this scenario

MAIN We have the polynomial resultant Res(P(x),Q(x)). We may use a power n truncation of each of P(x) and Q(x), then have _n(P(x),Q(x))=(_n(P(x)),_n(Q(x))) for example _3(ax^5+bx^3+x+1)=bx^3+x+1 however, this now means we can insert an analytic function Taylor series in place of the polynomials and still use the results. For increasing orders of truncation we then generate a sequence a(n) where n is the level of truncation. Then we have _n(f(x),g(x))= a(n) for n > 0, giving the following results f(x) & g(x) & a(n)\\ {(x+1)} & {(x+1)^2} & A000272 \\ {(x+1)^2} & {(x+1)^3} & A098721 \\ {(x+1)^3} & {(x+1)^2} & 1,2,19,444,19525,...\\ {(x+1)} & {(x+1)^2} & A060073 \\ {(x+1)} & {(x+1)^2} & {2^n} \\ {(x+1)} & {(x+1)^m} & {2^m} \\ {(x+1)} & {(x+1)^2} & A081215 \\ & {x+1} & 1,3,9,65,7051,11515203,...\\ For other functions we see a(n)→c some constant, for example _n(\exp(x),{x+1})=e\\ _n(\exp(-x),{x+1})=e^{-1}\\ _n(\exp(x^2),{x+1})=e^{-1}\\ _n(\exp(-x^2),{x+1})=e\\ _n(\log(x),{x+1})=? \\ _n(\sin(x),{x+1})={\sin(1)}\\ _n(\cos(x),{x+1})={\cos(1)}\\ _n(\tan(x),{x+1})={\tan(1)}\\ the exponential of 1. The log example above converges very slowly, we can try a power law extrapolation.

MAIN We may define sequences in the following format, a(1) = ^2 f(i_1) \\ a(2) = ^2 ^{f(i_1)} f(i_2) \\ a(3) = ^2 ^{f(i_1)} ^{f(i_2)} f(i_3) \\ \cdots \\ a(n) = ^2 ^{f(i_1)} ^{f(i_2)}\cdots^{f(i_{n-1})} f(i_n) for some f. This can also be extended with some g a(n) = ^2 ^{f(i_1)} ^{f(i_2)}\cdots^{f(i_{n-1})} f(i_n) for example, it would appear that for f(x)=x + 2 and g(n)=n, we have a(n) = ^2 ^{f(i_1)} ^{f(i_2)}\cdots^{f(i_{n-1})} f(i_n) = {n+2}{n+3} and for for f(x)=x + 1 and g(n)=n a(n) = ^2 ^{f(i_1)} ^{f(i_2)}\cdots^{f(i_{n-1})} f(i_n) = (n+1)^2+1 To keep this as general as possible define the following \Sigma(a,b,c,d;n)= ^{a(n)} ^{c(i_1)} ^{c(i_2)}\cdots^{c(i_{n-1})} d(i_n) Then for example we have \Sigma(1,1,n+1,n;n)= {(n+2)!(n-1)!} which is A000245. LIST \Sigma(1,1,1,1;n)=1 \\ \Sigma(n,n,n,n;n)=n \\ \Sigma(n,n,n,f(n);n)=f(n) \\ \Sigma(n,n-1,n,n;n)=n^2 \\ \Sigma(n,n-2,n,n;n)={2} \\ \Sigma(1,n,n+1,n;n)=n \\ \Sigma(1,1,n,n;n)=1 \\ \Sigma(1,1,n+1,n;n)= {(n+2)!(n-1)!} \\ \Sigma(n+1,n+1,n+1,n+1;n) = 3,13,55,231,966,\cdots I have found a very special conjecture \Sigma(1,1,\pi(n+1)+1,n;n) = 3(2(n-1))+0^{n-1} which is true for the first 17 terms. Where π(x) is the prime counting function. ^{a(n)} ^{c(i_1)} ^{c(i_2)}\cdots^{c(i_{n-1})} i_n = 3(2(n-1))+0^{n-1} this most likely is due to the sub linear growth of pi(n). INTEGRALS We also may write a series of integrals I_n = ^{a(n)} ^{c(i_1)} ^{c(i_2)}\cdots^{c(i_{n-1})} d(i_n)\;di_n\;\cdots di_2 di_1 If we let. a(n)=1, b(n)=0, c(n)=n + 1, d(n)=n + 1, This appears to give A000272/(n + 1)!. If we let. a(n)=1, b(n)=0, c(n)=n + 1, d(n)=n, This appears to give A213326/(n + 1)!. If we let. a(n)=n, b(n)=0, c(n)=n + 1, d(n)=1, This appears to give A193678/n! If we let. a(n)=2n, b(n)=0, c(n)=n, d(n)=1, This appears to give A062971/n! If we let. a(n)=n, b(n)=0, c(n)=n, d(n)=1, This appears to give nn/n!

Aside from Bell’s inequality, entanglement and local real model have other aspects which are expected in experiments. Analysis on a) the physics concept of entanglement and b) precise interpretation of experiments shows that 1) In a reported loophole-free violation of Bell inequality, the transition of wave function from odd parity to even parity reveals that the experiment is performed on the spin of a pair of local real nitrogen vacancy (NV) centre. 2) The equivalence between rotating spin by θ and rotating measurement basis by −θ is not applicable in entanglement case, thus in long range entanglement setups for closing locality loophole, the operation of rotating spin followed by measurement puts the entanglement in question. 3) Fair sampling assumption arises when a finite sample is used to represent the entire population space, thus it is a basic requirement of statistical experiment, fair sampling loophole can not be closed.

MAIN Define \Upsilon_k(s)=^\infty U(1,k+1,n^s) where U(a, b, z) is a confluent hypergeometric function. This must be extended for integer k. But we have, \Upsilon_1(s)=\zeta(s)\\ \Upsilon_2(s)=\zeta(s)+\zeta(2s)\\ \Upsilon_3(s)=\zeta(s)+2\zeta(2s)+2\zeta(3s)\\ \cdots \\ \Upsilon_k(s)=\zeta(s)+(k-1)\zeta(2s)+(k-1)(k-2)\zeta(3s)+(k-1)(k-2)(k-3)\zeta(4s)+\cdots We then have that \Upsilon_k(\infty)=^{k-1} {i!} Essentialy we then write \zeta(s)=\Upsilon_k(s)\\ \zeta(s)=^\infty U(1,2,n^s) use the formal identity U(a,b,z)={\Gamma(a-b+1)}M(a,b,z) + {\Gamma(a)}z^{1-b}M(a-b+1,2-b,z) for Kummer hypergeometric functions M, however the gamma functions break for the integer b = 2. Can write U(a,b,z)={\Gamma(a)}\int_0^\infty e^{-zt}t^{a-1}(1+t)^{b-a-1}\;dt\\ \zeta(s)=^\infty \int_0^\infty e^{-n^st}\;dt\\ \zeta(s)=\int_0^\infty ^\infty e^{-n^st}\;dt so then we have some kind of function for some value of s p(t) = ^\infty e^{-n^st} which may or may not relate to the derivative of the zeta function. Does this show any interesting behaviour in the regions around zeta zeros? Is it even valid to make the above assumptions. The continuous equivalent of p is \pi(t) = \int_1^\lambda e^{-n^s t} \; dn = {s}\left( E_{{s}}(t)-\lambda E_{{s}}(\lambda^s t) \right) for λ → ∞, where En(x) is the exponential integral E_n(x) = \int_1^\infty }{t^n} \; dt then if we take the integral of π(t) over t we get Z(s) = \int_0^\infty \pi(t) \; dt = (-\lambda+\lambda^s)}{s-1} which in the limit λ → ∞ goes to .... Interestingly if we replace the sum in Υ with an integral, we get ^\infty U(1,k+1,n^s) \; dn = {s-1}, \;\; Re[s]>1 DIRECT EVALUATION OF p INTEGRAL We may directly evaluate the integral of a truncation of p. For integer s, we have n!\int_0^\infty ^n \exp(-nt)\;dt = (n) n!^2\int_0^\infty ^n \exp(-n^2t)\;dt = A001819 n!^3\int_0^\infty ^n \exp(-n^3t)\;dt = A066989 for general s, observing the harmonic numbers H_{n,s}=^n {k^s} which apparently (Mathword), are also H_{n,s}= n^{-r} + (n)}{\Gamma(r)} + \zeta(r) for certain odd values of r > 3, which then shows a connection to zeta, directly. If in this formula we vary s about a zero in the zeta function, and look at the absolute value of the formula, we see oscillations about another function that go to zero! Then we can say that the values s, that seem to be zeroes are those that locally minimise curvature in this function of n.

_Additional notes:_ Authors: - Paul Hockett, National Research Council of Canada, 100 Sussex Drive, Ottawa, K1A 0R6, Canada - Tim Ingleby, Department of Architecture and Built Environment, Northumbria University, Ellison Place, Newcastle upon Tyne, NE1 8ST, UK Links: - Online version: Authorea - Repository for videos and files, Figshare, DOI: 10.6084/m9.figshare.c.3470907 - Arxiv version (1610.04281) - Ongoing work: femtolab.ca

ABSTRACT I briefly show how a whole set of formula mimicking the Gauss-Salamin formula can be derived associated with singular values of the complete elliptic integral K(k²), (with elliptic modulus k). MAIN From the Gauss–Salamin formula for π, it can be easily shown that {2}=\left(1,{}\right)K\left({}\right) where K is a complete elliptic integral of the first kind with modulus $k={}$ and agm(x, y) is the arithmetic geometric mean function, which is the shared limit of a∞ or g∞ of the iterative formula a_0={2}(x+y)\\ g_0=\\ a_n={2}(a_{n-1}+g_{n-1})\\ g_n=g_{n-1}} Then if we define F(x)=\left(1,{}\right)K\left({}\right)={4}}{}{}\right)}{K\left(-1}{+1}\right)} we have F\left(2\right)={2} but if we search for more relations we also find that F\left({8}\left(4+3 \right)\right)=\pi\\ F\left(17+12\right)=F\left(\left(1+\right)^4\right)={4}\\ F\left(\left(1+\right)^2\right)={2}\\ F\left(2-\right)=\kappa-{2}i\\ F\left(1+{}\right)=\kappa where κ = 1.6999683849605364588... this gives the identity F\left(2-\right)=F\left(1+{}\right)-iF(2) we see that these identities are in fact independent of π as the values cancel from equation 3. We discover from these various identities about the complete elliptic integral of the first kind, }{}}\right)}{K\left(}-4}{}+4}\right)}=}}{1+{2}{2}}} and \right)}{K\left({}\right)}={4}(2+) and as K\left({}\right)=}{\Gamma(-{4})^2} we have the closed form K\left(3-2\right)=)\pi^{3/2}}{\Gamma\left(-{4}\right)^2}=1.58255... but by letting a=}{8} {}\right)}{K\left(-1}{+1}\right)}=}{1+} It would appear that ({2},{})={\Gamma(1/4)} this directly leads to K(2-3)=)\Gamma(1/4)}{8\Gamma(3/4)}

© 2016, Kevin J. Black. This work is licensed under a Creative Commons Attribution 4.0 International License.

_Publication history_ - Original document (Authorea) - arXiv 1702.00744 (Feb. 2017) - J. Chem. Phys. special issue “Developments and Applications of Velocity Mapped Imaging Techniques”, accepted (March 2017) - Data and analysis scripts (OSF), DOI: 10.17605/OSF.IO/RRFK3.