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Shortly before reaching a power of 2, the algorithm behind the Collatz conjecture reaches an odd integer k so that the last few integers reached are k, 3*k+1 = 2^i, 2^(i-1),..., 1

Let n = 5*k + 2. I conjecture that the last few powers of (n+1)/2 mod n are precisely k, 3*k+1 = 2^i, 2^(i-1),..., 1
Definition: An exit point in relation to the Collatz conjecture is the last odd integer reached before reaching a power of 2. Example: Let s = 10 then the Collatz exit point is 5 since 10/2 = 5 and 3*5 + 1 = 16, which is a po...
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Jürgen Weiß's profile photo
 
I don't a lot , but you must handle this conjecture like you have in the ' quersumme ' of numbers .there you multiply 3 a lot of times .for the ' quersumme' of the result you you always get 9 .conjecture this ! It's informatically ......
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vzn

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math 2016 celebration: pythagorean triples problem solved by supercomputer running Satisfiability problem code ie empirical attack! other big math news eg mochizuki proof, math profiles, student/edu highlights, essays by mathematicians, reactions to the ramanujan movie, other misc math advances.
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Deep learning has been making it possible for powerful machines to approximate and imitate abilities and techniques once thought to be uniquely human. Mathematicians have struggled to explain how they work so well and may now get some answers by looking outside mathematics and into the nature of the universe.
 
Artificial Intelligence
Deep learning can be understood as modeling high-level abstractions using data and a set of algorithms, with a deep graph and multiple processing layers of linear and non-linear equations. While it is largely a mathematical tool, the things that deep neural networks can do have surprised mathematicians. How can networks arranged in layers be quick to perform human tasks like face and object recognition if it has to go through layers of computations? This has perplexed mathematicians for sometime.

However, what mathematics cannot makes sense of, physics explains simply.

According to Henry Lin of Harvard University and Max Tegmark at MIT, the answer lies in understanding the nature of the universe, for which physics is the best tool.

Mathematically, object recognition goes through multiple possibilities. As explained by MIT Technology Review, to determine whether a megabit greyscale image shows a cat or a dog, it has to go through 2561000000 possible images. Not only that, it has to compute whether each shows a cat or a dog. Neural networks do this, however, with ease.
Neural networks approximate complex mathematical functions with simpler ones. In theory, this mean that neural networks have to go through a magnitude of mathematical functions more than what is possible for them to approximate. Lin and Tegmark believe that it doesn’t have to go through all the probability of mathematical functions but just a tiny subset of them.

And this is because the universe, in all its complexity, is governed by a tiny subset of all possible functions. Mathematically, the laws of physics can be described using functions with a set of simple properties. Neural networks approximate how nature works.
- read more:
http://futurism.com/the-key-to-understanding-ai-may-be-buried-in-the-laws-of-physics/









Deep learning has been making it possible for powerful machines to approximate and imitate abilities and techniques once thought to be uniquely human. Mathematicians have struggled to explain how they work so well and may now get some answers by looking outside mathematics and into the nature of the universe.
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Bruce Mincks's profile photo
 
Sure do hope that look "outside mathematics and into the nature of the universe" reveals new "abilities and techniques" that are intelligent rather than artificial.
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Two integer sequences that I recently submitted to the Online Encyclopedia of Integer Sequences have been approved:

1.The sequence of primes p such that (p+1) reaches p under the Collatz conjecture has only 9 terms and if a 10th one exists, then it must be greater than 10^9.

2.The sequence of products of two primes p and q such that (p*q + 1) reaches p or q under the Collatz conjecture is a lot denser. 
Two integer sequences that I recently submitted to the OEIS have been approved. The first sequence is the sequence of Collatz primes(A276260) , which I talked about here. http://oeis.org/A276260 The mesmerizing thing about th...
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A Collatz product is the product n of 2 odd primes p and q such that n+1 reaches p or q when put through the algorithm behind the Collatz conjecture.The first few such products of odd primes are:

25,35,55,65,77,85,95,115,133,143,145,155,161,185,203,205,209,...

Unlike the sequence of Collatz primes, this sequence looks way denser.
Definition: A Collatz product is the product n of 2 odd primes p and q such that n+1 reaches p or q when put through the algorithm behind the Collatz conjecture. As I already mentioned here, such products seem to appear much...
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A very interesting proposal to link the eigenvalues of an operator with the zeros of an analytic function through a Mellin transform to make the Hilbert Polya conjecture can be applied to the Riemann hypothesis
https://www.researchgate.net/publication/260591754_The_Hilbert-Polya_conjecture_a_generalization_proposal_and_an_interesting_example
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Kevin OConnor's profile photo
 
Hi Manuel,
Your obviously busy but wonder your take on new '5th ' controlling force binding the Unified Field and Quantum theories. Since Higgs first found there now seem to be variations on the particle suggested by findings at CERN2 where quantum gravity sits in theory easier .
Am out of my depth on this but find it fascinating.
If you not into this - don't wish to take your time up.
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What happens when the algorithm behind the Collatz Conjecture is applied to ((n+1)/2)^2 mod n where n is a product of distinct odd primes?
In my previous blog entry I talked about how The Collatz Conjecture might have something to do with generating powers of (n+1)/2 mod n, especially when n is the product of distinct primes. I conjecture that the algorithm be...
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Varun Prasad

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Does anyone know of ways to partition sets (of even cardinality) into subsets of odd cardinality?
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Babak Farhang's profile photo
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+Babak Farhang​ my solution is only for partitions with subsets of the same size (cardinality). Maybe it's a start..
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Can topological arguments be used to show that the set of non negative reals minus the set of possible degrees of L-functions from the Selberg class has infinitely many connected components?
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Sylvain JULIEN's profile photoBarrank Obama's profile photo
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+Sylvain JULIEN we just need that set to be not dense in R, then for our purpose we can assimilate it N even if it has accumulation points (also infinitely many)
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At 20 years of age, Stefan Buijsman is probably the youngest in Sweden ever to complete a Ph.D.
Whether mathematics is real, or a story we agreed upon, has long been debated by philosophers. A new dissertation from Stockholm University shows that philosophers failed to include non-experts in the theories. Stefan Buijsman recently defended his thesis in philosophy of mathematics, as Sweden's youngest Ph.D ever.
read more and video here:
http://www.su.se/english/research/research-news/youngest-ph-d-wants-to-understand-the-mathematics-of-non-experts-1.279473
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julio zuben's profile photoIwuchukwu Obieze's profile photo
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does he want it becus they're better than him? i'm not saying i'm proud 'o'r anything. if anyone has a job in sweden with research, lemmeno!!!
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kamel Aouadi

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The theory can be considered a form of Pythagoreanism or Platonism in that it posits the existence of mathematical entities; a form of mathematical monism in that it denies that anything exists except mathematical...
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Bruce Mincks's profile photonytom4info's profile photo
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Mathematics is a tool....a theory is an opinion! ;)
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Equivalence to the original problem statement XVI Hilbert. One step to solve the problem 16 by infinite algebraic equations of infinite variables (Fourier series). Exhibited in Russia on the 70th anniversary of V. I. Arnold. https://www.researchgate.net/publication/260163204_An_equivalency_to_the_second_part_of_Hilberts_16th_problem?ev=prf_pub 
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Article: UNEXPECTED BIASES IN THE DISTRIBUTION OF CONSECUTIVE PRIMES : http://arxiv.org/pdf/1603.03720v4.pdf (This paper proposes a
conjectural explanation for this phenomenon, based on the Hardy-Littlewood conjectures. The conjectures are then compared to numerical data, and the observed fit is very good.)
 
A clear rule determines exactly what makes a prime: it’s a whole number that can’t be exactly divided by anything except 1 and itself. But there’s no discernable pattern in the occurrence of the primes. Beyond the obvious — after the numbers 2 and 5, primes can’t be even or end in 5 — there seems to be little structure that can help to predict where the next prime will occur.

As a result, number theorists find it useful to treat the primes as a ‘pseudorandom’ sequence, as if it were created by a random-number generator.
- Article pdf: UNEXPECTED BIASES IN THE DISTRIBUTION OF CONSECUTIVE PRIMES : http://arxiv.org/pdf/1603.03720v4.pdf
Maths whizz solves a master’s riddle

But if the sequence were truly random, then a prime with 1 as its last digit should be followed by another prime ending in 1 one-quarter of the time. That’s because after the number 5, there are only four possibilities — 1, 3, 7 and 9 — for prime last digits. And these are, on average, equally represented among all primes, according to a theorem proved around the end of the nineteenth century, one of the results that underpin much of our understanding of the distribution of prime numbers. (Another is the prime number theorem, which quantifies how much rarer the primes become as numbers get larger.)

Instead, Lemke Oliver and Soundararajan saw that in the first billion primes, a 1 is followed by a 1 about 18% of the time, by a 3 or a 7 each 30% of the time, and by a 9 22% of the time. They found similar results when they started with primes that ended in 3, 7 or 9: variation, but with repeated last digits the least common. The bias persists but slowly decreases as numbers get larger.
The k-tuple conjecture

The mathematicians were able to show that the pattern they saw holds true for all primes, if a widely accepted but unproven statement called the Hardy–Littlewood k-tuple conjecture is correct. This describes the distributions of pairs, triples and larger prime clusters more precisely than the basic assumption that the primes are evenly distributed.

The idea behind it is that there are some configurations of primes that can’t occur, and that this makes other clusters more likely. For example, consecutive numbers cannot both be prime — one of them is always an even number. So if the number n is prime, it is slightly more likely that n + 2 will be prime than random chance would suggest. The k-tuple conjecture quantifies this observation in a general statement that applies to all kinds of prime clusters. And by playing with the conjecture, the researchers show how it implies that repeated final digits are rarer than chance would suggest.
- read more : http://www.nature.com/news/peculiar-pattern-found-in-random-prime-numbers-1.19550?WT.mc_id=GOP_NA_1603_NEWSPRIMENUMBERS_PORTFOLIO






Last digits of nearby primes have ‘anti-sameness’ bias.
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Bruce Mincks's profile photo
 
When I use the "bc" program in Linux, where it's easy to get decimal forms of irrational numbers at any degree of accuracy, there seems to be a pattern between the remainders from the next integer and the rate of increase for the integers under the radical. But I can't really identify it. Start with the identity 2/sqrt3) = 2/3*sqrt(3) and see how sqrt(2) figures into other ratios. Then recognize that sqrt(5) can either be a hypotenuse for the {3,4,5} triangle or the diagonal of a 2 x 1 rectangle to see the difference. So if 2 = sqrt(4) the progression seems to reflect the divergence in formal terms.

I was trying to compare these values with the ratios (a-b)/(b-c) = a/c and (a-b)/(b-c) = b as expressions for the harmonic and proportional means of numbers where a > b > c.
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Una formulación para avanzar hacia la solución de la conjetura de Goldbach. En estas notas aparece un bello lema que formulé. (el 3) https://www.researchgate.net/publication/260035451_la_conjetura_de_Goldbach_parte_1
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Un ejemplo de como los eigenvalores de un operador son los ceros de una función analítica kernel de las eigenfunciones. La conjetura de Hilbert-Polya para solucionar la hipótesis de Riemann. https://www.researchgate.net/publication/255908850_Operador_asociado_al_reciproco_de_gamma_Un_ejemplo_de_la_conjetura_de_Hilbert-Polya?ev=prf_pub
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Western counting ends at #Sextillion or upto 21 zeroes... While Indian counting goes well ahead upto 28 zeroes or Maha Ashohini...
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KING SOBIESKI's profile photoAntonio Noack's profile photo
16 comments
 
Just try Cinicinicinicini...cinillions :P
(I hope it was written like this buth each cini should be another 3000 zeros)
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A Collatz prime is an odd prime p such that p+1 reaches p when put through the algorithm behind the Collatz conjecture.
Although I found a counter example to the last claim from my previous Collatz adventure, which means that it is false in general, I decided to look for which products of primes it is true. It turns out that from all products ...
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Snail Erato's profile photo
 
I found 2 bigger Collatz primes after 251.
That gives the begining of the sequence :
5,13, 17, 53, 61, 107, 251, 283, 1367 ...
and if a next one exists it must be bigger than 67867967 !
If you do not restrict to primes and search for any numbers n so that n+1 reach n using Collatz algorithm the sequence has more numbers ...
The begining of this sequence is:
2,4,5,8,10,13,16,17,38,40,46,53,56,58,61,70,80,88,106,107,160,251,283,377,638,650,958,976,1367,1438,1822,2158,2429,2734,3238,4102,4616,4858,6154,7288,9232 ..
And, If there is another one it must be bigger than 1100000 !
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I recently stumbled onto the Collatz Conjecture, which states that any n iterated through the algorithm below will eventually be reduced to 1: Given an integer n, then either divide n by 2 if n is even or multiply n by 3 and ...
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Happy Powers of 2 Day!  The following algorithm generates powers of 2 mod n backwards.
In a previous post I described an algorithm for generating powers of 2 mod n when n is an odd integer. It required the generation of an array of multiples of 2 mod n that relied on consecutively adding 2 to an initial integer...
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Nearly four years after Shinichi Mochizuki unveiled an imposing set of papers that could revolutionize the theory of numbers, other mathematicians have only made modest progress in understanding his work.
Conference on Shinichi Mochizuki’s work inspires cautious optimism.
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