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	<title>Comments on: (Research Thread III) Determinstic way to find primes</title>
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	<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/</link>
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	<item>
		<title>By: Michael Anthony</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-1777</link>
		<dc:creator><![CDATA[Michael Anthony]]></dc:creator>
		<pubDate>Sun, 14 Mar 2010 13:10:00 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-1777</guid>
		<description><![CDATA[Here is the simple formula you are looking for:  
arctan(tan(Pi*bernoulli(2*n)*GAMMA(2*n+1)))*(2*n+1)/Pi= 1 for Prime or zero for non-prime!
In log form: 
i*(n+1/2)*log((1-tan(Pi*bernoulli(2*n)*GAMMA(2*n+1)))*(2*n+1)/Pi)/(1+tan(Pi*bernoulli(2*n)*GAMMA(2*n+1)))*(2*n+1)/Pi)  )= 1 for Prime or zero for non-prime!]]></description>
		<content:encoded><![CDATA[<p>Here is the simple formula you are looking for:<br />
arctan(tan(Pi*bernoulli(2*n)*GAMMA(2*n+1)))*(2*n+1)/Pi= 1 for Prime or zero for non-prime!<br />
In log form:<br />
i*(n+1/2)*log((1-tan(Pi*bernoulli(2*n)*GAMMA(2*n+1)))*(2*n+1)/Pi)/(1+tan(Pi*bernoulli(2*n)*GAMMA(2*n+1)))*(2*n+1)/Pi)  )= 1 for Prime or zero for non-prime!</p>
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		<title>By: Michael Anthony</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-1776</link>
		<dc:creator><![CDATA[Michael Anthony]]></dc:creator>
		<pubDate>Sun, 14 Mar 2010 12:50:32 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-1776</guid>
		<description><![CDATA[Please correct typo in Gamma...Gamma(2*n)&gt;&gt; Gamma(2*n+1), In my paper primes and the Riemann Hypothesis, I show that the primes obey, the relation
frac(theta)=arctan(tan(theta)). You see the primes obey this rule for 
theta = Pi*bernoulli(2*n)*Gamma(2*n+1), only if 2*n+1 is a prime. The function theta is a consequence of the Riemann Hypothesis: s=1/2+i*tan(theta)/2.
 In fact, I show that:
Zeta(s) = sum(-2^nu*bernoulli(nu)*GAMMA(1-s)/(factorial(nu)*GAMMA(-s-nu+2)), nu = 0 .. infinity); and that when the function vanishes, s/(1-s) = (1/2+(1/2*I)*tan(theta))/(1/2-(1/2*I)*tan(theta))= exp(theta) iff RH is true. The connection with primes is obvious, since when s=-2*n and 2*n+1 is a prime, the function takes an exponential form, and von-Staudt Clausen makes it clear the primes can be counted.]]></description>
		<content:encoded><![CDATA[<p>Please correct typo in Gamma&#8230;Gamma(2*n)&gt;&gt; Gamma(2*n+1), In my paper primes and the Riemann Hypothesis, I show that the primes obey, the relation<br />
frac(theta)=arctan(tan(theta)). You see the primes obey this rule for<br />
theta = Pi*bernoulli(2*n)*Gamma(2*n+1), only if 2*n+1 is a prime. The function theta is a consequence of the Riemann Hypothesis: s=1/2+i*tan(theta)/2.<br />
 In fact, I show that:<br />
Zeta(s) = sum(-2^nu*bernoulli(nu)*GAMMA(1-s)/(factorial(nu)*GAMMA(-s-nu+2)), nu = 0 .. infinity); and that when the function vanishes, s/(1-s) = (1/2+(1/2*I)*tan(theta))/(1/2-(1/2*I)*tan(theta))= exp(theta) iff RH is true. The connection with primes is obvious, since when s=-2*n and 2*n+1 is a prime, the function takes an exponential form, and von-Staudt Clausen makes it clear the primes can be counted.</p>
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		<title>By: Michael Anthony</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-1775</link>
		<dc:creator><![CDATA[Michael Anthony]]></dc:creator>
		<pubDate>Sun, 14 Mar 2010 12:32:52 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-1775</guid>
		<description><![CDATA[Note also that the above prime counting formula is also given in log form as:

N= sum(i(n+1/2)*log( (1-i*tan(Pi*bernoulli(2*n)*gamma(2*n))/(1-i*tan(Pi*Bernoulli(2*n)*GAMMA(2*n+1))/Pi , n=3..m); which for large n reduces to the Prime Number theorem.]]></description>
		<content:encoded><![CDATA[<p>Note also that the above prime counting formula is also given in log form as:</p>
<p>N= sum(i(n+1/2)*log( (1-i*tan(Pi*bernoulli(2*n)*gamma(2*n))/(1-i*tan(Pi*Bernoulli(2*n)*GAMMA(2*n+1))/Pi , n=3..m); which for large n reduces to the Prime Number theorem.</p>
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		<title>By: Michael Anthony</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-1774</link>
		<dc:creator><![CDATA[Michael Anthony]]></dc:creator>
		<pubDate>Sun, 14 Mar 2010 12:26:39 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-1774</guid>
		<description><![CDATA[I submitted a proposal that the prime counting function is given by:
N=sum( arctan(tan(Pi*Bernoulli(2*n)*GAMMA(2*n+1))) *(2*n+1))/Pi, n=3..M), M=m/2 for m even and M =m/2-1/2 for m odd. This formula yields the exact number of primes less than m. It was derived from the Riemann hypothesis and a a solution that I submitted that was never looked at. If this formula is right, then I have solve Riemann&#039;s hypothesis! I invite a review of my paper.]]></description>
		<content:encoded><![CDATA[<p>I submitted a proposal that the prime counting function is given by:<br />
N=sum( arctan(tan(Pi*Bernoulli(2*n)*GAMMA(2*n+1))) *(2*n+1))/Pi, n=3..M), M=m/2 for m even and M =m/2-1/2 for m odd. This formula yields the exact number of primes less than m. It was derived from the Riemann hypothesis and a a solution that I submitted that was never looked at. If this formula is right, then I have solve Riemann&#8217;s hypothesis! I invite a review of my paper.</p>
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		<title>By: Gil Kalai</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-716</link>
		<dc:creator><![CDATA[Gil Kalai]]></dc:creator>
		<pubDate>Sat, 29 Aug 2009 09:01:48 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-716</guid>
		<description><![CDATA[This is interesting. I suppose the conjecture would be that a random k-digit number is a young twin prime with probability k^{-2}; but this just like Cramer&#039;s conjecture are well well beyond reach.

So for integer programming in bounded dimension when the random subset is of density 1/2, does your argument gives average polynomial running time or even O(poly) running time almost surely. And what precisely the random sampling statement for primes you would need to extend it?

I suppose the next problem to play with is Frobenius coin problem. Ravi Kannan proved that there is a polynomial algorithm to determine the largest amount you canot pay with relatively prime k coins. We want to say that we can also determine the largest integer in a random subset of integers (or the largest prime) that we cannot pay. This does not look automatic from the problem but maybe follows from Kannan&#039;s algorithm.]]></description>
		<content:encoded><![CDATA[<p>This is interesting. I suppose the conjecture would be that a random k-digit number is a young twin prime with probability k^{-2}; but this just like Cramer&#8217;s conjecture are well well beyond reach.</p>
<p>So for integer programming in bounded dimension when the random subset is of density 1/2, does your argument gives average polynomial running time or even O(poly) running time almost surely. And what precisely the random sampling statement for primes you would need to extend it?</p>
<p>I suppose the next problem to play with is Frobenius coin problem. Ravi Kannan proved that there is a polynomial algorithm to determine the largest amount you canot pay with relatively prime k coins. We want to say that we can also determine the largest integer in a random subset of integers (or the largest prime) that we cannot pay. This does not look automatic from the problem but maybe follows from Kannan&#8217;s algorithm.</p>
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		<title>By: Nicolai Hähnle</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-709</link>
		<dc:creator><![CDATA[Nicolai Hähnle]]></dc:creator>
		<pubDate>Fri, 28 Aug 2009 13:33:48 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-709</guid>
		<description><![CDATA[I believe you can modify Lenstra&#039;s algorithm for integer linear programming to get a probabilistic algorithm for &quot;random subset of the integers linear programming&quot; by randomly sampling the lower-dimensional slices in the recursive step. The density of the random subset will clearly affect the required number of samples and thus the running time.

Applied to prime numbers, the algorithm&#039;s strategy for finding primes in an interval would just reduce to randomly sampling integers in the interval and checking primality. Same goes for what the algorithm would do to find twin primes. Is anything is known about random sampling and twin primes?]]></description>
		<content:encoded><![CDATA[<p>I believe you can modify Lenstra&#8217;s algorithm for integer linear programming to get a probabilistic algorithm for &#8220;random subset of the integers linear programming&#8221; by randomly sampling the lower-dimensional slices in the recursive step. The density of the random subset will clearly affect the required number of samples and thus the running time.</p>
<p>Applied to prime numbers, the algorithm&#8217;s strategy for finding primes in an interval would just reduce to randomly sampling integers in the interval and checking primality. Same goes for what the algorithm would do to find twin primes. Is anything is known about random sampling and twin primes?</p>
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		<title>By: (Research Thread IV) Determinstic way to find primes &#171; The polymath blog</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-705</link>
		<dc:creator><![CDATA[(Research Thread IV) Determinstic way to find primes &#171; The polymath blog]]></dc:creator>
		<pubDate>Fri, 28 Aug 2009 01:44:56 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-705</guid>
		<description><![CDATA[[...] previous research thread for the &#8220;finding primes&#8221; project is now getting quite full, so I am opening up a fresh [...]]]></description>
		<content:encoded><![CDATA[<p>[...] previous research thread for the &#8220;finding primes&#8221; project is now getting quite full, so I am opening up a fresh [...]</p>
]]></content:encoded>
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		<title>By: Nathan McKenzie</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-703</link>
		<dc:creator><![CDATA[Nathan McKenzie]]></dc:creator>
		<pubDate>Thu, 27 Aug 2009 10:02:36 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-703</guid>
		<description><![CDATA[Wow - amazing synchronicity.  I stumbled across this page trying to find if anyone, anywhere, knew of faster ways to count sums of divisor functions, and it turns out that you folks look like you&#039;re not only talking about that, but for very similar reasons to me.

A few years ago, I wrote some code in C to calculate the count of primes &lt; n by calculating the various sum of divisor functions and then applying Linnik&#039;s identity.  I tried to do everything I could to speed it up, but ultimately I never could get it fast enough to be competitive.  I think the current version runs in something like O(n^4/5) time (at least from eyeballing it) and O(epsilon) space, but because the various loops for the sums of divisor functions are particularly sped up by aggressive use of wheels, I think its constant time factors are pretty good - or at least for an algorithm with O(epsilon) for memory.  It calculates 10^12 in about a second on my reasonably decent laptop with a wheel culling out primes up to 19.

The code is here

http://www.icecreambreakfast.com/math/mckenzie_math.pdf     (the code is at the bottom of this pdf, along with descriptions of what I&#039;m doing above it)

if anyone wants to look over it or play with it.  I keep trying to find some way to rearrange things in the algorithm to find some clever way to cache... something... to massively speed things up, but so far I&#039;ve had no real luck.  I&#039;ve tried a lot of things, but I clearly haven&#039;t stumbled on the right idea yet.  I could list some of those if anyone is interested.

At any rate, if any of you feel like taking a look at the code and seeing if you have any brilliant ideas, I would be overjoyed.  While I wasn&#039;t looking, this whole topic (counting sums of divisors quickly) has turned into my Ahab-style White Whale.]]></description>
		<content:encoded><![CDATA[<p>Wow &#8211; amazing synchronicity.  I stumbled across this page trying to find if anyone, anywhere, knew of faster ways to count sums of divisor functions, and it turns out that you folks look like you&#8217;re not only talking about that, but for very similar reasons to me.</p>
<p>A few years ago, I wrote some code in C to calculate the count of primes &lt; n by calculating the various sum of divisor functions and then applying Linnik&#039;s identity.  I tried to do everything I could to speed it up, but ultimately I never could get it fast enough to be competitive.  I think the current version runs in something like O(n^4/5) time (at least from eyeballing it) and O(epsilon) space, but because the various loops for the sums of divisor functions are particularly sped up by aggressive use of wheels, I think its constant time factors are pretty good &#8211; or at least for an algorithm with O(epsilon) for memory.  It calculates 10^12 in about a second on my reasonably decent laptop with a wheel culling out primes up to 19.</p>
<p>The code is here</p>
<p><a href="http://www.icecreambreakfast.com/math/mckenzie_math.pdf" rel="nofollow">http://www.icecreambreakfast.com/math/mckenzie_math.pdf</a>     (the code is at the bottom of this pdf, along with descriptions of what I&#039;m doing above it)</p>
<p>if anyone wants to look over it or play with it.  I keep trying to find some way to rearrange things in the algorithm to find some clever way to cache&#8230; something&#8230; to massively speed things up, but so far I&#039;ve had no real luck.  I&#039;ve tried a lot of things, but I clearly haven&#039;t stumbled on the right idea yet.  I could list some of those if anyone is interested.</p>
<p>At any rate, if any of you feel like taking a look at the code and seeing if you have any brilliant ideas, I would be overjoyed.  While I wasn&#039;t looking, this whole topic (counting sums of divisors quickly) has turned into my Ahab-style White Whale.</p>
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		<title>By: Kristal Cantwell</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-698</link>
		<dc:creator><![CDATA[Kristal Cantwell]]></dc:creator>
		<pubDate>Wed, 26 Aug 2009 21:28:41 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-698</guid>
		<description><![CDATA[I think I can solve the toy problem in o(n^1/2) which is less than n^1/3 but doesn&#039;t need a factoring oracle. It uses at least n^1/2 in memory so while it breaks the barrier in terms of time it pays the price in storage. It is based on the sieve of Atkins:

http://en.wikipedia.org/wiki/Sieve_of_Atkin

This computes the primes up to n in O(n/log log n)
time. The idea is as follows we first use this sieve to find the number of square free numbers up to n^1/4  and save the value for each value less than n^1/4 then for each of these numbers i we find the number of primes in the region  n/p^2 is k plus a fractional value then for each prime bigger than n^1/4 we look at those primes for which n/p^2 =k rounded down has the same value then we used the data from the algorithm for lower values to see how man of the values one through k are square free then we have an accurate count of the number of numbers removed by the primes and the whole process should be o(n^1/2) although the memory used is at least O(n^1/2).]]></description>
		<content:encoded><![CDATA[<p>I think I can solve the toy problem in o(n^1/2) which is less than n^1/3 but doesn&#8217;t need a factoring oracle. It uses at least n^1/2 in memory so while it breaks the barrier in terms of time it pays the price in storage. It is based on the sieve of Atkins:</p>
<p><a href="http://en.wikipedia.org/wiki/Sieve_of_Atkin" rel="nofollow">http://en.wikipedia.org/wiki/Sieve_of_Atkin</a></p>
<p>This computes the primes up to n in O(n/log log n)<br />
time. The idea is as follows we first use this sieve to find the number of square free numbers up to n^1/4  and save the value for each value less than n^1/4 then for each of these numbers i we find the number of primes in the region  n/p^2 is k plus a fractional value then for each prime bigger than n^1/4 we look at those primes for which n/p^2 =k rounded down has the same value then we used the data from the algorithm for lower values to see how man of the values one through k are square free then we have an accurate count of the number of numbers removed by the primes and the whole process should be o(n^1/2) although the memory used is at least O(n^1/2).</p>
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		<title>By: Ernie Croot</title>
		<link>http://polymathprojects.org/2009/08/13/research-thread-iii-determinstic-way-to-find-primes/#comment-680</link>
		<dc:creator><![CDATA[Ernie Croot]]></dc:creator>
		<pubDate>Mon, 24 Aug 2009 14:56:25 +0000</pubDate>
		<guid isPermaLink="false">http://polymathprojects.org/?p=111#comment-680</guid>
		<description><![CDATA[Actually, I see that there is an issue with the fact that $latex \sum_{d \geq 2} (1/d^2)^{0.49}$ doesn&#039;t converge, meaning that one needs to be a little more careful in order to get a $latex x^{0.49}$ algorithm to find the parity of $latex \pi(x)$.  Surely it can be done somehow...]]></description>
		<content:encoded><![CDATA[<p>Actually, I see that there is an issue with the fact that <img src='http://s0.wp.com/latex.php?latex=%5Csum_%7Bd+%5Cgeq+2%7D+%281%2Fd%5E2%29%5E%7B0.49%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;sum_{d &#92;geq 2} (1/d^2)^{0.49}' title='&#92;sum_{d &#92;geq 2} (1/d^2)^{0.49}' class='latex' /> doesn&#8217;t converge, meaning that one needs to be a little more careful in order to get a <img src='http://s0.wp.com/latex.php?latex=x%5E%7B0.49%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x^{0.49}' title='x^{0.49}' class='latex' /> algorithm to find the parity of <img src='http://s0.wp.com/latex.php?latex=%5Cpi%28x%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='&#92;pi(x)' title='&#92;pi(x)' class='latex' />.  Surely it can be done somehow&#8230;</p>
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