I just want make thing clear for myself. Others may have asked before in different ways. Does Yitang Zhang's paper prove that for any given length gap $g_n > N$ there is a prime $p_n$ for which there are infinitely many prime numbers $p_k > p_n$ which have $g_k = g_n$? Or, is it something else? If I do have this correctly, is p_n equal to the first prime of gap $N$? Does this mean that for any gap greater than g_n there is no "last one" only "last know use"?
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4$\begingroup$ You should give a proper citation to the paper you are interested in. The way your question is asked it's not clear if you're looking for anything more than the abstract of the paper, available here:annals.math.princeton.edu/2014/179-3/p07 $\endgroup$– Ryan BudneyCommented Apr 10, 2014 at 14:56
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2 Answers
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Yitang Zhang's paper proves that there is a gap less than or equal to 70 million that occurs infinitely often. More precisely, the paper proves that if you take 3.5 million integers that do not form a complete residue system modulo any integer greater than 1, then one of the pairwise distances among these integers occurs infinitely often as the difference of two distinct primes. We do not know a single value, written down as a concrete number like 2014, that provably occurs as a difference of two distinct primes infinitely often.
P.S. Due to the efforts of Maynard, Pintz, Tao and other members of the PolyMath8 group, the 70 million above has been improved to 252, and the 3.5 million above has been improved to 51.
Added. The current record is now 50 for the size of the tuple in the method, yielding the bound 246 for a gap size that occurs infinitely often among the primes. The relevant papers by the Polymath8 group have appeared:
New equidistribution estimates of Zhang type, Algebra & Number Theory 8 (2014), 2067-2199.
Variants of the Selberg sieve, and bounded intervals containing many primes, Res. Math. Sci. 1 (2014), no. 12, 83 pp.
answered Apr 10, 2014 at 15:00
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$\begingroup$ So using the improved numbers of 252 and 51, you are saying that any number from 2 to 51 that do no complete a residue system modulo, which mean that the dividing number is greater than $\ceilimg{\sqrt{51}} = 8$ has a pair with another prime < 252? $\endgroup$ Commented Apr 10, 2014 at 18:55
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$\begingroup$ The difference of the pair $k$ happens for an infinite number of other pairs > 252 with a difference of $k$? $\endgroup$ Commented Apr 10, 2014 at 19:07
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$\begingroup$ @John: To your first comment, I am saying that if you take 51 integers that do not form a complete residue system modulo any integer greater than 1, then one of the pairwise distances among these integers occurs infinitely often as the difference of two distinct primes. In particular, there is a gap less than or equal to 252 that occurs infinitely often. $\endgroup$ Commented Apr 10, 2014 at 20:08
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$\begingroup$ @John: To your second comment, it was proved by János Pintz that there is a positive integer $C$ such that among any $C$ consecutive positive integers there is a number that occurs infinitely often as a difference of two primes. This follows from the quoted result of Zhang. On the other hand, we only know the existence of such a $C$, and in fact the quoted results of Zhang/Maynard/Tao/PolyMath8 do not allow to specify this $C$. $\endgroup$ Commented Apr 10, 2014 at 20:14
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2$\begingroup$ My understanding is that $H_1$ has been lowered from 252 to 246: see arxiv.org/abs/1409.8361. $\endgroup$ Commented Feb 15, 2015 at 14:43
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Zhang's paper proves that there are infinitely many prime pairs whose difference is less than 70,000,000 (this has since been reduced). What that means is that there is some $k \in \{2, 4, 6, \dots, 70,000,000\}$ such that, for infinitely many $n$, $p_{n+1} - p_n = k$. It doesn't tell us where these pairs are and it doesn't tell us whether there are any $k_0 > k$ such that there are infinitely many prime pairs with gap $k_0$ (although Pintz has proved that this does, in fact, happen (see the paper "On the difference of primes" by Janos Pintz))
I hope this answers the question.
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$\begingroup$ OK, so Zhang proof does not touch or cover the values greater than 70 million, but the Pintz paper does. I guess with the current reduction to 252 it is the Pintz paper doing the work for 252 to 70 million. $\endgroup$ Commented Apr 10, 2014 at 16:30
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$\begingroup$ Oops, I referenced the wrong Pintz paper, it should have been "Polignac numbers, conjectures of Erdős on gaps between primes, arithmetic progressions in primes, and the bounded gap conjecture" which is on the arXiv, but that says that there is an ineffective constant $C$ such that the interval $[M, M+C]$ always contains some number k with the property that primes will infinitely often differ by precisely k. So we don't know if there is some Polignac number between 252 and 70 million but there is certainly a Polignac number above 252. $\endgroup$– StijnCommented Apr 10, 2014 at 17:28
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1$\begingroup$ @Stijn: Actually there are weaker but effective versions of Pintz's result. So probably we know that there is a number between 252 and 70 million that occurs infinitely often as a difference of two primes. For that, all we need is an admissible 51-tuple whose pairwise distances fall between 252 and 70 million. $\endgroup$ Commented Apr 10, 2014 at 20:18
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$\begingroup$ That makes sense. If we insist on a lower bound on the distance between the elements of the admissible sets, say that $h_{n+1} - h_n \geq k$, are there any known general results for the diameter of the minimal admissible sets, that you're aware of? Something asymptotic, maybe? $\endgroup$– StijnCommented Apr 10, 2014 at 21:58
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$\begingroup$ 251 is the 54th prime, so this 51-tuple will have most of the primes included. This seems similar to A165959 at the OEIS. This sequence does not fix k, but fixes the number of primes in the interval $[p_{i-n}, p_k]$ to n and allows one to find the next prime $p_i < 2*p_{i-n}$. What I am really wondering is if there is a way to combine these two ideas? It would be really cool if someone could prove that there are an infinite number of 3 in the sequence because this would prove the twin prime conjecture. The prime $p_i$ is then next prime after $R_n$ and $p_{i-n}$ is the next one after $R_n/2$. $\endgroup$ Commented Apr 10, 2014 at 22:01