A group of seasoned mathematicians including Terence Tao are working to improve on Yitang Zhang’s exciting recent proof that there are infinitely many pairs of consecutive primes separated by no more than 70 million. A glance at the polymath8 wiki shows that the gap has now been shrunk to a measly 250,000. Although this is still nowhere near the gap of two required by the twin primes conjecture, one fact stands out to me as a junior faculty. As Mario Livio recently stated in his new book “The road to triumph [is] paved with blunders.” And on the path to furthering their (and our) knowledge, those contributing to polymath8 are recording all progress, including mistakes! It’s exciting to see research happening in real time, and even as I wrote this, I noticed new World Records being established for the smallest bound on the gap. The current record-holder is MIT Computational Number Theorist Andrew Sutherland.

Terrence Tao’s June 3^{rd} post gives a great introduction to recent events as he clearly lays out the definition of H admissible k-tuples of integers and how they relate to the twin primes conjecture. A k-tuple of integers (a list of k distinct integers) is admissible if it avoids at least one residue class modp for every prime p. These are the k-tuples that have a fighting chance at possessing the desirable characteristic that infinitely many translates consist exclusively of primes. For instance, an example of a NON-admissible 3-tuple is {-2, 0, 2} since each residue class mod 3 is represented. Notice that every translate of {-2,0,2} must contain a multiple of 3, and therefore there are only finitely many translates containing only primes. Filtering all k-tuples to find these special ones is accomplished by creating a “sieve”, thus a series of posts on sieve theory can be found on Tao’s blog and a quick encapsulation of many sieves can be found on another polymath wiki.

The famous Hardy-Littlewood conjecture asserts that all H-admissible k-tuples have infinitely many translates consisting exclusively of primes. But this over-arching theorem is extremely difficult (the twin primes conjecture is a corollary where k=2). Tao proceeds to explain various dilutions and progress made including the 2005 breakthrough by Goldston-Pintz-Yildirim. Apparently, finding narrow H-admissible k-tuples leads to a reduction in the gap between consecutive primes, thus improving on Zhang’s result. This is just one approach to improving the result.

As a non-number theorist, it’s great to be able to understand at least the introductory notions used to explore these questions and to see some of the relationships between older and newer results. It seems that Zhang used many common techniques, but applied them in novel ways, and presented his work so clearly that it was quickly confirmed as valid. While many articles have dwelled on Zhang’s relative obscurity, it’s worth noting that Zhang clearly has a taste for problems that are easy to state and hard to solve as he tackled The Jacobian Conjecture for his dissertation work. Zhang’s dissertation was in the field of Algebraic Geometry (not number theory), and that his advisor remembers him as “a free spirit” who did not even request letters of recommendation for jobs! From now on it seems that Zhang need only his own accomplishments to recommend him.

The current record is close to 60,000.

A question from a mathematical dabbler. The consequtive sequence 8,9,10 is bounded by the primes 7 and 11. Call this a prime free sequence (PFS) of length 3. The consequtive sequence 24,25,26,27,28 is bounded by the primes 23 and 29. It is a PFS of length 5. The question is: what is the longest PFS?

Great question! Can you think of any ways to create a PFS?

But, there are arbitrarily large gaps in the series of primes. In other words, given any positive integer k, there exist k consecutive composite integers. (This would mean that there is no “longest” PFS). Proof (the proof is v well known): Consider the following integers : (k+1)!+2, (k+1)!+3, ……, (k+1)!+k, (k+1)!+k+1. Every one of these is composite because j divides (k+1)!+j, if j lies in between 2 and (k+1) (both inclusive). What do you think? Am I right ? Regards, NP.