 First, I would like to thank the organizers for this invitation. It is a big honor for me to give a talk at this seminar. There is a beautiful seminar, a beautiful idea. And I must apologize because I talked not about recent results and some people have commissioned me before, commissioned my talk on this subject. Yes, I hope that in the nearest future I will prepare this paper for publication. So I will talk about the problem of hearts and details. They had a conjecture that for any integers x and y, a greater than or equal to 2, we have this inequality. As usual, pi of x is the number of primes after x. And it is used to check that this inequality holds if x or y. x is smaller, y is small. And it has been verified for minimum of x and y up to this guy, 1371. It was under damage 1998. But hints and visions show that this conjecture is not compatible with the prime critical conjecture. Now, what does it mean? And actually we need the prime critical conjecture in some particular case. So we have the following conjecture. So let me give you a definition. As I said, the formula is admissible. If for any prime p, this set misses at least one concurrent class model p. And the prime critical conjecture says that for any admissible sequence, b1 is at the bk. There is infinitely many integers such that all numbers n plus b1 and n plus bk are primes. So actually it is related to the question what conditions do we need for a finite number of a finite set of integers b1, etc. bk such that there is infinitely many n such that n plus b1 and bk are primes. Of course, we can have local obstacles. What are the local obstacles? If for some prime p, for any n, one of these numbers is divisible by p. Then, of course, we can't have large answers with all these guys are primes. And to avoid it, we require that this sequence is admissible. If it is admissible, then we don't have any local obstacles. So for any p, there is n such that all these numbers are not divisible by p. And the conjecture says that if there are no local obstacles, then we can expect that all these numbers are primes. And we use the representation of Rostar of X. It is the maximum number of integers in the interval y from y to y plus x, excluding y and including y plus x. And you see that this interval contains exactly x, exactly x integers. And this guy is the maximum number of integers in this interval, which are related to the prime to all positive integers, integers, objects. Yes, it is such that if y is greater than or equal to x, then we have this equality because y is just a number of primes in this interval. And if a number of primes in this interval is a prime, then it must be related to all positive integers, objects. And where is the simplification between this quantity of Rostar of X and the admissible frequency? So if you have some admissible frequency, you don't necessarily decrease such that the difference between the largest and the smallest integer from the frequency frequency is less than x. Then by the China's Mediterranean, we can translate this difference such that after the translation, all numbers will be related to prime to p, for any prime p, for any prime p objects, we can do it. Okay, and if you have such translation that all translated numbers are related to prime to p, then we can put them in some interval y and y plus x. And we see that this interval contains at least k numbers related to prime to all positive integers of the objects. Or equivalent to all primes objects. So we see that the real prime of x is greater than or equal to k. On the other hand, let us take k equal to Rostar of X. Then by definition, there is an integer z. And there are numbers, given a set to be k such that all numbers are positive integers up to x. And another of these numbers z plus b1, z plus bk, is divisible by any prime p objects. So we see that the condition to check that this sequence is admissible is satisfied for all primes p objects. But if p is greater than x, then the condition is trivially satisfied because here we have k numbers. k is not greater than x, and so these numbers cannot complete all raise the classes for the p if p is greater than x. And if we assume that the prime k capable condition holds, then we have this equality. Also the classes are equal. Indeed, it is true that the first number does exceed the second number, the second number does exceed Rostar of X. But on the other hand, we have seen that if you have Rostar of X, then we have admissible sequence of k numbers in the interval from 0 to x, and using the k-taple conjecture, we see that we can find the infinitely many x such that this difference is equal to Rostar of X, and we have this equality. Also we will use this notation, log of X, log of X, log of X, log of X, and so on. And I have mentioned about the result of Henson and Richards. What is the proof that for large x, Rostar of X is actually greater than p. And then we have this as a political inequality. So it has been checked by Baflar and Gernrich in 2003 that this difference holds for x equal to this guy. So we see that it's human that the prime k-taple conjecture, we can conclude that for any large x, this number is positive, and the maximum over all y greater than or equal to x, or this expression, minus pfx plus y, minus pfx minus pfy, can be estimated from below by this expression. For example, we will have this inequality, and so this shows that the prime k-taple conjecture is not compatible with the conjecture, it's very difficult to disprove the conjecture of hardly a little bit unconditionally. In principle, we could try to find the number y for this x equal to 4,916. We can try to find the x actually, this difference is big, but if you want to find it by computer, it is a hopeless problem because we can expect that the least such number y is very, very huge, and there is no way to find it, only we can try to catch it by a random truth, but let me say again that from a computational point of view, it is a hopeless problem. And now I'll talk about the arguments of Hansp and Richards. What do they suggest? They suggest to consider the function b, b consists of plus minus p over p, runs over all primes from some number y to one-half of x. And we assume that the x is odd, it's not principal, but if x is odd, then the set b is contained in the interval of x consecutive integers. And we have this equal to the cardinality of the set b. Now I want to compare this guy to power of one-half of x with power of x. We know it is a simple asymptotic estimate for the function power of x. For power of x, we have this asymptotic equality. And if we plug in x, then we have these asymptotics after small transformations. We have the following asymptotic expression for power of x over two. And so the pi of x over two minus power of x has the asymptotic estimate. The asymptotic estimate. So I want to show that for some appropriate y, this set y is an admissible set, but also the pi of y is small, and the pi of x over two is close to the cardinality of the set b. So actually if y satisfies this supposition, then power of y is small over x over small over x. And from this equality, we conclude the following asymptotic equality for the cognitive b minus power of x. And now we have to understand why set b for some appropriate y is an admissible set. And one can show that this holds for this number y, for some percent c. And to show that this set is an admissible set, we can see that if p is less than or equal to y, then by our construction, all elements of b are plus or negative, and prams greater than y. So now element of the set b is divisible by any p up to y. On the other hand, if p is less than p is greater than the cardinality of b, then the set b can't contain the cardinality of b against the case of model p, and not all the cases of model p. And the most difficult case is the case when p is between y and the cardinality of b. And to consider such prams, we use the classical result of Erdogan's invention. So I can also reckon when they proved the existence of big gaps between concept of prams, they called default lemma, default result, for large post-effects and for why satisfying this inequality. There is an interview such that all these numbers, such that any numbers from this info from u plus 1 to u plus 1, so they have a very consecutive numbers and each number from this interval has a protector of takes. And now it is known that this is true for slightly large values of y. So we can put this bound and it was done several years ago by Stephen Ford and Green, a minor tower in Massive. But this is not essential for our purpose so the result can slightly increase the size of an admissible set, but it's not essential for us. And now how to check the condition of admissibility. So we have some number p, p greater than small y, but p does not exceed the coordinate of the set b. And we define capital X and we define capital Y. We can estimate from above the size of capital Y. And we see that this is equal to holds and if you take capital C equal to 3 over small c, then for this x and for this y we can use the result of overframing. Next we define the capital P and the capital P is the product of all primes up to capital X. And after that we take an integer d, satisfying this quadrant's condition. You see with this, this is compared to u times p, modular capital P. And also, you can take such a number v in any interval of size p and in particular, we can take v in this interval. So we have this quadrant. We can use this quadrant and so we see that this number is divisible by some primes q objects because u times g by the result of further thinking is divisible by some primes q objects. And now we take what we want to say, what we want to show. What we want to show, we want to show that if you have this quadrant's class modular p, v modular p then any number from this quadrant's class from the interval from minus sexual object to sexual object to any number is divisible by some primes q objects. So we take this number n and it's v plus j p, what is g? What is g? v is smaller than the number of hertz and if one belongs to these intervals we see that g must be a positive integer. On the other hand, we can write an upper estimate for j. It is bounded by x plus capital p or small p and this is just the most x or y plus one by the the shape of capital y is at most is at most capital y. But we can prove that for any sub g this number n plus n is v plus j p is divisible by some primes q objects. And you see if n is divisible by q objects then n n cannot be an element of the set b because all the elements p are primes plus minus primes all primes are greater than p. So, we have checked with our set which is an admissible set but you see that using this contraction this contraction we get a set b such that its resonant is bounded by this guy to k f over two. And one more question maybe this is true that Rostarovix is bounded by this guy or a wicked person is it true that we have this equality and the Shinsen Shinsen offers the construction against these conditions. Actually it is not known it is not known whether just the construction of the Shinser works why does it give an admissible set but one can use it for an American experiment and Gordon showed that this inequality doesn't hold for this x. Now let me talk about the construction of Shinsen what is this? Sorry to interrupt you Igor Spalinski has a question regarding your previous slides Igor, can you please ask away? Okay, yes. Could you come back to the previous slide please? The previous slide. Why is v is less than minus x over two? I wonder if the first slide contains a type of v should be minus x. Yes. It is a mistake. Minus x over two. Okay. Thank you very much Igor. So we belong to minus x over two minus p minus x over two. Thank you very much. And let me turn to the construction of Shinsen let y is greater than square root of x and we use the hard see where it is finished to the interval where next we may do all multiples of the problems p up to y and yes if I stop here so this is just the but we do it for all odd p or the odd prime p for odd prime p but for p for p equal to two we eliminate the conference class one model one model so after the first thing we have only even numbers and and and with the not remaining set by you you depends on why it is easy to see what what is the structure of the set you you you can see so four numbers two to elf over p so elf is a positive integer p is greater than y okay so it is easy to see because because any element of you cannot divide by any prime up to y it it can be done easily by many prime up to square root of x because we eliminated all primes up to y but if a number doesn't exceed x and all prime factors are greater than square root of x it must be prime so p must be prime and prime p is multiplied by some power two and now we will compare the set u of y u0 u0 is the set of all such primes up to x and p is an unrestricted of prime and it is easy to to estimate the coordinate of all the set u0 because for any elf the number of primes of the prime two to elf times p it is just this number and we have to take the sum over all elf and it is yes we can add this in the form of p for 5x over 2 to elf and after taking the sum we see that we have this estimate for the 0 and the idea was to choose an appropriate number of y such that the difference between the sets u0 and u is small this difference can be estimated by by this by this sum and the issue of this is equal to 9 for y and it is not difficult to to show that that we can estimate the coordinate of this set u0 minus u by this guy and so we have this is equal to provided or whatever this is equal to provided that y is small y is smaller than this guy x divided by logarithms divided by logarithms and so we can take this number y we have a required estimate for the coordinate of the set u but is this set admissible we don't know probably is and more Washington suggested the generalization of his construction so this is small let me assume that m is small and it is bounded by this guy square root of log of x but actually we we think about m as a number going to infinity sx tends to infinity and we use again the artisans but now for all the primes p p1 is set with p sub m we eliminate the Hong Kong's class equal to 1 mod p sub i instead of 0 mod p i and u is again the residual set so we can assume we can take it far away and again it is easy to understand the structure of the set u the element of the element small u of the set capital u can be used by some primes up to p sub m so we we don't know of such primes of such primes other primes up to p sub m divergence u by capital m this set is not empty because any element u must be divisible by 2 and the set s is the complement to the set r in the set of m least primes and so the structure of the set u and again if we skip if we skip the condition greater than y then we can estimate we can take another set u0 and after summer rather long by standard calculations one can show this asymptotic formula for the set u for the content of the set u0 and if you look at this sum this sum tends to infinity if m tends to infinity and so the content of u0 minus primes is asymptotically x divided by log x squared but after that we multiply by some number of log m which tends to infinity and now we want to compare the set u0 with the set u and and our to have this asymptotic formula for the set u rather than for the set u0 and it's easier to show that the content of the difference u0 minus u can be estimated from above by this guy and so we have the required asymptotic formula provided that y superspires this condition you see that if you increase the m this condition becomes a stronger one so you find this condition we have this inequality but the question is whether the set u0 is admissible or not probably it is admissible but now the discussion looks almost hopeless so so you see that that when I was talking about the result of Hinsley and Richard to prove that that it was admissible I used some local arguments and here the local argument should be that for this for this y there is an interview such that each number from this interval has a perfect uptake so we should have the stronger version of by but but actually we believe that that in this form the result doesn't hope that this is not true if it was written to so probably it is highly likely that we can't use local arguments to prove that our set is admissible set but we don't know how to use other arguments yes and so there is a problem but we want to overcome this problem so we want to correct the set constructed by Hinsley and then we prove the form here we have this upper estimate for rostar of x-perfects and the whole said that if this indicates a type of conjecture then we have default local estimate for power of x y-perfects okay so far yes actually if you use the arguments of we can prove default lemma you see that I give a I look for this lemma and I give on this sketch of proof of the rest because it's not simple but but as I have mentioned to prove this lemma it's not to use the arguments of rz and rm and the use of this lemma we prove another lemma we prove lemma 2 what is lemma 2 so s and p are positive integers it is small it's at most log x over 2 p p is a prime it's in square root of x and x over log log x x is sufficiently large post-adventures and then we claim the form that there exists the a-sets and the form condition the number of post-adventures up to x congruent to a mod p such that all prime divisors often are either less than or equal to 2 greater than square root of x can be estimated by this number so so most numbers in congruent to a mod p up to x have some prime divisor between p and square root of x and we will use this property to do the form so we have a construct maybe it's not admissible but for any p we assume some numbers from this to some congruent to a mod p and we do it for for all prime from t to square root of x from t to some appropriate number and for any p to p to remove some numbers from some congruent class in such a way to improve a small number of numbers and so we in such a way we correct the set constructed by Schindsel there will be a few numbers from it and we get an admissible set how are we so we use number one we take with capital x and capital y and number one we take some new and the corresponding set of positive integers such that u plus r has no prime divisors from this interval from p to x yes and and this the result of such number n doesn't exceed this guy this is okay just one question please from Kevin Kevin can you please unmute and ask away one moment yes I'm here the question is about lemma one the hypotheses on t since since you're sieving out the interval t comma x you certainly cannot conclude this if t is close to x t equals x so I he must be very small okay okay one moment okay one moment okay actually we are using this lemma for t less than a log log of x so you can take here log log of capital x okay okay all right I can't very much I have to think more about the amount but we can issue that t is very small because we are using for very small t that's much to log log of x so yeah and here okay the idea is that t is smaller so it's less than up okay now let the log log of x now there appear to the arguments I was talking about you the proof of tensile and reach of result we take this number capital Q and again I will take this number a zero and and then for any integer g and for any n from 1 to y this number is minus g to percent because no prime divisors from t to x only if n belongs to capital so for most numbers then it has prime divisors from this okay and now we will use the arguments of minus matrix method so we take this number g g is minimum of disgrace so since Q is essentially at most square root of x so we have capital G is also close to square root of x and for any n the number of g from this interval such that is zero minus g to percent has no prime divisors from x to square root of small x can be bounded by this guy we use simple sieve arguments and and this number is bounded by by this number okay and so you see that that if you are interested of of this guy is zero minus g to percent p we have to consider only n from script n yes but for for any such n this number has no prime divisor from t to square root of the number of pairs of nj such that this guy has no prime divisor from t to square root of x can be divided by can be estimated by this this number so this is as I told you for any number n the number of j is bounded by by this number and now we use the pigeonhole principle we can we can take j so that the number of fern such that this number has no prime divisor from t to square root of x is small is bounded by this guy and fern but fern has this form and fern is bounded by x when n has can be represented in this form for some n up to capital Y and so this actually proves a limit 2 and we can we can compute corollary 2 that if p satisfies this condition then there is supposed to be nj a a is called in 10 modulo p and n is supposed to be nj not x is nothing x and n is such a masquerade divisor from capital p to square root of x so to prove corollary 2 we we use limit 2 and then we look at the estimate of limit 2 we see that the number of exceptional n if p satisfies this condition the number of exceptional n is smaller than 1 if this number is smaller than 1 then we see that such numbers n don't exist and now let me look at the construction of Schindsel we take the form the form value for m so this is y and and then we know then that the set you constructed by Schindsel specifies this is a asymptotical asymptotical equality and so this is about times this guy so so before one an extra further log-log-log effects log-log-log effects so but the definition of u for any prime p up to y for any p up to y there are some components corresponding to p which contains none of elements of u and and now we and now we we denote by p this guy p sub m so p p is less than log-log up to square root of log-log effects and z is this guy by color I'm greater than z there is this a from z such that that for for any n any such n has a a prime divisor from t to square root of x so this number n can belong to the set constructed by Schindsel to the set u so the condition of admissibility is shown for prime p greater than z and we have to yes but let me say again that we can't we can't check we don't know how to check the condition of admissibility for the set u for medium prime u for medium prime p greater than t and less than z and now for such p we use limit to to remove some numbers for any prime p we do not by this guys we can estimate by limit to we can estimate your p by by this guy u p actually we had an extra factor log 4 of x but log log log log x in the later but but we ignore it and we have the different up rest of it and we can certainly use that u prime u prime is u minus the even of of all such u of p and for any p we remove some model p yes and the prime is in doubt where the prime from this interval from y to z I think from t to z okay it should be and for y to z because if p is less than y our construction the condition of admissibility holds for y so the primes in doubt are primes between y and z and they removed some numbers the number of such numbers is bounded by this sum the sum of the sum of u p the sum of u p it can be estimated by this guy it can be done by best standard calculations by the choice of prime we have the business quality here we use that m squared is less than log log of x and we get the result we get the result thank you very much