Erick Wong

Assuming that the degeneracies that Eric Towers pointed out are avoided for the range of $a,b,c$ in question, it seems that these would be fairly smooth objects which could be uniformly approximated by polynomials on $[0,1]$ (or really, any reasonable family of Stone-Weierstrass basis functions). Then you don't get any combinatorial explosion when adding the 100000 approximants together. Seems like you would have some freedom to adjust the degree to trade accuracy vs cost, and you could use different degrees for each $i$ to account for higher variability depending on the coefficients.

@mick You said you had no idea how to look for a covering set :).

Have you looked at the example at rieselprime.de/ziki/Sierpiński_problem? That should give you some sense of what shape a covering set is likely to have: look for small arithmetic progressions in the exponent $n$. The useful prime divisors will tend to be ones that divide $2^n-1$ where $n$ itself has small prime divisors, e.g. $73$ divides $2^9-1$.

@TurlocTheRed Since we know the infinite sum, the tail is well-approximated by the telescoping series (or Euler-Maclaurin). You can get both upper and lower bounds of the form $\pi^2/6 - 1/(2020+c)$ for some small values of $c$.

It's certainly possible to calculate the exact sum, since the numerator and denominator have only about 1700 digits each (and at least half that many, even with cancellation). However, that would make it a purely computational question with very little mathematical content.

@user2284570 Fair, this is before I realized these factorials are a bit too large for Karatsuba multiplication. I agree it is unlikely you’d find a decimal-based large integer package that implements Strassen

@JairTaylor Yes, it is feasible, but you are right that it is not really obvious. I think it has about 600 million digits and fits into about 30 million 64-bit words. So Schönhage-Strassen multiplication would be warranted, which is $O(n \log n \log \log n)$ in the number of words. It sounds like something you could do within a trillion operations, and it doesn’t take inordinate amounts of memory by modern standards. A feasible but by no means insignificant effort.

If integer division is the most expensive step, can’t you do the computation with a library that represents large integers in base $10^n$ instead of the typical binary word size?

Also, base conversion can be done much faster than the naive approach of iteratively dividing to extract a constant-size block of digits. See for instance cs.stackexchange.com/questions/21736/…

@JairTaylor The question indicates that they did it by computing the factorial explicitly and then splitting it into digits.

So basically you’re getting into the rudiments of sieve theory, which was first developed by Brun a century ago to study twin primes, and remains an active area of research (recent breakthroughs by Zhang and Maynard have come closer to proving twin primes than I’d dreamed of). The basic idea follows the same pattern of inclusion-exclusion, with alternating adding and subtracting diffrernt

@DerekPenaoza This technique looks like a form of inclusion-exclusion counting. When you add back $1/(3\cdot 5)$ and $1/(3 \cdot 7)$ you may be double-counting $1/(3\cdot 5 \cdot 7)$, so superficially it has the appearance of being an upper bound, not a lower bound. Of course once you subtract out the third-order terms you probably get something that goes to $-\infinity$. But this is the heart of what makes sieve theory challenging :).

@DerekPenaoza What conjecture do you refer to? I am confident that this is still a very long way from proving twin primes conjecture, if that’s what you mean. For instance, if this is an estimate for the fraction of twin primes in a certain range, the fact that it eventually becomes $>1$ means that the approximation must get worse and worse as $p$ increases.

As written, I’d say your sum diverges to infinity as $p\to\infinity$, since it’s roughly as large as $\frac12 (\log \log p)^2$. However $\log \log n$ diverges extremely slowwwwwly, so it may be hard to observe this computationally (one can make a strong case that if you add reciprocals of every prime ever computed by human civilization, it would still be less than $5$).

@DerekPenaoza Your sum can be simplified significantly. The majority of terms come from pairwise products such as $ab+ac+ad+bc+bd+cd$. These can always be rewritten as $\tfrac12 [(a+b+c+d)^2 - (a^2 + b^2 + c^2 + d^2)]$, turning a double sum into single sums. Try to rewrite $G(p)$ this way and I think it will be more legible, plus we know a lot about sums like $\sum 1/p$ and $\sum 1/p^2$.

A limit of the form $\infty/\infty$ does not allow you to conclude that it does not exist. It simply means more effort is needed to determine whether it exists. Notice that very easy-to-understand limits like $\lim_{x\to \infty} x/x$ fit the same pattern.

For the negative case it seems like one could make a good deal of headway considering the factorizations of denominators (possibly combined with size considerations).

The problem is an interesting problem, but the question lacks all sorts of context that has been raised in the comments. Please edit.

The claim is that if x in an integer such that x^2 is divisible by 3, then x is also divisible by 3, You will find that the example where x = \sqrt{3} is completely irrelevant to the truth or falsehood of this claim.

@user58865 No problem, this is very easy once you focus on the right thing. In order for $x^2 - 3$ to be divisible by $3^n$ it must be divisible by $3$, but for this to happen, $x$ itself must be divisible by $3$ (this follows from uniqueness of prime factorization). However, then $x^2$ is in fact divisible by $9$, so $x^2 - 3$ is not a multiple of $9$. This proves that $x^2 - 3$ cannot be divisible by $9$ or any higher power of $3$.

@user58865 I mean that the problem, as you have stated it, is false and therefore cannot be proven. It can be adjusted so that it is true, but we are not mind readers so it's your job to go back and check if you omitted any details from the question. One reasonable possibility is to require that $a$ is not divisible by $p$, and that $p$ be a prime $>2$. But again, not mind readers, and note that the assumptions that you're missing are almost as long as the question itself.

$x^2 = 3$ is solvable mod $3$, not solvable mod $3^n$ for any $n>1$. You should adjust the statement of the problem.

Seems quite unlikely to me. For a given $n$, the value of $\log a - \log r$ can span a very wide range. I expected it will be impossible to prove $\log a - \log r < n - s_2(n) -2$ for all $a,r$, and it will be impossible to prove $\log a - \log r > n - s_2(n) - 2$ for all $a,r$, because neither one is true. You will need to exclude the specific value $n - s_2(n) - 2$, and you will need to use arguments beyond size considerations (i.e. number theory).

Sure, sounds sufficient to me. But you haven't even defined what $a$ and $r$ are. Are they all possible decompositions of the odd part of $n!$ into two factors? It certainly seems unlikely that $\log a - \log r > n$ will uniformly hold over all choices of $a$ and $r$. It also is certainly unlikely that $\log a - \log r < n-s_2(n)-2$ will uniformly hold. So what makes you believe there is any inductive structure on which to establish the inequality?

It wouldn't hurt for you to read (and do exercises) from Tom Apostol's Introduction to Analytic Number Theory, say Chapter 3. However, be aware that it is highly unlikely anything so elementary will lead to a proof of Brocard's conjecture (in most cases if that were possible then Erdős would have already worked it out :).

$2^O(\log n)$ is literally the same as $n^{O(1)}$, which matches a great number of functions: the RHS could be $\sqrt n$, or $100000/n^{100}$ or even $\exp(-\sqrt{n}) + n^{1 + \sin n}$, there would be no contradiction.

In 1.2, you define $k=4t-1$, then you rewrite $4t-1 = a$. Can you see that all you are doing is manipulating in circles with no progress? There is nothing useful about 1.2. After you fix equation 1.8, the $(e/2)^n$ term will become $(e/e)^n$ or $(2/2)^n$ and you will not have a contradiction. Which is as it should be because Stirling's formula is internally consistent.

I would guess that the perfect squares are spaced closely enough together to well-approximate $n!$ up to the error term of Stirling's infinitely often.

If size alone was enough to do the trick, then there wouldn't be real-valued solutions.

But if all you are doing is expanding out Stirling's approximation, then nothing good can come out of it because you have essentially removed all but a sliver of the number-theoretic content. There are obviously real-valued solutions to $k^2 = n! + 1$, and you haven't yet done anything substantial to capture the unusualness of $k$ and $n$ being integers.

Once you correct the error in approximation, no contradiction will be possible without introducing some new idea. Your current argument boils down to applying Stirling's approximation (in $\log(n!)$ form) and then comparing it to Stirling's approximation (in $\sqrt{2\pi n} (n/e)^n$ form). This is circular: when one of the two Stirling's was transcribed incorrectly, there was a specious contradiction;

Elliptic curves (more generally, Siegel's theorem) is truly the best way to understand the finiteness of integer points. Calculus methods are completely inadequate for the task. If you wrote out your proof formally you would see that. But instead you posted a sketchy version with nothing defined properly.

The reason people mention those is because your line of argument makes absolutely no sense, so that part gets ignored.

elliptic curves, divisibility

In general, your question should include as much context as possible so that responders understand what you are interested in (and you don't have to rudely tell people not to write about x)

Then why didn't you mention it in your question?

Are you denying that this is related to math.stackexchange.com/questions/1214044/… ?

You can say that because of MVT, not because of the reason you claimed

If two functions f(x) and g(x) are everywhere differentiable and f'(x) != g'(x), then f(x)=g(x) cannot have more than 1 solution. It cannot have 2 solutions. This is a trivial consequence of MVT

the k-=1 doesn't make sense either (but is harmless because k isn't used anywhere inside the loop)

Why is there a k-=1, n-=1 at the end?

Just helping to debug :)

Main advice is that restricting to distributions with formulas is very restrictive. E.g. the quantity $\int_x^1 F(t) dt$ can be bounded below by $(1-x) F(x)$ and above by $(1-x)$, and there will exist continuous distributions that come arbitrarily close to those bounds. That's the type of thinking that you need here.

For this problem it's ok to restrict to everywhere differentiable functions. Any bounds on $U(x)$ should extend to arbitrary distributions by taking limits (so some $>$ signs might turn into $\ge$ signs, etc.)

By impulse I mean: your calculation is assuming $f$ is differentiable, which isn't true for this case. We can approximate the bump function by something differentiable, but the new function would have to decrease very sharply at $x=1/4$, so that there is a very large negative value of $f_t$ inside the integral.

I think it would help to go back to the definition of $U(x)$ for this.

The problem is that your integral contains a large negative impulse at $1/4$ which isn't accounted for.

@user2966729 Thanks for checking, but did you take into account the fact that $f(x) = 0$ for $x > \tfrac14$ so that $\int_x^1 = \int_x^{1/4}$?.

@user2966729 You are mistaken. I said uniform distribution on $[0,\tfrac14]$ where $f(x) = 4$ on the support of $f$. This is $F(m) = \min(1,4m)$, which you have not tested.

I think this is false for the uniform distribution on $[0,\tfrac14]$ but the calculations are a bit messy, so please double-check this. In general (if I haven't botched something) you want a distribution which has little or no mass on $[\tfrac12,1]$ and reasonably large values of $f(x)$ for some small $x$ where $F(x)$ is close to $1$.