#HELP NUMBER THEORY PROBLEM.
171 messages · Page 1 of 1 (latest)
I would imagine yes. If we consider the factorial to be a product of values of i, as in $\prod_{i=1}^n i = n!$, then each 3rd i will contribute one factor of 3, each 9th will contribute an additional factor and so on. Similarly with 5 and 7. 2n choose n is $(2n)!/(n!)^2$ so we only need to find an n such that it is only slightly larger than a power of 3, 5, and 7, so that no extra powers of these factors are contributed from a new prime power.
However, proving that you can always find such a number might be tricky. One way might be to prove that you can generate one of these numbers from one that already exists, but the action of multiplying by (for instance) 105 would compound differences between the number in question and each of the prime powers. Try playing around with this idea though.
OmnipotentEntity
Well, I don't know the answer, and I suspect it might be an open problem
Yes, it is open. This paper may be of interest https://arxiv.org/abs/2201.11274
We show that for every $r \geq 1$, and all $r$ distinct (sufficiently large) primes $p_1,..., p_r > p_0(r)$, there exist infinitely many integers $n$ such that ${2n \choose n}$ is divisible by these primes to only low multiplicity. From a theorem of Kummer, an upper bound for the number of times that a prime $p_j$ can divide ${2n \choose n}$ is ...
@vapid fractal
lol how did you know
It had that feeling
anything to do with prime numbers bro
I mean
as long as n * n+1 *.....2n
doesn't have a multiple of 5, 11, 105
but then
idk
it seems more and more unlikely as n gets bigger
More precisely. Let $n = \sum_i a_i \cdot 3^i$ be the expansion of $n$ in base 3. Similarly $n = \sum_i b_i 5^i$ and $n = \sum_i c_i 7^i$.
We want to avoid introducing an extra prime power when we double. This means we want to avoid a carry. In other words, for an $n$ to be admissible we require $a_i \in {0, 1}, b_i \in {0, 1, 2}, c_i \in {0, 1, 2, 3} \forall i$
In other words, we can use this criteria to hunt down these values. I just manually started with 50 (i.e. one of the first non-trivial values that might not be coprime to 105) and just kept heading up using the base expansions above until I found an example: $n = 756 = 2130_7 = 11011_5 = 1001000_3$
As you observe, these examples are more and more unlikely as n gets large because for a $k$-digit integer in a particular base selected uniformly at random each digit has $2/3$ chance of being admissible in base 3, a $3/5$ chance in base 5, and a $4/7$ chance in base 7 (these probabilities are not actually independent though, and, of course, the number has differing counts of digits in different bases).
OmnipotentEntity
,w gcd((2*756) choose 756, 105)
(however, I still think it is likely that there are infinitely many examples, because although the proportion of numbers that are admissible for a given base go to zero, as k gets large, the absolute number of values for a given number of digits actually increases)
Here is a list of all number n satisfying the requirement up to a googol
what is tis
@vapid fractal it's an example of a value of n (n = 756) such that 2n choose n is coprime to 105. The document uploaded contains many more.
It doesn't help with the proof. I was just uploading it because @bleak dew seemed skeptical
What is your current level of mathematical attainment?
The answer will probably require machinery from algebraic number theory, so you'll probably want to start heading towards that direction
Maybe?
Anyway, it's much more difficult than my own level of math. This would be something where you might be eligible for a field's medal if solved. But you would not be solving specifically this problem, this would likely just be a corollary.
hm ok
can we work together by any chance
im js confused like what makes THIS PROBLEM so difficult??
It is feasible that for any finite set of odd primes { a_i } there are infinitely many n such that 2n choose n is coprime to prod_i a_i
As mentioned above, it has quite a bit to do with how numbers are represented in certain bases.
BUT I WILL BE THE ONE TO SOLVE THIS PROBLEM
Don't joke about that
and change mathematics for once
who, me or cor?
@bleak dew gonna need you to tell me that's a joke
yeah its a joke
im just 13.5
ask me any undergrad easy question
first year
i prove
guys trust me
i will be the one to solve this, im not gonna give up !!!!
even though people who are phd level are fumbling
Best of luck.
Yeah if terence tao can't how can you
im final year 🤣
bro uses ai to solve erdos problems 🥀
also, its not like terence has dedicated a lot of time to these particular unsolved problems
yeah cos he's not stupid
if he did, then theres a possibility he could crack it
wym
he's not gonna try and solve a problem that lacks the required groundwork and machinery
It's like trying to build a GPU when all you have is sand
its not impossible though
solving these maths problems isnt impossible
no one knows if the theorems are even correct
just because mathematicians cant crack it, doesnt mean noone can
you'd just be wasting your time, terence knows that, and knows that he only has very little time on earth, that is why he is working on so many problems that are hard but can be solvable
there is just not the required machinery for some of these questions
damn
very little time on earth 🤣
doesnt mean you can dodge it, ignore it and give up
actually, you can, like all of quantum theory is based on the unproven zeta function, the rieman hypothesis
which isn't proven
but some problems could be revolutionary
right but some problems are revolutionary and can be solved
and we are in no shortage of those problems
i think he would agree with me
@vapid fractal the required machinery to solve a problem can come decades or centuries after the problem is proposed, such as FLT. If Andrew Wiles lived in the 1600s no shot he would have proven it
Doesn't matter how smart you are
no, it can.
bro you clearly aren't final year then
a genius could crack it, intelligence is profound
how smart do you have to be to be this sanguine
better on being enthusiastic and driven with dedication rather than being dismissive
right OmnipotentEntity can you remove this guys undergraduate role, because this guy has never done a single proof in his life

im just being realistic
im just being optimistic
yes but you have to be realistic as well,
bros literally first year and demands only easy undergrad problems, you cant be talking rn...
i am, im merely kneeling towards a more hopeful aspect
I mean, you are simply wrong. Math is developed piece by piece, and to figure out FLT we needed to develop the modularity theorem. However, these two pieces of math are only related at a deep level, and if you're looking at all of math in the 1600s you're not going to start exploring elliptic curves (which had not been seriously explored up to this point) for the answer to your problem in number theory.
optimistic, ok, but it has to be you who solves it right?
says a lot...
There are no requirements to have an undergrad role. We take these by the honor system. If a user is found to be abusing their role to post nonsense in topic channels then rather than removing the role we assign the emdash role to prevent them from disturbing those areas of the server
fairs
We do review the postgrad role though.
why
well i wish to?
because next year when you want the post grad and you say nonsense like this...
idek what to say to you lol. Its literally like trying to build a cpu from scratch by yourself, not gonna happen mate
CPU from scratch is honestly easier.
simply underestimating intelligence, ofc there are limits and boundaries to these certain aspects in maths however, intelligence from knowledge is boundless. this could have a significant impact in the working of these unsolved maths problems. who knows, i don't think it's accurate to be making definitive statements against this belief. but ofc, i do understand what @wispy panther has been discussing about...
We have processes and schematics and the CPU doesn't need to be good.
Making pure silicon is achievable in a small lab
Etc
have you ever actually done a single proof in your life
like ever
because this has gotta be ragebait
I think it is honestly
you are ragebait
what does doing a proof got anything to do with this topic of conversation
So I'll just stop feeding the troll, and if the troll continues saying he's hungry I'll need to take other actions
unreasonable...
troll?! 😭
hop on vc bro, takes you a month to respond and im not understanding
I should probably stop engaging as well, but literally the whole conversation was about proving questions that have stumped every mathematician, like ever
not every
@vapid fractal given that a) you knowingly posted an open problem, b) you first attempted to draw people into DMs to hide this, c) gave up after you realized that I knew it was open, d) started up again when you realized I did a little work on this, it's pretty obvious that you're just trying to be disruptive.
yes because it hasn't been proven lol?
honestly i dont think these erdos problems have received much attention, and bare in mind that only 41% have been solved and this is only one
I've been tolerant up to this point because you were being a little bit circumspect about it.
gave up?
also btw when i received this problem from someone, i didnt even know it was an open problem until ages after so i started to want other people's perspective on it
no shot sherlock
Anyway, I will be closing this help thread. It's an open problem. You need postgrad math to effectively attack it, and as you don't have postgrad math you'll need to work up to it.
.close
Post marked as solved by @wispy panther.
Use .unsolved if this was a mistake.
.unsolved
Post marked as unsolved by @vapid fractal.
Use .solved to mark as solved.
bro
Yes?
what do u cover in university math
ill leave it as unsolved and ill come back to this channel stronger...even if i won't crack this, i'd at least have some useful research upon it...
Try #math-discussion
how to goon
.close
it says no access, i think it's cus my role is pre university math
Post marked as solved by @wispy panther.
Use .unsolved if this was a mistake.
Poke me in #discussion