Open Problems
I’ve gathered some problems I haven’t been able to completely resolve.
If you decide to try these, or if you find that they’ve been solved in literature somewhere, I’d be interested to hear about them. I’ll include a brief summary of progress.
Problem 1
Suppose you have two sets
and of natural numbers.
Writeif the sets have zero natural density, whenever are different elements of . In other words, has nearly disjoint -dilations.
Then the upper density ofis at most
See the following posts for context: 1 2
The first post basically proves it for nice
This case extends to
The second post proves it for the case in which
I’ve proved it by a few arguments for the cases (for example)
The general case, as well as the special case
Problem 2
Fix a base
and a natural number .
Letwhere is the sum of the base- digits of .
Similarly create the sequencestarting from a natural number .
Then there exists an integersuch that for all large iff .
In the binary case, there always exists such an integer.
It seems like there’s some simple proof that I’m just totally overlooking here.
Let tetration be written as
Here’s a proof (minus computer calculation) that, in the binary case, any starting values
Fix
Any integer in this range will eventually be mapped to the range
From there it gets a little uglier, this range maps into
Continuing on, you get ranges of the form
Now, that doesn’t reduce the problem to one integer, rather a small range (seemingly linear in
My super old Facebook post on this says that every starting value under
Problem 3
Say
is an algebraic number.
Is there always a finite sequence of polynomialsall of which having integer coefficients and rational zeros1 which, when composed, has ?
If
What if
Then the sequence
I posted this some years ago on the actually good math problems Facebook group. There, Jordi Ribes provided a working sequence for the cubic case:
Any cubic x^3+ax^2+bx+c divides the polynomial (x+a/3)^2((x+a/3)^2-(a^2-3b)/3)^2-((2a^3-9ab+27c)/27)^2. This comes from computing the depressed cubic t^3+dt+e (where d,e depend on a,b,c) through the change t=x+a/3, and noting that t^3+dx+e divides t^2(t^2+d)^2-e^2. So f_1(x)=(x+a/3)^2, f_2(x)=x*(x-(a^2-3b)/3)^2, f_3(x)=x-((2a^3-9ab+27c)/27)^2.
Seems interesting. The cases with degree
-
Edit added 7/3/2024. Oops! ↩