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Jun
16
comment Criterion for sum/difference of unit fractions to be in lowest terms
I think this should be the answer. It also tells you how to construct such pairs $(a, b)$. Pick any number $d$, and pick any pair of coprime numbers $a', b'$. If you want the pair $(a, b)$ to be "easy": make sure that $a' + b'$ is coprime to $d$; if you want it not to be, make sure $a' + b'$ has something in common with $d$.
Jun
16
awarded  Good Answer
Jun
16
comment Could we define multiplication of “complex numbers” in this way?
I see no reason to downvote this question; it's wrong but that shouldn't be a reason to downvote here.
Jun
15
accepted The sum $\sum\limits_{n \ge 0} \binom{m-1}{n} \frac{n!}{m^n} + \binom{n+1}{2}\binom{m-1}{n-1}\frac{(n-1)!}{m^n}$ is asymptotic to $\sqrt{9πm/8}$
Jun
15
comment The sum $\sum\limits_{n \ge 0} \binom{m-1}{n} \frac{n!}{m^n} + \binom{n+1}{2}\binom{m-1}{n-1}\frac{(n-1)!}{m^n}$ is asymptotic to $\sqrt{9πm/8}$
It's very clear now, thank you very much once again!
Jun
15
comment Maximum length of sequence of non-coprimes of $N$ - least upper bound for Jacobsthal's function
The smallest factor of $N$ is not an upper bound: for $N = 6$, the smallest factor is $2$, but $g(6) = 3$ as there are three consecutive numbers in $F_6$, namely $2, 3, 4$.
Jun
15
comment The sum $\sum\limits_{n \ge 0} \binom{m-1}{n} \frac{n!}{m^n} + \binom{n+1}{2}\binom{m-1}{n-1}\frac{(n-1)!}{m^n}$ is asymptotic to $\sqrt{9πm/8}$
I'm having trouble with this part: "for every $t$ in $[0,1)$ there exists $c(t)$ such that for every $x$ in $[0,t]$, $1-x\geqslant\mathrm e^{-c(t)x}$" — I can follow it for $t \in [0, 1)$ as written, but later you take $t \to \infty$, so if I try to prove it for $t \in [0, \infty)$ instead, it's not so clear. (E.g. $x = 1$ would trip us up.) Could you help?
Jun
15
revised The sum $\sum\limits_{n \ge 0} \binom{m-1}{n} \frac{n!}{m^n} + \binom{n+1}{2}\binom{m-1}{n-1}\frac{(n-1)!}{m^n}$ is asymptotic to $\sqrt{9πm/8}$
fix bug
Jun
15
comment The sum $\sum\limits_{n \ge 0} \binom{m-1}{n} \frac{n!}{m^n} + \binom{n+1}{2}\binom{m-1}{n-1}\frac{(n-1)!}{m^n}$ is asymptotic to $\sqrt{9πm/8}$
This is beautiful, thank you very much. I'm still digesting and verifying this, but I've already learned a lot.
Jun
15
comment Could we define multiplication of “complex numbers” in this way?
@Vishal (and others): Oh oops, I didn't notice that it's $\mathbb{R}[t]/(t^2 - 1)$, rather than $\mathbb{R}[t]/(t^2 + 1)$ which is the complex numbers. Sorry. :-)
Jun
15
revised The sum $\sum\limits_{n \ge 0} \binom{m-1}{n} \frac{n!}{m^n} + \binom{n+1}{2}\binom{m-1}{n-1}\frac{(n-1)!}{m^n}$ is asymptotic to $\sqrt{9πm/8}$
an -> a
Jun
15
comment Second pair of matching birthdays
@Fixee: I've asked it as a question here.
Jun
15
asked The sum $\sum\limits_{n \ge 0} \binom{m-1}{n} \frac{n!}{m^n} + \binom{n+1}{2}\binom{m-1}{n-1}\frac{(n-1)!}{m^n}$ is asymptotic to $\sqrt{9πm/8}$
Jun
15
comment Second pair of matching birthdays
@Fixee: Yes, I don't know of a closed form either. BTW, after running (a C version of) that Python program for many days, going up to $2^{38}$, gave the impression that the limiting constant $c$ should be slightly less than 1.879973, and this is suspiciously close to $\sqrt{\frac98 \pi} \approx 1.879971$, but this may be just a coincidence.
Jun
14
comment What should be proved in the binomial theorem?
You have already accepted an answer on Jun 9; what does it mean for you to edit the question and ask something again? :-) Does it mean your question is not fully answered yet? Anyway the answer to the latest line in the question ("Is this all that should be done?") is no, this is not all that should be done. You have verified it for $n = 0$ and $n = 1$, but what you need to prove is that if (whenever) it's true for some $n = k$, then it's also true for $n = k +1$.
Jun
14
awarded  number-theory
Jun
14
revised Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$
avoid the "divides" notation, to make this require less knowledge
Jun
13
answered A numerical coincidence with continued fractions
Jun
13
revised Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$
order by complexity
Jun
13
comment numerically evaluate a continued fraction
@BillyJean: You don't need the $a_i$s (note the difference in notation between your question and this answer: what you call $p_n$ is $a_n$ here) to be positive integers, or even integers: they can be arbitrary real numbers, or even complex numbers or members of some field. :-) The proof just uses properties of addition, multiplication and division; it uses nothing about the signs of the numbers.