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Jun
16
comment How to efficiently generate five numbers that add to one?
Sure they add up to $1$, but you also need to show that the distribution is really uniformly random.
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 Conjecture: The following sum is asymptotic to $\sqrt{9πm/8}$
Jun
15
comment Conjecture: The following sum 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 Conjecture: The following sum 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 Conjecture: The following sum is asymptotic to $\sqrt{9πm/8}$
fix bug
Jun
15
comment Conjecture: The following sum 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 Conjecture: The following sum 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 Conjecture: The following sum 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
15
revised Ordering a set of infinite numbers to make it countable, with the constraint that we have no formula for generating those numbers.
better lead
Jun
15
comment Ordering a set of infinite numbers to make it countable, with the constraint that we have no formula for generating those numbers.
@AyushKhaitan: Maybe you're confused because the set $A$ in my answer is different from the set $A$ in Lord_Farin's answer? His $A$ is my $A \cup \{1, 2\}$.
Jun
15
comment Ordering a set of infinite numbers to make it countable, with the constraint that we have no formula for generating those numbers.
@AyushKhaitan: No, read what I wrote: there are two facts: (1) If a set $A$ is countable, then the definition of "countable" itself means that there exists an enumeration of the set as $a_1, a_2\dots$, or equivalently a mapping $n \mapsto a_n$. I'm making no assumption here, beyond what you wrote in the question: namely, that the set $A$ is countable. (2) If a set $A$ is countable then so is $A \cup \{1, 2\}$, and I've given a proof of this fact. The proof is by showing how any enumeration of $A$ (some such enumeration necessarily exists) can be converted into an enumeration of $A\cup\{1,2\}$.
Jun
15
answered Ordering a set of infinite numbers to make it countable, with the constraint that we have no formula for generating those numbers.
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 $n = k +1$.