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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 $n = k +1$.
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.
Jun
13
revised Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$
wow, not all permutations are valid, and it was "without lots of generality"
Jun
13
revised Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$
had missed one in the summary
Jun
13
answered Find $x,y,z \in \mathbb Q$ such that $x + \frac 1y, y + \frac 1z, z+ \frac 1x \in \mathbb Z$
Jun
13
comment numerically evaluate a continued fraction
@BillyJean: The pseudocode is in the answer: set p[0]=a[0]; q[0]=1; p[1]=a[1]*p[0]; q[1] = a[1]; and in a loop for $k \ge 2$ evaluate p[k] = a[k]*p[k-1] + p[k-2]; q[k] = a[k]*q[k-1] + q[k-2];. The fraction p[n] / q[n] is the numerical value of the continued fraction up to $n$ terms.
Jun
13
comment Binomial coefficients as multiple sums
We are here after all to help people asking the questions, and using notation unfamiliar to them (without telling them what it means) is not the best way to do so. Also I wish to clarify: I usually don't edit answers, but in this case it seemed evident that you were simply following the question in the choice of notation, and I assumed the reason was that you too wished to be helpful to the OP and use the same notation. (So since the notation in the question was not what the OP used, I changed both.) It seems this assumption was wrong, and I was mistaken. I won't edit your answers again.
Jun
12
comment Binomial coefficients as multiple sums
I was trying to be helpful to the OP, and use the same notation that he/she was familiar with: that's why I edited both the question and answer to use the same notation that was originally in the question, rather than the $C^{m}_{n-1}$ that was introduced by an edit, which I've never seen before, and which the OP would probably find confusing. I'm also unhappy about the question now using the notation $\binom{n-1}{m}$ which is not taught in many schools (am never happy about questions being edited into forms the asker wouldn't recognize), but sure, I won't edit again.
Jun
12
comment Is my proof correct? $2^n=x^2+23$ has an infinite number of (integer) solutions.
@awllower: See Erick Wong's answer; the statement in the question is definitely false, and we don't even need to use some heavy artillery like the abc conjecture to prove it. Anyway, the state of the abc conjecture itself is quite unclear; I recommend this quite nice piece of journalism by Caroline Chen: The Paradox of the Proof.
Jun
12
comment Show that $x=2\ln(3x-2)$ can be written as $x=\frac{1}{3}(e^{x/2}+2)$
Actually I think the question would have been better phrased as showing that $x = 2\ln(3y - 2)$ is equivalent to $y = \frac13(e^{x/2} + 2)$ and then pointing out the special $x = y$ case of it. :)
Jun
12
comment Why prime number theorem tends to one?
+1. Shouldn't the $1/n$ in the first line be $n$?