Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Computational Number Theory is for explicit calculations or algorithms involving anything of interest to number theorists.
10
votes
Is formula valid for relating $\pi$ with ALL of its OEIS A002485(n)/A002486(n) convergents?
We have
$$(|i|\cdot2^j)^{-1} \int_0^1 \big(x^l(1-x)^{2(j+2)}(k+(i+k)x^2)\big)/(1+x^2)\; dx $$
$$= (\operatorname{sgn}(i) \cdot2^j)^{-1} \int_0^1 \big(x^{l+2} (1-x)^{2(j+2)}\big)/(1+x^2)\; dx +\frac{ …
6
votes
Accepted
A weird property of odd positive integers $n$ with $\sigma(n)\sim2n $
There's two parts of this: $\sigma(n) - 2n$ being congruent to $2$ mod $4$ and being divisible by $3$. Since $n$ is odd, $2n$ is congruent to $2$ mod $4$, so the first part is explained by $\sigma(n)$ …
7
votes
Computational complexity of finding the smallest number with n factors
The number $\prod_{i} p_i^{e_i}$ where $e_i = \lfloor \frac{1}{p_i^\alpha-1} \rfloor$ for any real number $\alpha$ is locally optimal, i.e. no smaller number has a larger product of factors. (This use …
17
votes
Accepted
Using the Eichler-Selberg Trace formula to compute class numbers?
Generally speaking, formulas like the trace formula admit an uncertainty principle: To obtain an identity where one side is highly concentrated (e.g. a sum over a small number of class numbers, or eve …
18
votes
Accepted
Can we compute the first $n$ digits of $\pi$ in $F(n)$ time?
Mahler proved(1) that for any $p,q$, $\left | \pi -\frac{p}{q} \right| > \frac{1}{q^{42}}$. It follows that if one can compute $2^{ 42n} \pi$ to within an error of at most $1$, one can compute the fi …