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Search options not deleted user 5712

This tag is used if a reference is needed in a paper or textbook on a specific result.

2 votes
Accepted

On a sum involving Euler totient function

I have found a proof of more general formula in the book Postnikov, A. G. Introduction to analytic number theory American Mathematical Society, 1988, (section 4.2). This proof is simple but it has a s …
Alexey Ustinov's user avatar
10 votes
Accepted

Frobenius number for three numbers

Simple algoritm based on continued fractions was proposed by Rödseth, O. J. On a linear Diophantine problem of Frobenius J. Reine Angew. Math., 1978, 301, 171-178 All algorithm are described in Ramre …
Alexey Ustinov's user avatar
4 votes

Frobenius number for three numbers

Rödseth formula for Frobenius numbers is good not only for computation. It allows to find weak asymptotic for Frobenius numbers with three arguments and density function for normalized Frobenius numbe …
Alexey Ustinov's user avatar
20 votes

Conway's lesser-known results

Conway's Soldiers. And an interesting special case Reaching row 5 in Solitaire Army.
5 votes
1 answer
971 views

On a sum involving Euler totient function

Let $$S_a(N)=\sum_{n\le N}\frac{\varphi(an)}{n^2}.$$ The usual machinery gives an asymptotic formula $$S_a(N)=\frac1{\zeta(2)}\cdot\frac{a^2}{\varphi_+(a)}\log N+C(a)+O(N^{-1+\varepsilon}a^{1+\varepsi …
Alexey Ustinov's user avatar
9 votes

Is it possible to "get" quaternions without specifically postulating them?

They can be deduced via Dirac Belt Trick, see Understanding Quaternions and the Dirac Belt Trick by Mark Staley. This Trick is also known as Plate Trick.
Alexey Ustinov's user avatar
3 votes
1 answer
233 views

Addition law for elliptic curves of the form $x^2y^2+a(x+y)+b=0$

Did anybody consider addition law for elliptic curves of the form $$x^2y^2+a(x+y)+b=0\,?$$ Does this form have any specific name?
Alexey Ustinov's user avatar
1 vote
Accepted

Addition law for elliptic curves of the form $x^2y^2+a(x+y)+b=0$

It was done by Euler . Here you can see original text and english tranclation. $\smile$ Now it is known as Euler-Chasles correspondence.
Alexey Ustinov's user avatar
1 vote

Growth rate for the average of the entries in the fundamental period of the continued fracti...

Also a partial answer. Partial quotients in periods of quadratic irrationals satisfy the Gauss-Kuz’min law (in average over quadratic irrationals with the length $\le N$), see Spin chains and Arnold …
Alexey Ustinov's user avatar
1 vote

Is it possible to approximate irrational by fractions with denominator and numerator odd?

The formula $|\alpha-(2m+1)/(2n+1)|<\epsilon/(2n+1)$ is equivalent to $\|\alpha m+\frac{\alpha-1}{2}\|<\epsilon/2,$ where $\|x\|$ is a distance from $x$ to nearest integer. This problem is covered by …
Alexey Ustinov's user avatar
1 vote

Frequency of digits in powers of $2, 3, 5$ and $7$

It is not an answer but some numercal data for $A_2(N)$. Here digits are ordered according to their frecuences ($2$, $4$ and $6$ look like most frequent): $$ \begin{array}{rl} N=1000: & 2, 1, 4, 6, 8, …
Alexey Ustinov's user avatar
3 votes

Last term of repeating continued fraction expansion

This result is classical, but unfortunately many usual textbooks (e.g. Davenport's "The higher arithmetic") don't give a direct refference to the original paper. This theorem belongs to Évariste Galo …
Alexey Ustinov's user avatar
10 votes
Accepted

Counting points on elliptic curves

Corresponding sums of Legendre symbols are know as Jacobsthal sums $$J(u)=\sum_{x\mod p}\left(\frac {x^3+ux}p\right).$$ They are equal to $0$ for prime $p\equiv 3\pmod4$. If $p\equiv 1\pmod4 $, $\le …
Alexey Ustinov's user avatar
11 votes
2 answers
1k views

Yang–Baxter explanation

What are the most simple examples which can explain the meaning of Yang–Baxter equation? Is there any way to explain this mysterious object to a person who is not a professional in quantum groups? Il …
Alexey Ustinov's user avatar
4 votes

How close do partitions get to perfect squares?

Numerical results give random picture. If it is really the case then usual heuristic arguments confirm your conjecture. ListPlot[Table[FractionalPart[Sqrt[PartitionsP[i]]], {i, 1, 5000}]] Distance …
Alexey Ustinov's user avatar

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