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These are all questions I would never even have asked myself until that incident with Don. Every day, my friend Don and I would see who could trip each other the most times. But then one day I tripped him and he fell and broke his jaw. He looked up and, with slurred speech, said, “I guess you win.” But what did I win? I didn't win anything, and you know why? Because I forgot to make a bet with him. But something else was wrong, and I knew it. Why did I want to trip Don in the first place? To show how clever I was, or how brave, or how successful? Yes, all those things. So I guess that answers that. -- Jack Handey
Jul 31 |
answered | Lower bounding the probability that $\gcd(t,N)≤B$, for a random $t$ and fixed (large) $N$ |
Jul 28 |
comment |
Upper density of the set of $n$'s such that $p(n)$ is prime, where $p$ is polynomial
An alternative is to use some of the theory of Beurling primes. See alpha.math.uga.edu/~pollack/beurling.pdf |
Jul 28 |
comment |
Upper density of the set of $n$'s such that $p(n)$ is prime, where $p$ is polynomial
A quick comment: One can derive the estimate $\sum_{p \le x} \nu_F(p)/p = \log\log{x} + O(1)$ without worrying about zeros of $\zeta_K(s)$ on $\Re(s)=1$; e.g., apply the Tauberian theorem mentioned as Proposition 5 in alpha.math.uga.edu/~pollack/eulerprime.pdf to $\log \zeta_K(s)$. Everything needed to check the hypotheses is in Hecke's algebraic number theory book (and goes back to Dedekind, I think). |
Jul 17 |
awarded | Yearling |
Jun 25 |
comment |
Number of primes one larger than divisors of a fixed number, which is LCM of 1,2,3,…,L
I don't have an account on Math Stackexchange at the moment, but the conjecture of Wythagoras that motivated your question follows quickly from Theorem 1 in this paper of Erdos--Pomerance--Schmutz: math.drexel.edu/~eschmutz/PAPERS/lambda.pdf |
Jun 6 |
comment |
When is $(-1+\sqrt[3]{2})^n$ of the form $a+b\sqrt[3]{2}$?
Have you seen Proposition 3.8 and Example 3.16 in esc.fnwi.uva.nl/thesis/centraal/files/f310232185.pdf ? |
May 25 |
comment |
On the number of consecutive divisors of an integer
Here's a PDF: renyi.hu/~p_erdos/1978-26.pdf |
May 25 |
answered | On the number of consecutive divisors of an integer |
May 14 |
answered | On the natural density of almost perfect numbers |
May 10 |
revised |
Are There Infinitly Many $n$ Which $a\times n!+1$ Be Composite?
deleted extraneous plus sign |
May 9 |
answered | Are There Infinitly Many $n$ Which $a\times n!+1$ Be Composite? |
May 4 |
answered | Powers modulo a fixed integer |
Apr 18 |
comment |
When has the Borel-Cantelli heuristic been wrong?
Another example of (probable) failure in the presence algebraic structure: Pomerance has a heuristic, described at oddperfect.org/pomerance.html, predicting no odd perfect numbers. The same heuristic could be adapted to suggest there are only finitely many even perfect numbers --- but of course we "know better" here, by Euclid. |
Mar 26 |
awarded | Nice Answer |
Mar 26 |
comment |
Numbers $n$ such that the sum of the divisors of $n$ is a nontrivial power
Let me supplement my previous comment by saying that it's not realistic at this time to expect definitive results for fixed $a$. For example, consider the equation $\sigma(n) = 6^b$. Probably there are infinitely many solutions $n,b$, since if $p=2\cdot 6^m-1$ is prime, then $n=2p$ gives rise to a solution. On the other hand, it seems hopeless to prove at this time that there are infinitely many solutions. For then there are infinitely many $p$ with $p+1$ a $3$-smooth number. We don't know anything nearly this strong! The ``smoothest'' we can get $p+1$ is $p^{0.2931}$ (Baker-Harman)! |
Mar 25 |
answered | Does there exist an integer that is both solitary and almost perfect? |
Mar 25 |
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Numbers $n$ such that the sum of the divisors of $n$ is a nontrivial power
Here is a comment on the fixed $a$ problem: One can show that there are only finitely many proper prime powers $p^e$ for which $\sigma(p^e)$ is supported on a given finite set of primes. See, e.g., the book of Shorey and Tijdeman on Exponential Diophantine Equations. This means that the problem in some sense reduces to understanding the shifted primes $\sigma(p)=p+1$. For instance, $\sigma(n) = 2^b$ has infinitely many solutions $(n,b)$ if and only if there are infinitely many Mersenne primes. |
Mar 24 |
answered | Numbers $n$ such that the sum of the divisors of $n$ is a nontrivial power |
Feb 11 |
comment |
Waring problem for binomial coefficients (generalization of Gauss' Eureka Theorem)
As Gjergji has pointed out, this is a consequence of a general theorem of Kamke. A modern reference is Nathanson's Elementary methods in Number Theory; see Chapter 11, especially Theorems 11.10 and 11.12. |
Feb 1 |
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Surveys of the items of Erdős' “toolbox”
The recent volume that came out of the Erdős centennial conference might also be of interest to you: springer.com/new+%26+forthcoming+titles+%28default%29/book/… |