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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


1d
comment On the number of consecutive divisors of an integer
Here's a PDF: renyi.hu/~p_erdos/1978-26.pdf
1d
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
comment 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
comment 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/…
Dec
24
comment Ruth-Aaron triples, etc
It seems that, contra Hoffman, Erdos never claimed to prove the infinitude of Ruth--Aaron pairs. After the original Ruth--Aaron paper appeared in the Journal of Recreational Mathematics, Pomerance learned of Erdos's interest from a letter (not a phone call). In the letter, Erdos says he can prove that the Ruth--Aaron numbers have density $0$. Far from claiming a proof that there are infinitely many Ruth--Aaron numbers, Erdos says that problem "seems hopeless" ! The original letter is available online --- see the very first scan at cah.utexas.edu/collections/math_erdos.php
Nov
24
comment Siegel Walfisz Theorem for algebraic number fields
An analogue of of Siegel--Walfisz for Hecke characters can be found in the following paper: Goldstein, Larry Joel: A generalization of the Siegel-Walfisz theorem. Trans. Amer. Math. Soc. 149 1970 417–429.
Nov
10
comment A conjecture of Erdos on consecutive differences of primes
The problem is solved by recent work of Banks--Freiberg--Turnage-Butterbaugh. Using the method of Maynard--Tao, they show that the sequence of e_k's contains arbitrarily long runs of 0's, as well as arbitrarily long runs of 1's. Hence, the sum has a nonperiodic binary expansion and so represents an irrational number.
Oct
30
awarded  Nice Answer
Oct
25
comment Effective version of the Bombieri-Vinogradov theorem
For effective variants of Bombieri--Vinogradov, see e.g. this paper of Akbary and Hambrook: cs.uleth.ca/~akbary/Akbary-Hambrook.pdf or Lemma 5.2 of this paper of Lenstra and Pomerance: math.dartmouth.edu/~carlp/PDF/complexity12.pdf
Oct
25
awarded  Nice Answer