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2:12 PM
@Vepir Hi, I was actually able to arrive here with my computer, with which I have serious problems since about a week. Some files are damaged, and I do not know how to repair them. Did you notice my last question ?
 
@Peter I don't think so, link?
 
1
Q: Can I accelerate the trial division because of the large exponents?

PeterI want to find a factor of the number $$3^{3^{14}}+3^{3^{13}}+1$$ and I wonder whether the large exponents ($\ 3^{14}\ $ and $\ 3^{13}\ $) allow an acceleration of the trial division. A primality test and methods like pollard-rho or ECM are slow for this number because it has $$2\ 282\ 057$$ digi...

@TheSimpliFire Hi, do you want to search for a factor of a new large prime candidate ?
 
@Peter I am currently running your code.
 
2:32 PM
@TheSimpliFire Thank you !
 
@Peter Do you know if something is known about sums of combinations from consecutive primes?
 
You mean something like 101+103+107+109 ? I do not expect useful results for this.
 
More generally, I want to find a closed from for the following problem, when $A$ is the set of primes;
Define triangle entries (coefficients) $T(A;k,n)$, as the length of the longest consecutive run (streak) of sums, among all $k$-subset ($k$-combination) sums of first $n$ elements of set $A$. Specially, define $T(A;0,n)=1$. We have $k=0,\dots,n$ and $n=0,1,2,3,\dots$ natural numbers. Observe, by definition, $T(A;k,n)\le \binom{n}{k}$.


If $A=\mathbb N$, then we have a closed form:

$$ T(\mathbb N;k,n)= k(n-k)+1$$

If $A=\{2k+1,k\in\mathbb N\}\cup\{2\}=A_{\{2\}}$ is the set of odd numbers and number two (set of the first prime and all numbers not divisible by it), then we have the closed form:
 
How can I reactivate chat-jax ?
 
@Peter I have the following function saved in my browser:

javascript:(function(){if(window.MathJax===undefined){var script = document.createElement("script");script.type = "text/javascript";script.src = "https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.1/MathJax.js?config=TeX-AMS_HTML";var config = 'MathJax.Hub.Config({' + 'extensions: ["tex2jax.js"],' + 'tex2jax: { inlineMath: [["$","$"],["\\\\\\\\\\\\(","\\\\\\\\\\\\)"]], displayMath: [["$$","$$"],["\\\\[","\\\\]"]], processEscapes: true },' + 'jax: ["input/TeX","output/HTML-CSS"]' + '});' + 'MathJax.Hub.Startup.onload();';if (window.o
(For starting ChatJax)
 
@TheSimpliFire I just have to click at this link ?
 
@Peter There are instructions on that page that help activate Jax.
 
it works ! Thank you
@TheSimpliFire Did you pass $10^{10}$ ?
 
3:10 PM
@Peter Yes
5
Q: Can $4\cdot n!-4n+1$ be a perfect square when $n>4$?

NextWhen $n=4,4\cdot n!-4n+1=4\times24-4\times4+1=9^2.$ I wonder if $4\cdot n!-4n+1$ can be a perfect square when $n>4$? I searched $n$ from $5$ to $10000$ but no qualified number was found. I know that: (1) $n$ is even, since $-4n+1 \equiv1 \mod 8$; (2) if $p|4n-1$ and $p\leq n/2$ then $...

@Mathphile @Haran as well
 
3:21 PM
@TheSimpliFire No further square upto $n=10^5$
 
3:43 PM
I passed $n=370\ 000$ without finding another perfect square.
 
4:01 PM
@TheSimpliFire Where did the routine arrive ?
 
Hi]
 
@Peter It finished at 10^10 as I didn't add on any more increments
 
I have some progress
Assuming that $4n-1 \neq p,3p,5p,7p$ for some prime $p$, we can see that all prime factors of $4n-1$ will be lesser than $\frac{n}{2}$. Then, for sufficiently large $n$, we will have $(4n-1)^2 \mid n!$. Thus, we will have $$4n!-4n+1=(4n-1)(\frac{4n!}{4n-1}-1)$$ and the second factor is clearly coprime to $4n-1$ which forces $4n-1$ to be a perfect square. However, this is impossible as $4n-1 \equiv 3 \pmod{4}$
 
4:19 PM
Will read after dinner
 
@TheSimpliFire Why did you stop at $10^{10}$ ?
 

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