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

I 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: [["$","$"],["\\\\\\$","\\\\$$/extract_itex]"]], displayMath: [["",""],["\\\\[","\\$$"]], processEscapes: true },' + 'jax: ["input/TeX","output/HTML-CSS"]' + '});' + 'MathJax.Hub.Startup.onload();';if (window.o
(For starting ChatJax)

2:54 PM

@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

When $n=4,4\cdot n!-4n+1=4\times24-4\times4+1=9^2.$ 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}\$ ?