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English is not my native tongue. German is not my native tongue. French is not my native tongue.
1h

answered  Should one post a paper on the arXiv if it is not intended to be published? 
3h

comment 
number theory which is close to analysis
I second this. I thought of the same book, and I am pleased I am not alone! 
7h

answered  Tauberian theorem with better error term 
Mar 27 
comment 
Brocard's problem
No, you cannot say that all primes $<\sqrt{z}$ are in $z$ and $z1$, because this is usually not true, and it is definitely false when $z(z1)=n!/4$ and $n\geq 10$. This is because the largest prime factor of $n!/4$ is at most $n$, while $\sqrt{z}>(z(z1))^{1/4}=(n!/4)^{1/4}>2n$ for $n\geq 10$. You create a condition that contradicts Brocard's equation, and you will not get anything useful from that. Just take my word for it if you don't see this. 
Mar 27 
comment 
Brocard's problem
I see you included your motivation, namely Brocard's problem. If $x^2=n!+1$ and $z=(x+1)/2$ as in your post, it is not clear why $z(z1)$ is divisible by all primes less than $\sqrt{z}$. In fact it seems straightforward to prove that this condition holds true if and only if $n=7$ is the largest solution to Brocard's problem. This is because $\sqrt{z}>(n!/4)^{1/4}$, and this is much larger than the first prime after $n$ (unless $n$ is very small like $n=7$). So your condition on $z$ contradicts Brocard's equation unless $n$ is very small. 
Mar 27 
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Brocard's problem
@jim198810: I used SAGE to check all $z\leq 1763$. My original program focused on $A>1$, while my second (or third) program focused on $A<0$. At any rate, I don't have more time for this, and your question is definitely not of research level. 
Mar 27 
comment 
Brocard's problem
@jim198810: I did not read your proof, because it looked problematic from the very beginning. Finding a counterexample is easier than to explain what is wrong with the proof. At any rate, this site is not about discussing purported proofs, how to fix them etc. 
Mar 27 
revised 
Brocard's problem
added 335 characters in body 
Mar 27 
comment 
Brocard's problem
@jim198810: Actually, I did find three counterexamples: $z=4,10,16$. 
Mar 27 
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Brocard's problem
@jim198810: You said earlier the condition is that $z(z1)$ is divisible by all primes less than $\sqrt{z}$. For $z=18$ this says that $18*17$ is divisible by all primes less than $\sqrt{18}$, i.e. by the primes $2$ and $3$, which is true. The primes between $5$ and $17$ are irrelevant for your condition when $z=18$. 
Mar 27 
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Brocard's problem
@jim198810: Do not rewrite the question. Instead, explain what is wrong with my counterexamples. 
Mar 27 
comment 
Brocard's problem
@jim198810: It is not true for your special $z$'s. See my second added section. 
Mar 27 
revised 
Brocard's problem
added 244 characters in body 
Mar 27 
comment 
Brocard's problem
@jim198810: Your condition on $z$ implies that $z\leq 1763$. This is a small number of cases, hence not too interesting, and certainly not of research level. See my added section for details. 
Mar 27 
revised 
Brocard's problem
added 607 characters in body 
Mar 27 
revised 
Brocard's problem
added 166 characters in body 
Mar 27 
answered  Brocard's problem 
Mar 27 
comment 
Etale cohomology approach on $\tau(n)$
@JeremyRouse: You need to raise to the $24$th power to get $\sum_{n=1}^\infty\tau(n)q^{n1}$. But your reasoning is fine as $k(3k1)/2$ is a quadratic sequence. 
Mar 26 
revised 
Etale cohomology approach on $\tau(n)$
I fixed the big oh notation, the symbol $\sim$ was not right there. 
Mar 25 
revised 
Source for equations involving congruences of Fibonacci and Lucas numbers
added 37 characters in body 