Let $p$ be a prime number and $l$ be the greatest prime number less than $2p$. Moreover, let $m,t_i<p$ be positive integer numbers such that $\sum_{i=1}^mit_i=2p$. Is it possible that $1^{t_1}2^{t_2}\ldots m^{t_m}t_1!t_2!\ldots t_m!\mid(2p-l)!$
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$\begingroup$ Did you mean $\ell !& or something? $2p-1$ could be prime. $\endgroup$– alpogeAug 9, 2015 at 21:17
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$\begingroup$ @alpoge, I take the question to be, do there exist $p$, $m$, and $t_i$ such that .... You are pointing out that if $2p-1$ is prime then $m$ and $t_i$ don't exist, but what if $2p-1$ is not prime? $\endgroup$– Gerry MyersonAug 10, 2015 at 0:14
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$\begingroup$ @GerryMyerson the question asks if for some prime $p$ we express him as $\sum i\cdot t_i$ ....then is it possible that $1^{t_1}...t_m!\mid (2p-l)!$ One example can answer the question in a positive way. $\endgroup$– Konstantinos GaitanasAug 10, 2015 at 7:24
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6$\begingroup$ @nina The number-theory tag already indicates that your question is about number theory. In future, it would be better to use a more descriptive title. $\endgroup$– Joe SilvermanAug 10, 2015 at 11:42
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1 Answer
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It is possible. Take $p=61$ and $t_1=2$, $t_{120}=1$. Then $l=113$ and $1^{t_1}\cdot t_1!\cdot 120^{t_{120}}\cdot t_{120}!=240$ divides $(2p-l)!=9!=362880$.
Another example: $p=673$, $t_1=t_{672}=2$. Then $l=1327$ and $1^{t_1}\cdot t_1!\cdot 672^{t_{672}}\cdot t_{672}! = 1806336$ divides $(2p-l)!=19!$.
UPDATE. If all $t_1,\dots,t_m$ are required to be (strictly) positive, then there are likely no required examples. Numerically I tested this for primes $p$ below $10^5$. For large $p$ this can be assessed heuristically as follows.
By Cramer's conjecture, $2p-l$ is asymptotically bounded by $\log(p)^2$. We also have that $m!$ divides $1^{t_1}\cdot 2^{t_2}\cdots m^{t_m}$, which in turn divides $(2p-l)!$, and thus $m\leq 2p-l$. Furthermore, since $\sum_{i=1}^m it_i=2p$, at least one of $t_i$ must be at least $$\frac{2p}{1+2+\dots+m}\geq \frac{4p}{(m+1)^2}\geq \frac{4p}{(2p-l+1)^2}.$$ That is, $t_1!\cdots t_m!\mid (2p-l)!$ would asymptotically imply $\frac{4p}{\log(p)^4}\leq \log(p)^2$, which is not possible.
answered Aug 10, 2015 at 10:34
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2$\begingroup$ But the $t_i$ are supposed to be positive integers for $i$ from $1$ to $m$. If $t_{120}=1$, then $t_2,t_3,\dots,t_{119}$ have to be positive integers as well, and then $\sum it_i\dots$. $\endgroup$ Aug 10, 2015 at 13:29
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$\begingroup$ @GerryMyerson: Good point (if the "positive" is not in French meaning). For primes $p$ below $10^5$, there are no examples with $t_1, \dots, t_m$ all being positive. $\endgroup$ Aug 10, 2015 at 15:50