**QUESTION** Find all triples of odd natural numbers $\ a < b\ $ and $\ c\ $ such that $\ a+b = c-1\ $ and


$$ \frac {c!!}{a!!\cdot b!!}\ =\ \frac {P(c)}{P(b)} $$


where $\ P(x) \ $ is the product of all primes $\le x$.



> 
> The above fraction on the left looks somewhat similar to binomial coefficients but they are not integers in general; in the odd case of this double !! expression the numerator and the denominators are more balanced (or at least their totality).
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**EXAMPLE**


$$ \frac {45!!}{15!!\cdot 29!!}\,\ =\,\ 31\cdot 37\cdot 41\cdot 43\,\ 
=\,\ \frac{P(45)}{P(29)} $$



> 
> Perhaps there are only a finite number of such odd fractions which produce an exact prime product as in the question. Are all of them among fractions
>  $$ \frac {(6\cdot n-3)!!}{(2\cdot n-1)!!\,\cdot\,(4\cdot n-3)!!} $$
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**EDIT** It follows from @GerhardPaseman's comments in this thread that a necessary condition for a required triple $\ a<b\ $ and $\ c\ $ is following: if
$\ b<n\le c\ $ is not a prime then all prime divisors of $\ n\ $ are $\ \le\ a$, for all such $\ n$.


Thus there should be a lot of such triples. However, despite everybody's common sense, it is still not certain at this moment that there are infinitely many of such solutions.



Odd Chebyshev, part 1
Let me prove that there are only finitely many solutions for $a>1$. Assume the contrary, then for large $a$ we may suppose $b/a\to \lambda$, where $\lambda\geqslant 1$ is either finite constant or $+\infty$ (for fixed $a>1$ there are finitely many solutions of course). Using Stirling approximation (or its proof) and PNT, which says that the product of primes between $b$ and $c$ behaves as $e^{a+o(a)}$, we get the equation $\lambda\log(1+1/\lambda)+\log(1+\lambda)=2$, thus $\lambda=2.23\dots$. In this case consider a prime $p$ slightly greater than $a/3$. Its square divides $(b+2)\dots c$ (the product contains both $7p$ and $9p$), but does not divide $a!!$, so divides $c!!/a!!b!!$. A contradiction. 


If $\lambda$ tends to infinity, we use some upper estimate for the number of primes between $c$ and $b$. At first, we have $\log(b)/\log(a)\to 1$, since almost all numbers between $b$ and $c$ are composite and otherwise their product would exceed $a!!$. Next, by the result of Huxley (or Heath-Brown, or whatever, many results suffice, see [the references here](https://projecteuclid.org/download/pdf_1/euclid.mmj/1029003189)), we have $\pi(c)-\pi(b)\sim \pi(a)$ in this case. Thus the product of composite numbers between $b$ and $c$ is at least $b^{a/2-2a/\log a}>a^{a/2}>a!!$ for large $a$.


