Odd Chebyshev, part 1

Question

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).

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.

Answer 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), 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 $a$.

Answer 2

Let me complement the @FedorPetrov's general result by the cases of small value $a$. This will also clarify possible vague impressions from the thread. The general pattern should soon emerge even from cases with relatively small $\ a.\ $ Later, I may expand my list.

(My computations didn't use any computer).

Let a triple of odd natural numbers $\ (a\ b\ c)\ $ such that $\ a\le b\ $ and $\ a+b = c-1\ $ satisfy:

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

Then, and only then, I'll say that $\ (a\ b\ c)\ $ produces arPP (produces a right prime product)

Case $\ a=3:\ $ there are no triples $\ (3\ b\ c)\ $ which produce arPP;

Case $\ a=5:\ $ there are exactly three different triples $\ (5\ b\ c)\ $ which produce arPP, visually:

$\qquad \frac{15!!}{5!!\cdot 9!!}\ =\ 11\cdot 13 $

$\qquad \frac{17!!}{5!!\cdot 11!!}\ =\ 13\cdot 17 $

$\qquad \frac{19!!}{5!!\cdot 13!!}\ =\ 17\cdot 19 $

Case $\ a=7:\ $ there are exactly two different triples $\ (7\ b\ c)\ $ which produce arPP, visually:

$\qquad \frac{107!!}{7!!\cdot 99!!}\ =\ 101\cdot 103\cdot 107 $

$\qquad \frac{109!!}{7!!\cdot 101!!}\ =\ 103\cdot 107\cdot 109 $

Case $\ a=9:\ $ there are no triples $\ (9\ b\ c)\ $ which produce arPP.

Case $\ a=11:\ $ there are exactly two different triples $\ (11\ b\ 105)\ $ which produce arPP, visually:

$\qquad \frac{107!!}{11!!\cdot 95!!}\ =\ 97\cdot 101\cdot 103\cdot 107 $

$\qquad \frac{109!!}{11!!\cdot 97!!}\ =\ 101\cdot 103\cdot 107\cdot 109 $

To summarize it:

THEOREM There are exactly seven different triples $\ (a\ b\ c)\ $ which produce arPP for $\ a\ $ such that $\ 1<a\le 11$.