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The primorial $p_{n}\#$ is given by the product $p_n\# = \prod_{k=1}^n p_k$ (where $p_{k}$ is the $k$th prime) -- is there a natural (a la the gamma function $\Gamma(z)$) way of interpolating it for arguments not necessarily a natural number? (or in $\mathbb{C}$?)

I tried starting with the following definition of the gamma function:

$$\Gamma(z) = \lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)} = \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}$$

My first thought was to modify the Pochhammer symbol in the denominator:

$$\Gamma_{?}(z) = \lim_{n \to \infty} \frac{p_{n}\# \; p_{n}^z}{z \; (z+p_{1})\cdots(z+p_{n})}$$

But this clearly doesn't work, because the primes aren't regularly spaced.

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  • $\begingroup$ Maybe first you should decide what's "natural" about the Gamma function. $\endgroup$ – Gerry Myerson Feb 6 '12 at 10:08
  • $\begingroup$ Or consider what might a function that is the generating function for the primorial (or its reciprocal) might look like, and then apply Cauchy's differentiation formula. $\endgroup$ – J. M. is a poor mathematician Feb 6 '12 at 10:16
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    $\begingroup$ There's a bunch of computer code entitled "Variants of primorial and lcmultorial extended to be continuous over the positive reals" at phodd.net/gnu-bc/code/orialc.bc - maybe it will mean more to others than it means to me. $\endgroup$ – Gerry Myerson Jul 24 '12 at 22:52
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    $\begingroup$ @deoxygerbe it might be related,recently I asked about the possible existence of a primorial $p_i\#$ after two consecutive given factorials in the worst of cases, (n)! and (n+2)! ;Will Jagy and Daniel Fisher provided a very nice demonstration of it. If by interpolation in this question you mean an approximation, then it would be possible to approximate the value by the closer factorials before and after the primorial $\frac{(n+2)!-n!}{2}$ (of course the value is not accurate as in the answer of Draks).So just in case,please disregard if unrelated: math.stackexchange.com/a/1778478/189215 $\endgroup$ – iadvd May 17 '16 at 9:32
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Take the log of $p_n\# = \prod_{k=1}^n p_k$ to get $$ \log p_n\# = \sum_{k=1}^n \log p_n, $$ where you recognize the first Chebyshev function $\theta(n)$, which has an asymptotic behaviour of $\theta(n)\sim n$. Write the sum as integral and use $$ \int_2^x f(t) d(\pi(t))=f(t)\pi(t)\biggr|_{2}^{x}-\int_{2}^{x}f'(t)\pi(t)dt. $$ from here to get: $$ \begin{eqnarray} \sum_{k=1}^n \log p_n &=& \int_2^n \log k\; d\pi(k)\\ &=& \log(k)\pi(k)\biggr|_{2}^{n}-\int_{2}^{n}\frac1k \pi(k)dk. \end{eqnarray} $$ Now, put in your favorite representation for $\pi(x)$, like $ \pi(x) = \operatorname{R}(x^1) - \sum_{\rho}\operatorname{R}(x^{\rho}) , $ with $ \operatorname{R}(x) = \sum_{n=1}^{\infty} \frac{ \mu (n)}{n} \operatorname{li}(x^{1/n})\;$ and $\rho$ running over all the zeros of $\zeta$, to get $\log p_n\#\;$.

Choose, for example, the approximation $\pi(n)\sim \frac{n}{\log n}$, then you get $$ \log p_n\# \sim \log(k)\frac{k}{\log k}\biggr|_{2}^{n}-\int_{2}^{n}\frac1k \frac{k}{\log k}dk = (n-1)-\text{Li}(x) \;. \tag{$*$} $$ Exponentiate $(*)$ and you are done...

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