16
$\begingroup$

Do either of the two infinite products $~\displaystyle\prod_{p~\in~\mathbb P}\bigg(1+\frac{2^2}{p^2}\bigg)~$ and $~\displaystyle\prod_{p~\in~\mathbb P}^{p~>~2}\bigg(1-\frac{2^2}{p^2}\bigg)~$ possess a closed

form expression, where $\mathbb P$ represents the set of all primes ?


If not, do they at least possess a number-theoretical interpretation, like the Feller-Tornier constant, for instance, whose decimal expansion can be found on OEIS ?

The motivation behind this question can be found here.

The multiplicative inverse of $~\displaystyle\prod_{p~\in~\mathbb P}^{p~>~2}\bigg(1\pm\frac4{p^2}\bigg)~$ can be written as $~\displaystyle\sum_{n=0}^\infty\frac{(\mp~4)^{\Omega(2n+1)}}{(2n+1)^2}~,~$ where $\Omega(k)$ represents the total number of prime factors of k, counted with multiplicity.

  • 2
    Let $\frac{1}{p^2}=x$. We my notice that $$ 1+4x^2 = (1+x^2)^4 (1-x^4)^6 (1+x^6)^{20} (1-x^8)^{60} (1+x^{10})^{204}\cdots $$ where the exponents are the elements of the sequence [A027377][1] (number of irreducible polynomials with degree $n$ over $\mathbb{F}_4$). So by setting $x=\frac{1}{p}$ and multiplying over $p\in\mathscr{P}$ we have: $$ \prod_{p\in\mathscr{P}}\left(1+\frac{4}{p^2}\right)=\left(\frac{\zeta(2)}{\zeta(4)}\right)^4 \zeta(4)^{-6}\left(\frac{\zeta(6)}{\zeta(12)}\right)^{20}\zeta(8)^{-60}\left(\frac{\zeta(10)}{\zeta(20)}\right)^{204}\cdots $$ [1]: https://oeis.org/A027377 – 2017-02-09
  • 0
    I am not sure about closed form, but asymptotic definitely https://en.wikipedia.org/wiki/Mertens%27_theorems – 2017-02-18
  • 0
    @JackD'Aurizio: Could we at least ascribe a number-theoretical meaning to it, similar to that of the [Feller-Tornier constant](http://mathworld.wolfram.com/Feller-TornierConstant.html), for instance ? – 2017-02-18
  • 2
    In general, $~\displaystyle\sum_{n=0}^\infty\frac{(-1)^{\Omega(2n+1)}}{(2n+1)^k} ~=~ \Big(1+2^{-k}\Big)~\frac{\zeta(2k)}{\zeta(k)}.~$ – 2017-02-24
  • 0
    A [related](http://math.stackexchange.com/questions/1070645) question. – 2017-02-25

0 Answers 0