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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.

Lucian
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    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

     – Jack D'Aurizio Feb 09 '17 at 12:40
  • I am not sure about closed form, but asymptotic definitely https://en.wikipedia.org/wiki/Mertens%27_theorems – rtybase Feb 18 '17 at 22:28
  • @JackD'Aurizio: Could we at least ascribe a number-theoretical meaning to it, similar to that of the Feller-Tornier constant, for instance ? – Lucian Feb 18 '17 at 23:13
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    In general, $~\displaystyle\sum_{n=0}^\infty\frac{(-1)^{\Omega(2n+1)}}{(2n+1)^k} ~=~ \Big(1+2^{-k}\Big)~\frac{\zeta(2k)}{\zeta(k)}.~$ – Lucian Feb 24 '17 at 21:34
  • A related question. – Lucian Feb 25 '17 at 03:13

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