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Let $\ \mathbb N:= \{1\ 2\ \ldots\}\ $ be the set of natural numbers. Let $\ \mathbb P:=\{2\,\ 3\,\ 5\,\ 7\,\ 11\,\ \ldots\}\ $ be the set of primes. Then natural radical $\ rad(n)\ $ is

$$ rad(n)\ :=\ \prod\,\{p\in\mathbb P: p\,|\,n\} $$

for every $\ n\in\mathbb N$. Define

$$ \rho\ :=\ \sum_{N\in\mathbb N}\,\frac 1{n\cdot rad(n)} $$

hence $\ \rho > \frac{\pi^2}6.\ $ Furthermore, it'd be an exercise for 1' year college students (or advanced high school students) to show that $\ \rho<\infty.\ $ Don't tell them too early that

$$\ \rho\ =\ \prod_{p\in \mathbb P}\,\left(1+\frac 1{p\cdot(p-1)}\right) $$

QUESTION $\ $ Can you find (yourself or a reference) an algerbaic expression, possibly in terms of $\pi$ and/or $e$ for $\rho$.

Something like $\ \frac{\pi^{\frac 52}}6$ or similar.

$$ \rho(s)\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot (rad(n))^s} $$

for every complex $\ s\in\mathbb C\ $ such that it's real part is positive,$\ \Re(s)>0.\ $ Thus

$$ \rho(s)\ =\ \prod_{p\in\mathbb P}\, \left(1+\frac 1{p^s\cdot(p-1)}\right) $$

QUESTION A $\ $What would be the compact expressions for $\ \rho(n)\ $ for all or as many natural values of $\ n\ $ as possible?

QUESTION B $\ $ What are the other formulas for $\ \rho(s)\ $ in the style of multiplicative number theory.

QUESTION C $\ $ How does the complex analytic extension of $\ \rho(s)\ $ look like?

$\qquad\qquad\qquad$ NOTES

Here is my third personal proof (:-) of the infinitude of primes:

THEOREM (Euclid) $\ |\mathbb P| = \infty$

PROOF $\ $ We see from the product representation of $\ \rho\ $ that $\ \rho<\infty\ $ in general, and especially if $|\mathbb P|<\infty.\ $

On the other hand, when $\ |\mathbb P|<\infty$ then $\ r:=\prod\mathbb P\in\mathbb N.\ $ Then

$$ \rho\ >\ \sum_{k=1}^n\,\frac 1{k\cdot rad(k)}\ \ge\ \frac 1r \cdot\sum_{k=1}^n\,\frac 1k\ \longrightarrow\ \infty $$

when $\ n\rightarrow\infty.\ $ Assumption $\ |\mathbb P|<\infty\ $ has lead us to a contradiction.

END of proof

$\qquad\qquad\qquad$ ADDENDUM

Let me copy @Lucia's formula from their first comment below. You may also read Lucia's second comment related to a paper by Bateman. In the derivation below I applied @Wojowu's observation from a comment below, which has drastically simplified and shortened one of the steps.

THEOREM (@Lucia) $$ \rho\ =\ \frac {\zeta(2)\cdot\zeta(3)}{\zeta(6)} $$

PROOF

$$ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}\ = \ \frac{p^3+1}{p\cdot(p^2-1)} \cdot \frac{p^6}{p^6-1}\ = $$ $$ \frac{p^2-p+1}{p\cdot(p-1)} \cdot \frac{p^6}{p^6-1}\ = \ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1} $$

i.e. $$ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$

Thus $$ \prod_{p\in\mathbb P}\, \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \prod_{p\in\mathbb P}\,\frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$ or $$ \rho\cdot\zeta(6)\ =\ \zeta(2)\cdot\zeta(3) $$

and the theorem follows. END of proof

Let $\ \mathbb N:= \{1\ 2\ \ldots\}\ $ be the set of natural numbers. Let $\ \mathbb P:=\{2\,\ 3\,\ 5\,\ 7\,\ 11\,\ \ldots\}\ $ be the set of primes. Then natural radical $\ rad(n)\ $ is

$$ rad(n)\ :=\ \prod\,\{p\in\mathbb P: p\,|\,n\} $$

for every $\ n\in\mathbb N$. Define

$$ \rho\ :=\ \sum_{N\in\mathbb N}\,\frac 1{n\cdot rad(n)} $$

hence $\ \rho > \frac{\pi^2}6.\ $ Furthermore, it'd be an exercise for 1' year college students (or advanced high school students) to show that $\ \rho<\infty.\ $ Don't tell them too early that

$$\ \rho\ =\ \prod_{p\in \mathbb P}\,\left(1+\frac 1{p\cdot(p-1)}\right) $$

QUESTION $\ $ Can you find (yourself or a reference) an algerbaic expression, possibly in terms of $\pi$ and/or $e$ for $\rho$.

Something like $\ \frac{\pi^{\frac 52}}6$ or similar.

$$ \rho(s)\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot (rad(n))^s} $$

for every complex $\ s\in\mathbb C\ $ such that it's real part is positive,$\ \Re(s)>0.\ $ Thus

$$ \rho(s)\ =\ \prod_{p\in\mathbb P}\, \left(1+\frac 1{p^s\cdot(p-1)}\right) $$

QUESTION A $\ $What would be the compact expressions for $\ \rho(n)\ $ for all or as many natural values of $\ n\ $ as possible?

QUESTION B $\ $ What are the other formulas for $\ \rho(s)\ $ in the style of multiplicative number theory.

QUESTION C $\ $ How does the complex analytic extension of $\ \rho(s)\ $ look like?

$\qquad\qquad\qquad$ NOTES

Here is my third personal proof (:-) of the infinitude of primes:

THEOREM (Euclid) $\ |\mathbb P| = \infty$

PROOF $\ $ We see from the product representation of $\ \rho\ $ that $\ \rho<\infty\ $ in general, and especially if $|\mathbb P|<\infty.\ $

On the other hand, when $\ |\mathbb P|<\infty$ then $\ r:=\prod\mathbb P\in\mathbb N.\ $ Then

$$ \rho\ >\ \sum_{k=1}^n\,\frac 1{k\cdot rad(k)}\ \ge\ \frac 1r \cdot\sum_{k=1}^n\,\frac 1k\ \longrightarrow\ \infty $$

when $\ n\rightarrow\infty.\ $ Assumption $\ |\mathbb P|<\infty\ $ has lead us to a contradiction.

END of proof

$\qquad\qquad\qquad$ ADDENDUM

Let me copy @Lucia's formula from their first comment below. You may also read Lucia's second comment related to a paper by Bateman. In the derivation below I applied @Wojowu's observation from a comment below, which has drastically simplified and shortened one of the steps.

THEOREM (@Lucia) $$ \rho\ =\ \frac {\zeta(2)\cdot\zeta(3)}{\zeta(6)} $$

PROOF

$$ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}\ = \ \frac{p^3+1}{p\cdot(p^2-1)} \cdot \frac{p^6}{p^6-1}\ = $$ $$ \frac{p^2-p+1}{p\cdot(p-1)} \cdot \frac{p^6}{p^6-1}\ = \ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1} $$

i.e. $$ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$

Thus $$ \prod_{p\in\mathbb P}\,\left( \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\right)\ = \ \prod_{p\in\mathbb P}\, \left(\frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}\right)$$ or $$ \rho\cdot\zeta(6)\ =\ \zeta(2)\cdot\zeta(3) $$

and the theorem follows. END of proof

Let $\ \mathbb N:= \{1\ 2\ \ldots\}\ $ be the set of natural numbers. Let $\ \mathbb P:=\{2\,\ 3\,\ 5\,\ 7\,\ 11\,\ \ldots\}\ $ be the set of primes. Then natural radical $\ rad(n)\ $ is

$$ rad(n)\ :=\ \prod\,\{p\in\mathbb P: p\,|\,n\} $$

for every $\ n\in\mathbb N$. Define

$$ \rho\ :=\ \sum_{N\in\mathbb N}\,\frac 1{n\cdot rad(n)} $$

hence $\ \rho > \frac{\pi^2}6.\ $ Furthermore, it'd be an exercise for 1' year college students (or advanced high school students) to show that $\ \rho<\infty.\ $ Don't tell them too early that

$$\ \rho\ =\ \prod_{p\in \mathbb P}\,\left(1+\frac 1{p\cdot(p-1)}\right) $$

QUESTION $\ $ Can you find (yourself or a reference) an algerbaic expression, possibly in terms of $\pi$ and/or $e$ for $\rho$.

Something like $\ \frac{\pi^{\frac 52}}6$ or similar.

$$ \rho(s)\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot (rad(n))^s} $$

for every complex $\ s\in\mathbb C\ $ such that it's real part is positive,$\ \Re(s)>0.\ $ Thus

$$ \rho(s)\ =\ \prod_{p\in\mathbb P}\, \left(1+\frac 1{p^s\cdot(p-1)}\right) $$

QUESTION A $\ $What would be the compact expressions for $\ \rho(n)\ $ for all or as many natural values of $\ n\ $ as possible?

QUESTION B $\ $ What are the other formulas for $\ \rho(s)\ $ in the style of multiplicative number theory.

QUESTION C $\ $ How does the complex analytic extension of $\ \rho(s)\ $ look like?

$\qquad\qquad\qquad$ NOTES

Here is my third personal proof (:-) of the infinitude of primes:

THEOREM (Euclid) $\ |\mathbb P| = \infty$

PROOF $\ $ We see from the product representation of $\ \rho\ $ that $\ \rho<\infty\ $ in general, and especially if $|\mathbb P|<\infty.\ $

On the other hand, when $\ |\mathbb P|<\infty$ then $\ r:=\prod\mathbb P\in\mathbb N.\ $ Then

$$ \rho\ >\ \sum_{k=1}^n\,\frac 1{k\cdot rad(k)}\ \ge\ \frac 1r \cdot\sum_{k=1}^n\,\frac 1k\ \longrightarrow\ \infty $$

when $\ n\rightarrow\infty.\ $ Assumption $\ |\mathbb P|<\infty\ $ has lead us to a contradiction.

END of proof

$\qquad\qquad\qquad$ ADDENDUM

Let me copy @Lucia's formula from their first comment below. You may also read Lucia's second comment related to a paper by Bateman. In the derivation below I applied @Wojowu's observation from a comment below, which has drastically simplified and shortened one of the steps.

THEOREM (@Lucia) $$ \rho\ =\ \frac {\zeta(2)\cdot\zeta(3)}{\zeta(6)} $$

PROOF

$$ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}\ = \ \frac{p^3+1}{p\cdot(p^2-1)} \cdot \frac{p^6}{p^6-1}\ =\

\frac{p^2-p+1}{p\cdot(p-1}} \cdot \frac{p^6}{p^6-1}\ = \ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ $$

i.e. $$ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$

Thus $$ \prod_{p\in\mathbb P}\, \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \prod_{p\in\mathbb P}\,\frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$ or $$ \rho\cdot\zeta(6)\ =\ \zeta(2)\cdot\zeta(3) $$

and the theorem follows. END of proof

END of Proof

Let $\ \mathbb N:= \{1\ 2\ \ldots\}\ $ be the set of natural numbers. Let $\ \mathbb P:=\{2\,\ 3\,\ 5\,\ 7\,\ 11\,\ \ldots\}\ $ be the set of primes. Then natural radical $\ rad(n)\ $ is

$$ rad(n)\ :=\ \prod\,\{p\in\mathbb P: p\,|\,n\} $$

for every $\ n\in\mathbb N$. Define

$$ \rho\ :=\ \sum_{N\in\mathbb N}\,\frac 1{n\cdot rad(n)} $$

hence $\ \rho > \frac{\pi^2}6.\ $ Furthermore, it'd be an exercise for 1' year college students (or advanced high school students) to show that $\ \rho<\infty.\ $ Don't tell them too early that

$$\ \rho\ =\ \prod_{p\in \mathbb P}\,\left(1+\frac 1{p\cdot(p-1)}\right) $$

QUESTION $\ $ Can you find (yourself or a reference) an algerbaic expression, possibly in terms of $\pi$ and/or $e$ for $\rho$.

Something like $\ \frac{\pi^{\frac 52}}6$ or similar.

$$ \rho(s)\ :=\ \sum_{n\in\mathbb N}\,\frac 1{n\cdot (rad(n))^s} $$

for every complex $\ s\in\mathbb C\ $ such that it's real part is positive,$\ \Re(s)>0.\ $ Thus

$$ \rho(s)\ =\ \prod_{p\in\mathbb P}\, \left(1+\frac 1{p^s\cdot(p-1)}\right) $$

QUESTION A $\ $What would be the compact expressions for $\ \rho(n)\ $ for all or as many natural values of $\ n\ $ as possible?

QUESTION B $\ $ What are the other formulas for $\ \rho(s)\ $ in the style of multiplicative number theory.

QUESTION C $\ $ How does the complex analytic extension of $\ \rho(s)\ $ look like?

$\qquad\qquad\qquad$ NOTES

Here is my third personal proof (:-) of the infinitude of primes:

THEOREM (Euclid) $\ |\mathbb P| = \infty$

PROOF $\ $ We see from the product representation of $\ \rho\ $ that $\ \rho<\infty\ $ in general, and especially if $|\mathbb P|<\infty.\ $

On the other hand, when $\ |\mathbb P|<\infty$ then $\ r:=\prod\mathbb P\in\mathbb N.\ $ Then

$$ \rho\ >\ \sum_{k=1}^n\,\frac 1{k\cdot rad(k)}\ \ge\ \frac 1r \cdot\sum_{k=1}^n\,\frac 1k\ \longrightarrow\ \infty $$

when $\ n\rightarrow\infty.\ $ Assumption $\ |\mathbb P|<\infty\ $ has lead us to a contradiction.

END of proof

$\qquad\qquad\qquad$ ADDENDUM

Let me copy @Lucia's formula from their first comment below. You may also read Lucia's second comment related to a paper by Bateman. In the derivation below I applied @Wojowu's observation from a comment below, which has drastically simplified and shortened one of the steps.

THEOREM (@Lucia) $$ \rho\ =\ \frac {\zeta(2)\cdot\zeta(3)}{\zeta(6)} $$

PROOF

$$ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}\ = \ \frac{p^3+1}{p\cdot(p^2-1)} \cdot \frac{p^6}{p^6-1}\ = $$ $$ \frac{p^2-p+1}{p\cdot(p-1)} \cdot \frac{p^6}{p^6-1}\ = \ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1} $$

i.e. $$ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$

Thus $$ \prod_{p\in\mathbb P}\, \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \prod_{p\in\mathbb P}\,\frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$ or $$ \rho\cdot\zeta(6)\ =\ \zeta(2)\cdot\zeta(3) $$

and the theorem follows. END of proof

Let me copy @Lucia's formula from their first comment below. You may also read Lucia's second comment related to a paper by Bateman. (My proof below is certainly overly pedantic. I just wanted to avoid any mistakes).

PROOF

Equivalently: $\ \rho\cdot\zeta(6)\ =\ \zeta(2)\cdot\zeta(3).\ $ Thus it is enough to prove

$$ \left(1+\sum_{k=2}^\infty p^{-k}\right)\cdot\sum_{k=0}^\infty p^{-6\cdot k}\ = \ \sum_{k=0}^\infty p^{-2\cdot k}\cdot\sum_{k=0}^\infty p^{-3\cdot k} $$

for every $\ p\in\mathbf P$.

Consider the formal equality:

\begin{equation}\label{a} 1 +\sum_{n=2}^\infty\, a_n\cdot x^n\ =\ \left(1+\sum_{k=2}^\infty x^k\right)\cdot\sum_{k=0}^\infty x^{6\cdot k} \end{equation}

Obviously, for $\ n\in\{2\ 3\ 4\ 5\}\ $ we get

$$ a_2=a_3=a_4=a_5= 1$$

For the remaining $\ n\ge 6 $ we may write $\ n=6\cdot t+r\ $ where $\ (t\ r)\ $ is a unique pair of integers such that

$$ r\in\{0\ 7\ 2\ 3\ 4\ 5\} $$

(yes, $\mathbf 7$--not a typo). All contributions from the above righthand product to the summand $\ a_n\cdot x^n\ $ on the left are:

$$ x^{r+6\cdot k} \cdot x^{6\cdot (t-k)} $$

for $k=0\ldots t,\ $ meaning that there are $\ t+1\ $ contributions, i.e. $\ a_n=t+1$.

Now consider the formal equality

\begin{equation}\label{b} 1 +\sum_{n=2}^\infty\, b_n\cdot x^n\ =\ \sum_{k=0}^\infty x^{2\cdot k}\cdot\sum_{k=0}^\infty x^{3\cdot k} \end{equation}

All contributions from the above righthand product to the summand $\ b_n\cdot x^n\ $ on the left are of the form

$$ x^{2\cdot u}\cdot x^{3\cdot v}\ =\ x^{2\cdot(3\cdot k+r)}\cdot x^{3\cdot (2\cdot m+s)} $$

where $\ r=0\ or\ 1\ or\ 2\ $ and $\ s=0\ or\ 2.\ $ Obviously, $\ s\equiv n\ \mod 2,\ $ thus $\ s\ $ is the same for all said contributions for a fixed $\ n.\ $

Also, $\ r\equiv -n\mod 3\ $ is the same for all contributions for a fixed $\ n.\ $

Since $\ r\ $ and $\ s\ $ are fixed for a fixed $\ n,\ $ the value $\ w:=k+m\ $ is fixed too. Indeed, $\ w\,=\, \frac 16\cdot (n-2\!\cdot\! r - 3\!\cdot\! s).\ $ We obtain all the contributions by varying $\ k\ $ from $\ 0$ to $w,\ $ i.e. there are $\ w+1\ $ of them--in other words $\ b_n=w+1$.

Clearly, $\ w=0\ $ for $\ n=2\ 3\ 4\ 5,\ $ hence $\ b_n=a_n=1\ $ for $\ n=2\ 3\ 4\ 5.\ $ Furthermore $$ w = \left\{ \begin{array}{ll} \lfloor\frac n6\rfloor & \mbox{ $r<2\,$ or $\,s<1$} \\ \ & \mbox{ } \\ \lfloor\frac n6\rfloor-1 & \mbox{ $r=2\,$ and $\,s=1$} \end{array}\right . $$

Thus $\ w=t\ $ (see above about $t$ related to $n$), thus $\ b_n=a_n\ $ for every $\ n\ge 6\ $ too.

Let me copy @Lucia's formula from their first comment below. You may also read Lucia's second comment related to a paper by Bateman. In the derivation below I applied @Wojowu's observation from a comment below, which has drastically simplified and shortened one of the steps.

PROOF

$$ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}\ = \ \frac{p^3+1}{p\cdot(p^2-1)} \cdot \frac{p^6}{p^6-1}\ =\

\frac{p^2-p+1}{p\cdot(p-1}} \cdot \frac{p^6}{p^6-1}\ = \ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ $$

i.e. $$ \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$

Thus $$ \prod_{p\in\mathbb P}\, \left(1 + \frac 1{p\cdot(p-1)}\right) \cdot \frac{p^6}{p^6-1}\ = \ \prod_{p\in\mathbb P}\,\frac{p^2}{p^2-1} \cdot \frac{p^3}{p^3-1}$$ or $$ \rho\cdot\zeta(6)\ =\ \zeta(2)\cdot\zeta(3) $$

and the theorem follows. END of proof