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Excluding 1 being neither prime nor composite, I found the following closed forms for infinite products of composite numbers (happy to share how I derived these).

$$\displaystyle \prod _{composites}^{\infty } \left( \dfrac{{c}^{2}} {{c}^{2}-1}\right) = \frac{12}{\pi^2}$$

and

$$\displaystyle \prod _{composites}^{\infty } \left( \dfrac{{c}^{2}} {{c}^{2}+1}\right) = \frac{30}{\pi \sinh(\pi)}$$

Easy to see that multiplying both gives:

$$\displaystyle \prod _{composites}^{\infty } \left( \dfrac{{c}^{4}} {{c}^{4}-1}\right) = \frac{360}{\pi^3 \sinh(\pi)}$$

The latter is a product that very rapidly converges.

Are there any other closed forms for these "Euler-type" products of composites known?

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Still not sure that I understand the question clearly, but for any integer values $k>1$,

Products of the form: $$\prod_{c\text{ composite}}\frac{c^k}{c^k-1}$$

Can be evaluated in terms of a product of Gamma functions evaluated at roots of unity, a zeta constant, and a suitable rational scalar.

For example the case for when $k=2n$ is an even integer is given by,

$$\prod_{c\text{ composite}}\frac{c^{2n}}{c^{2n}-1}=\frac{(2n)!(4n)(-\pi i)^{n-1}}{\text{B}_{2n}(2\pi)^{2n}}\prod_{m=1}^{n-1}\csc(\pi e^{i\pi m/n})$$

For all natural numbers $n$, where $\text{B}_{n}$ are the Bernoulli numbers.

And for the special case $n=1$ the product on the right is simply assumed to be empty.

The key idea in obtaining this and similar products is to take note upon two ideas,

One, that given:

$$\prod_{n=2}^\infty\frac{n^s}{n^s-1}=f(s)$$

Can be evaluated in terms of the gamma function when $\text{s}$ is an integer greater then one.

And two, that through the use of the Euler product representation of the Riemann zeta function:

$$\prod_{p\text{ prime}}\frac{p^s}{p^s-1}=\zeta(s)$$

We can take the corresponding ratios of the two functions so that the primes occurring in $f$ are canceled out by the primes occurring in $\zeta$, leaving us with a product ranging over the composite numbers, that is:

$$\prod_{c\text{ composite}}\frac{c^s}{c^s-1}=\frac{f(s)}{\zeta(s)}$$

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  • $\begingroup$ Thanks, Ethan. A while ago I found parts of your evaluation (mathoverflow.net/questions/53266/…), but have never seen it explained so clearly! This time I used the relation $\prod_{n=1}^\infty \left(1- \frac{s}{a + i n} \right) \left(1- \frac{s}{{a - i n}} \right) = \frac{\xi_{int}(0-a+s)}{\xi_{int}(0-a)}$ with $\xi_{int}(z) = \frac{\sinh(\pi z)}{z}$ and split the $n$-factors into primes and composites. $a=1$ gives the middle closed form above. $a=0$ its inverse. Could it work for other $a$? $\endgroup$ – Agno Oct 22 '13 at 9:13
  • $\begingroup$ Forgot to say in the comment above, that $s=1$ to give the middle closed form in my question. $\endgroup$ – Agno Oct 22 '13 at 16:38

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