Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

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


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

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?

share|cite|improve this question

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

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

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,

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

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:


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:


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:


share|cite|improve this answer
Thanks, Ethan. A while ago I found parts of your evaluation (…), 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 $a$? – Agno Oct 22 '13 at 9:13
Forgot to say in the comment above, that $s=1$ to give the middle closed form in my question. – Agno Oct 22 '13 at 16:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.