Euler product for $\eta(z)^2 \eta(2z) \eta(4z) \eta(8z)^2$

I was looking up a modular forms online: $S_3^{new}\big(\chi_8(3, \cdot)\big)$ it can be written as an Eta product:

$$f(z) = \eta(z)^2 \eta(2z) \eta(4z) \eta(8z)^2 = q \prod_{n=1}^\infty (1 - q^n)^2 (1 - q^{2n}) (1 - q^{4n}) (1 - q^{8n})^2$$

However, $f(z)$ is a newform, and a cusp form, and a Hecke eigenform. The L-function also has an Euler product. If I take the coefficients and turn it into an L-function:

\begin{eqnarray*}L(f , s) &=& 1^{-s} - 2 \cdot 2^{-s} - 2 \cdot 3^{-s} + 4 \cdot 4^{-s} + 4 \cdot 6^{-s} - 8 \cdot 8^{-s} + 5 \cdot 9^{-s} + \dots \\ \\ &=& \; ?\;?\;?_3\times \big( 1 + 5^{-s}\big)^{-1} \times \big( 1 + 7^{-s}\big)^{-1} \times \dots \end{eqnarray*}

Can we write down the Euler product up to a larger prime like 13 or 31 ?

The $f(q)$ is the generating function of the OEIS sequence A030207 which has more information including the multiplicativity of the coefficients. $\;a(n)$ is multiplicative with $\;a(2^e) = (-2)^e,\;$ $a(p^e) = \frac12(1+(-1)^e) p^e\;$ if $\;p \equiv 5, 7 \pmod 8,$ $a(p^e) = a(p) a(p^{e-1}) - p^2 a(p^{e-2})$ if $p \equiv 1, 3 \pmod 8\;$ where $\;a(p) = 4 x^2 -2 p\;$ and $\;p = x^2 + 2 y^2.$
• I am not sure what you mean. You can search on the first few coefficients (such as $1,-2,-2,4,0,4,0,-8$ of the q-series and you will find it. I also have a lot of experience with eta-product/quotients and have contributed many of them to the OEIS. This particular example is one I am familiar with. – Somos Oct 15 '17 at 2:36