Today I was reading LMFDB (the L-functions and Modular Forms DataBase), and I came across something that confused me. When discussing degree 3 L functions on [this](https://www.lmfdb.org/L/degree3/) page, they assert that all the ones found so far have Euler products of the form


$$L(s)=\prod\_{p|N}\left(1-a\_np^{-s}+\left(a\_p^2-a\_{p^2}\right)p^{-2s}\right)^{-1}\prod\_{p\nmid N}\left(1-a\_pp^{-s}+\chi(p)\overline{a\_p}p^{-2s}-\chi(p)p^{-3s}\right)^{-1}$$


What I do not understand is why they chose to write $a\_p^2-a\_{p^2}$ instead of simply $a\_{p^2}$. There is not other reference to $a\_{n}$ anywhere else on the page and no information is given about $a\_n$, and so I assume they are meant as arbitrary complex numbers, and so writing $a\_{p^2}$ instead of $a\_p^2-a\_{p^2}$ would be just as complete and lose no generality. This hints to me that perhaps there are some restrictions (say, $\Re(a\_n)>0$) that are not being stated which calls for such a statement. Perhaps there is also "moral" reason to write it this way.


I note also that this is not an isolated phenominon on the LMFDB website. On the page for degree 4 L functions [here](https://www.lmfdb.org/L/degree4/) they assert that all known L functions of degree four have Euler products of the form


$$L(s)=
\prod\_{p|N} \left(1-a\_p\, p^{-s} + (a\_p^2 - a\_{p^2})\, p^{-2s} - (a\_p^3 - 2 \, a\_{p^2} \, a\_p + a\_{p^3} ) \, p^{-3s}\right)^{-1}\cdot\prod\_{p\nmid N} \left(1-a\_p\, p^{-s} + (a\_p^2 - a\_{p^2})\, p^{-2s} - \chi(p) \, \overline{a\_p} \, p^{-3s} +\chi(p) \, p^{-4s}\right)^{-1}
$$


which one again uses the notation $a\_p^2-a\_{p^2}$, but now also uses the expression $a\_p^3-2a\_{p^2}a\_p+a\_{p^3}$ instead of $a\_{p^3}$ which would lose no generality. Any insights are appreciated.



Why does LMFDB refer to L functions having coefficients of type $a_p-a_{p^2}$ instead of just $a_{p^2}$?
This seems to be related to taking a power-series truncation in a reciprocal.


The $p$th Euler factor can be written as
$$\sum\_{k=0}^\infty {a\_{p^k}\over p^{ks}}$$


The reciprocal of this is in general a polynomial of degree less than or equal to the degree of the $L$-function. When the prime is good it is equal, and it is less when $p|N$.


When $d=3$ and $p|N$ the above sum is thus $(1+A/p^s+B/p^{2s})^{-1}$ for some $A$ and $B$ (which could possibly be 0). This in turn is
$$1-A/p^s+(A^2-B)/p^{2s}+O(p^{-3s}),$$
and equating coefficients with the above sum gives the expression, as $-A=a\_p$ and $A^2-B=a\_{p^2}$.


