Why does LMFDB refer to L functions having coefficients of type $a_p-a_{p^2}$ instead of just $a_{p^2}$?

Question

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 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 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.

Answer 1

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 $A^2-B=a_{p^2}$.