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Nov 23, 2020 at 18:34 comment added Milo Moses @Wojowu ah, that makes more sense. Thank you for clarifying
Nov 23, 2020 at 10:50 comment added Wojowu @MiloMoses That's true, I phrased myself poorly. Rather what I meant is that this is what allows the Euler factors to be contractible into a form you see for many common L-functions, namely $P(p^{-s})^{-1}$ for some polynomial $P$.
Nov 23, 2020 at 6:25 vote accept Milo Moses
Nov 23, 2020 at 5:47 answer added Manipulator timeline score: 1
Nov 23, 2020 at 3:30 comment added Milo Moses @Wojowu do you mean that these relations are what allows them to admit functional equations? Any Dirichlet series that is totally multiplicative should have an Euler product
Nov 22, 2020 at 22:19 comment added Kimball See also: mathoverflow.net/a/287761/6518
Nov 22, 2020 at 21:21 comment added Wojowu Such relations between coefficients are precisely what enable the Dirichlet series to admit an Euler product, and near all L-functions have them. Perhaps the most famous of such is given by the Ramanujan conjectures on the $\tau$ function.
Nov 22, 2020 at 20:39 comment added Milo Moses @Wojowu I assumed this wasn't the case since this implied too beautiful of a connection between $a_p$, $a_{p^2}$, and the higher order coefficients $a_{p^n}$. L functions never cease to amaze me.
Nov 22, 2020 at 20:23 comment added Milo Moses Thank you! This makes a lot of sense.
Nov 22, 2020 at 20:22 comment added Wojowu $a_n$ is the $n$-th coefficient of the Dirichlet series, i.e. we write the product as $\sum a_n/n^s$. With this notation, the coefficient in the product is indeed $a_p^2-a_{p^2}$.
Nov 22, 2020 at 20:07 history asked Milo Moses CC BY-SA 4.0