Does this sequence associated to the prime numbers have a name and/or where can I learn more about it? If I understand correctly, the following are true:


*

*Over a finite field, two elliptic curves are isogeneous iff they have the same number of elements.

*If $p$ is prime, then over $\mathbb{F}_{p^2}$, all supersingular elliptic curves are isogeneous.

*(I think that) for all primes $p$, it holds that there's a supersingular elliptic curve over $\mathbb{F}_{p^2}.$


Putting these together, I surmise that there's a map
$$\mathbb{P} \rightarrow \mathbb{N}$$
that maps each prime $p$ to the number of elements of any (and hence every) supersingular elliptic curve over $\mathbb{F}_{p^2}$. This yields a sequence of natural numbers.

Question. Does this sequence have a name and/or where can I learn more about it?

 A: As cited in Constructing supersingular elliptic curves by Broker, a theorem of Waterhouse implies that a supersingular elliptic curve with trace of Frobenius $t$ (hence $p^2 + t + 1$ points) exists over $\mathbb{F}_{p^2}$ iff either


*

*$t = \pm 2p$

*$t = \pm p$ and $p \not \equiv 1 \bmod 3$, or

*$t = 0$ and $p \not \equiv 1 \bmod 4$.


This contradicts your claim. The problem with your argument is that Tate's isogeny theorem implies that two elliptic curves are isogenous over $\mathbb{F}_q$ iff $|E_1(\mathbb{F}_q)| = |E_2(\mathbb{F}_q)|$, but when people say that the supersingular isogeny graph is connected they are referring to isogenies over $\overline{\mathbb{F}_q}$. If such an isogeny is defined over $\mathbb{F}_{q^n}$ then we can only conclude that $|E_1(\mathbb{F}_{q^n})| = |E_2(\mathbb{F}_{q^n})|$.
Explicitly, consider over $\mathbb{F}_{p^2}$ a supersingular elliptic curve $E_1$ with $t = 2p$, so $(p + 1)^2$ points, and another supersingular elliptic curve $E_2$ with $t = -2p$, so $(p - 1)^2$ points. In the first case the eigenvalues of Frobenius over $\mathbb{F}_{p^2}$ are $p, p$ and in the second case they are $-p, -p$, from which it follows that
$$|E_1(\mathbb{F}_{p^4})| = |E_2(\mathbb{F}_{p^4})| = p^4 + 2p^2 + 1 = (p^2 + 1)^2$$
so $E_1$ and $E_2$ are isogenous over $\mathbb{F}_{p^4}$ but not $\mathbb{F}_{p^2}$. 
Similarly, when $t = \pm p$ the eigenvalues of Frobenius are $\pm p \omega, \pm p \omega^2$ where $\omega$ is a primitive third root of unity so we get an isogeny to $E_1$ or $E_2$ over $\mathbb{F}_{p^6}$ and to both over $\mathbb{F}_{p^{12}}$, and when $t = 0$ the eigenvalues of Frobenius are $pi, -pi$, so we get an isogeny to $E_1$ and $E_2$ over $\mathbb{F}_{p^8}$. 
