How does reduced homology make these formulas more concise? In rotman's Algebraic Topology, it is stated that using reduced homology these two statements:
 $H_p(S^0) = 
\begin{cases}
\Bbb Z \oplus \Bbb Z,  & \text{if $p = 0$ } \\
0, & \text{if $p \gt 0$ }
\end{cases}$
$H_p(S^n) = 
\begin{cases}
\Bbb Z,  & \text{if $p = 0$ or $p=n$ } \\
0, & \text{otherwise }
\end{cases}$
can be stated more concisely as
$\tilde H_p(S^n) = 
\begin{cases}
\ \Bbb Z,  & \text{if $p = n$ } \\
0, & \text{otherwise }
\end{cases}$
I don't necessarily understand why this is, what exactly is the relationship between the first two equations and the third one?  How is the third one a more concise statement of the first two?
The first two differ from the third when $n=0, p=0$ where $\tilde H_0(S^0) = \Bbb Z$ and $H_0(S^0) = \Bbb Z \oplus \Bbb Z$ and at $n \neq 0, p=0$ where $\tilde H_0(S^n)=0$ and $H_0(S^n) = \Bbb Z$.
What exacly is the relationship between these here?
 A: As you've noted, for all $n$ we have $H_0(S^n)=\tilde H_0(S^n)\oplus\Bbb Z$. Therefore, again by what you've already written (i.e. considering the cases $n=0$ and $n\neq0$ separately), you see that $\tilde H_0(S^0)=\Bbb Z$, and $\tilde H_p(S^0)=H_p(S^0)=0$ for $p>0$. Furthermore, you have $\tilde H_0(S^n)=0$, and for $p>0$, $\tilde H_p(S^n)=H_p(S^n)$, which is $\Bbb Z$ when $p=n$ and $0$ otherwise.
Putting these two cases together, you see that for all $n$ we have $\tilde H_p(S^n)=\Bbb Z$ if and only if $p=n$, and $\tilde H_p(S^n)=0$ otherwise. This is precisely what the "concise" form of the statement says.
Am I mistaken about what's confusing you here? Even after typing my answer I don't feel like I've added much to what you already said.
A: So the reason the third is more concise than the first two is simply numerical: two formulas vs one formula. The second equation is written down for all $p$ and $n$, without having to single out any special cases.
More specifically the zeroth homology counts the number of paths components of a space, $H_0(S^n)=\Bbb{Z}$ for $n>0$ because the spheres are path connected, while for $n=0$, $S^0=\{\pm1\}$ it is not path connected and has two components and so $H_0(S^0)=\Bbb{Z}\oplus \Bbb{Z}$.
Since a non-empty space has at least one path component, there will be at least one factor of $\Bbb{Z}$ for $H_0$. The reduced homology $\widetilde{H}$ takes this into account essentially by adding an extra $\Bbb{Z}$ into the chain complex, so before we had:
$$\ldots H_2(X)\to H_1(X)\to H_0(X)\to 0$$
we now have with exactness preserved:
$$\ldots \widetilde{H}_2(X)\to \widetilde{H}_1(X)\to \widetilde{H}_0(X)\to \Bbb{Z}\to 0$$
and that $H_0(X)=\widetilde{H}_0(X)\oplus \Bbb{Z}$, and $H_n(X)=\widetilde{H}_n(X)$ for $n>0$.
I haven't supplied all the details but Hatcher has a nice explanation of this, you can just ctrl+F Reduced homology. I think there's various reasons why you do this other than it looks more concise for spheres, as in it may allow you to state results without special cases, but maybe someone else can comment on that.
