Some computations here in wolfram alpha which i did show that :$$\sigma(4^n) \bmod 4=3$$ is true at a least from $n=1$ to $5000$ , really i don't know how do i can factorise $4^n$ over $\sigma $ sum of power divisor function .
My question here is: Is this true for $n \geq 1:\sigma(4^n) \bmod 4=3$ and how do i show it if it is?
Note: I checked that for some studies about periodicity of divisibilty among power of sum divisor function .
The divisors of $4^n=2^{2n}$ are $2^k$ for $k=0,1,\dots,2n$, hence if $n\geq 1$ then $$ \sigma(4^n)=\sum_{k=0}^{2n}2^k=2^{2n+1}-1\equiv 3\mod 4$$ since $4$ divides $2^{2n+1}$ when $n\geq 1$.