$$\frac{2(4^n-1)}{3(2n+1)}$$ it's all in the title.. I have just seen this on a forum, and I wondered whether it is true, and why. I'm not well versed in number theory.
I tried to find a counterexample using c++, but the number becomes very large very fast.
Suppose $2n+1$ is prime. By Fermat's Little Theorem: $$2^{2n}\equiv 1 \pmod {2n+1}$$
this is the key, since this gives $(2n+1)\mid (4^n-1)$.
In $\dfrac {2(4^n-1)}{3(2n+1)}$, $2$ in the numerator is useless.
Also since $3$ and $2n+1$ are coprime (for $n\ne 1$), and $3\mid (4^n-1)$, the result follows, (except for $n=1$ where the expression equals $\frac23$.)
