<s>[INST] Is the numerator of $\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}$ a power of $2$? I stumbled on something numerically, and was just starting to work on it, but it seemed fun enough to share.

Let $$f(n)=\sum_{k=0}^{n} \frac{(-1)^{k}}{2k+1}\binom{n}{k}$$
It appears, from the first few values, that $f(n)$ always has numerator equal to a power of $2$. Is this true? If so, why?

The first values:
$$\frac{1}{1},
\frac{2}{3},
\frac{8}{15},
\frac{16}{35},
\frac{128}{315},
\frac{256}{693},
\frac{1024}{3003},
\frac{2048}{6435},
\frac{32768}{109395},\\
\frac{65536}{230945},
\frac{262144}{969969},
\frac{524288}{2028117},
\frac{4194304}{16900975}
$$
Alternative ways you can see this value: $$f(n)=\int_0^{1}(1-x^2)^n\,dx=\int_{0}^{\pi/2}\cos^{2n+1} t\,dt=\frac{1}{2^{2n}}\sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{2n+1}{n-k}$$

To summarize some of the comments, the sequence above appears to match:
$$\begin{align}
f(n)&=\frac{(2n)!!}{(2n+1)!!} \\
&= \frac{2}{3}\cdot \frac{4}{5}\cdot \frac{6}{7}\cdots \frac{2n}{2n+1}
\end{align}$$
Thus, if this is correct (and the answer below proves it is,) we have $$f(n)=f(n-1)\cdot \frac{2n}{2n+1}=f(n-1)\left(1-\frac{1}{2n+1}\right).$$
There might be a proof of this recursion using integration by parts for one of the integral forms above.
 [/INST] $$\begin{eqnarray*} f(n) = \sum_{k=0}^{n}\frac{(-1)^k}{2k+1}\binom{n}{k}&=&\int_{0}^{1}\sum_{k=0}^{n}(-1)^k x^{2k}\binom{n}{k}\,dx\\&=&\int_{0}^{1}(1-x^2)^n\,dx\\&=&\frac{1}{2}\int_{0}^{1}z^{-1/2}(1-z)^{n}\,dz\\&=&\frac{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}{2\,\Gamma\left(n+\frac{3}{2}\right)}\\&=&\frac{4^n\,n!^2}{(2n+1)!}\\&=&\frac{2^{2n}}{(2n+1)\binom{2n}{n}}\end{eqnarray*}$$
hence it is enough to check that $\nu_2\left(\binom{2n}{n}\right)<2n$ to prove your conjecture.
As a matter of fact,
$$\nu_2\left(\binom{2n}{n}\right) = \sum_{k\geq 1}\left(\left\lfloor\frac{2n}{2^k}\right\rfloor-2\left\lfloor\frac{n}{2^k}\right\rfloor\right)\leq 2+\log_2(n),$$
since the terms of the last sum can be only $0$ or $1$, and they are zero as soon as $k$ is big enough.
As stated by the OP in the comments, a careful analysis of the previous formula reveals that $\nu_2\left(\binom{2n}{n}\right)$ is just the number of non-zero bits in the binary representation of $n$.
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