I'm playing with p-adic valuations, and find that the odd harmonic sums, $\tilde{H}_k=\sum_{i=1}^{k}\frac{1}{2i-1}$, has 2-adic valuation $||k^2||_2=2||k||_2$.
E.g.) $\tilde{H}_4=\frac{176}{85}$ has $||\tilde{H}_4||_2=||176||_2=4=2||4||_2$.
$\tilde{H}_6=\frac{6508}{3465}$ has $||\tilde{H}_6||_2=2$. $||\tilde{H}_8||_2=||\frac{91072}{45045}||_2=6$. $\cdots$
How can I prove this?
(Some simple observation I have tried: As $||H_k||_2 =-r$ for $2^r\leq k <2^{r+1}$, $||H_k||_2=||H_{2k}||_2+1=||\frac{1}{2}H_k||_2$. Thus ultrametric ineq for $\tilde{H}_k=H_{2k}-\frac{1}{2}H_k$ doesn't help at all, yielding trivial result $\tilde{H}_k\in \mathbb{Z}_2$.)
(That $||\tilde{H}_k||_2=2||k||_2$ for odd $k$ is almost trivial, so the problem can be reduced to show $||\tilde{H}_{2k}||_2=||\tilde{H}_k||_2+2$, still have no idea)
(I've checked this holds up to $\sim 1000$ using Mathematica, and Mathematica spits the result almost immediately for $n\sim 1000$ and takes about a minute for $n\sim 10000$. )
For anyone interested : I've used the following Mathematica code, try it yourself
a = Table[
IntegerExponent[
Numerator[HarmonicNumber[2 n] - HarmonicNumber[n]/2], 2]/2, {n,
10000}];
b = Table [IntegerExponent[n, 2], {n, 10000}];
NonNegative[Min[a - b]]
Only a partial result. I note $v_2$ the $2$-adic valuation.
Let $\displaystyle Q_k(x)=\prod_{i=1}^{k}(x-j)=x^{k}+b_{k-1}x^{k-1}+\cdots+b_0$. We have $b_l\in \mathbb{Z}$. Note that $b_{k-1}=-\frac{k(k+1)}{2}$. Put $P_k(x)=2^kQ_k(\frac{x}{2})$. Then $P_k(x)=x^{k}+2b_{k-1}x^{k-1}+\cdots+2^{k-1}b_1 x+2^k b_0=\prod_{j=1}^{k}(x-2j)$.
Now $$\frac{P_k^{\prime}(x)}{P_k(x)}=\sum_{j=1}^k \frac{1}{x-2j}$$ Putting $x=1$, we get that $v_2(\tilde{H_k})=v_2(P_k^{\prime}(1))$.
Now suppose that $k$ is odd. Then $P_k^{\prime}(1)=k+2q$ with $q\in \mathbb{Z}$. Hence your result is true in this case (as you have noted).
Suppose that $k=2m$ with $m$ odd.Then the first two terms in $P_k^{\prime}(1)$ give $ (2m-(2m(2m+1)(2m-1))=4m(1-2m^2)$, and it is easy to see that $8$ divide the other terms. Hence your result is true in this case.