Denote $$f(s;s_1,s_2,\ldots,s_k)=\sum_{n=1}^\infty\frac1{4^n}\binom{2n}n\frac{H_n^{(s_1)}\cdots H_n^{(s_k)}}{n^s}$$ Can $f(\cdots)$ always be represented as $\mathbb Q$-linear combination of alternating Euler sum of weight $W$? Here $s$ and $s_i$ are positive integers and $W=s+\sum_i s_i$. Alternating Euler sums are defined as $$\sum_{n=1}^\infty (-1)^{l(n-1)}\frac{H_n^{(s_1)}\cdots H_n^{(s_k)}}{n^s},$$ where $l=0,1$ and $s,s_i\in\mathbb Z^+$.
By Editor: Examples and related things: Here, here, here, here, here, here, here.
By Editor again: Harder examples and generalizations: Here, here, here, here.
Numerical Evidence and Motivation
I did some calculation of this binomial sum of small $n$'s. The table in the appendix is a part of that.
I also did some research on alternating Euler sum. The constants involved in the alt. Euler sums are exactly the same when $n\le4$. I randomly chose and evaluated some sums of weight $5$ and found that their constants are again same.
Possible Conversion
Although I noticed $\frac1{4^n}\binom{2n}n=\frac2\pi\int_0^1(4x-4x^2)^{n-1/2}dx$, I'm unsure about how to convert it into the whole sum into an integral and then (possibly making some substitution?) convert it into the form of alternating Euler sum.
Appendix
The number in the blanks are the coefficients.
\begin{array}{|c|c|} \hline
&\ln2\\\hline
f(1)&2\\\hline
\end{array}
\begin{array}{|c|c|c|}\hline
&\zeta(2)&\ln^22\\\hline
f(2)&1&-2\\\hline
f(1;1)&2&0\\\hline
\end{array}
\begin{array}{|c|c|c|c|} \hline
&\zeta(3)&\zeta(2)\ln2&\ln^32\\\hline
f(3)&2&-2&\frac43\\\hline
f(2;1)&\frac92&-4&0\\\hline
f(1;2)&\frac32&0&0\\\hline
f(1;1^2)&\frac{21}2&0&0\\\hline
\end{array}
\begin{array}{|c|c|c|c|c|c|} \hline &\zeta(4)&\zeta(3)\ln2&\zeta(2)\ln^22&\ln^42&\operatorname{Li}_4(1/2)\\\hline f(4)&\frac94&-4&2&-\frac23&0\\\hline f(3;1)&-\frac{13}4&-2&2&\frac13&8\\\hline f(2;2)&3&-3&0&0&0\\\hline f(2;1^2)&-14&7&-8&\frac43&32\\\hline f(1;3)&\frac{37}4&-7&2&-\frac13&-8\\\hline f(1;2,1)&\frac{49}4&-7&2&-\frac13&-8\\\hline f(1;1^3)&\frac{115}4&35&-10&\frac53&40\\\hline \end{array}