Asymptotic vlaue of $ f(n)=\sum_{i=0}^n\lfloor \sqrt{i}\rfloor\binom{n}{i} $ Inspired by this question I tried to find an asymptotic formula for
$$
f(n)=\sum_{i=0}^n\lfloor \sqrt{i}\rfloor\binom{n}{i}
$$
With the observation:
$$
f(n)=\sum_{i=0}^n\frac{\lfloor \sqrt{i}\rfloor+\lfloor \sqrt{n-i}\rfloor}{2}\binom{n}{i}
$$
And the a bit naive approximation 
$$
\frac{\lfloor \sqrt{i}\rfloor+\lfloor \sqrt{n-i}\rfloor}{2}=\frac{1}{n}\sum_{i=1}^{n}{\lfloor \sqrt{i}\rfloor}=\frac{n\lfloor \sqrt{n}\rfloor-\frac{1}{6}\lfloor \sqrt{n}\rfloor\left(\lfloor \sqrt{n}\rfloor-1\right)\left(2\lfloor \sqrt{n}\rfloor+5\right)}{n}\approx\frac{2}{3}\sqrt{n}
$$
We obtain $f(n)\approx\frac{2}{3}\sqrt{n}2^n=:g(n)$. The approximation doesn't seem that bad; I computed with wolfram alpha:
$$
$$
\begin{array}{c|c|c|c}
n & f(n) & g(n) & f(n)/g(n) \\
\hline
1 & 1 & 1.333 & 0.75 \\
2 & 3 & 3.771 & 0.795 \\
3 & 7 & 9.238 & 0.758 \\
4 & 16 & 21.333 & 0.75 \\
5 & 37 & 47.703 & 0.776 \\
10 & 1882 & 2158.782 & 0.872 \\
50 & 5.128\cdot10^{15} & 5.308\cdot10^{15} & 0.996 \\
100 & 8.391\cdot10^{30} & 8.451\cdot10^{30} & 0.993 \\
200 & 1.531\cdot10^{61} & 1.515\cdot10^{61} & 1.011 \\
350 & 2.929\cdot10^{106} & 2.86\cdot10^{106} & 1.024 \\
\end{array}
$$
$$
The error term fluctuates a bit so it seems natural to ask:
Is it true that:
$$
\lim_{n\to\infty}{\frac{f(n)}{g(n)}}=1
$$
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{\mathrm{f}\pars{n}} & =\sum_{k = 0}^{n}\left\lfloor \root{k}\right\rfloor{n \choose k}\ =\
1\times\overbrace{\bracks{{n \choose 1} + {n \choose 2} + {n \choose 3}}}
^{\ds{\color{#f00}{3}\ terms}}
\\[3mm] & +
2 \times\overbrace{\bracks{{n \choose 4} + {n \choose 5} + {n \choose 6} + {n \choose 7} + {n \choose 8}}}
^{\ds{\color{#f00}{5}\ terms}}
\\[3mm] & +
3 \times\overbrace{\bracks{{n \choose 9} + {n \choose 10} + {n \choose 11} + {n \choose 12} + {n \choose 13}+ {n \choose 14}+ {n \choose 15}}}
^{\ds{\color{#f00}{7}\ terms}}\ +\ \cdots
\\[5mm] &=
1 \times \sum_{k = 1}^{3}{n \choose k} +
2 \times \sum_{k = 4}^{8}{n \choose k} +
3 \times \sum_{k = 9}^{15}{n \choose k} + \cdots
\\[3mm] & =
\sum_{m = 1}^{\infty}m\sum_{k = m^{2}}^{m^{2} + 2m}{n \choose k} =
\sum_{m = 0}^{\infty}m\sum_{k = 0}^{2m}{n \choose k + m^{2}} =
\sum_{k = 0}^{\infty}\sum_{m\ \geq\ k/2}^{\infty}m{n \choose k + m^{2}}
\end{align}

I stop here. Some of the binomial sum seems to be related to Hypergeometric Functions. So far, I was able to get rid of the $\ds{\left\lfloor\cdots\right\rfloor}$ function and along the way to find a pattern.

