<s>[INST] Smallest $n$ such that $H_{n,2^{-1}} \geq n$ where $H_{n,2^{-1}}$ are generalized Harmonic numbers Consider the generalized harmonic numbers evaluated at $2^{-1}$ $$H_{n,2^{-1}} = 1+ {1 \above 1.5pt \sqrt{2}}+ \ldots+ {1 \above 1.5pt \sqrt{n}}$$ The table below lists some initial values:
$$\begin{array}{nc|cccccc}
n &1&2&3&4&5&6&7 \\ \hline
H_{n,2^{-1}} & 1.00 & 1.71 &2.28&2.78&3.23&3.64&4.01\\ 
\end{array}$$
Let $$s =min\{n\text{ }|\text{ }H_{n,2^{-1}} \geq n \}$$ For example the smallest $n$ such that $H_{n,2^{-1}}\ge 1$ is $1$. Similarly the smallest $n$ such that $H_{n,2^{-1}}\ge 2$ is $3$. We have the following sequence for $s$
 $$\mathfrak a(s) =1,3,5,7,10,14,18,22,\ldots$$ I am asking if the following claim is true - 

$$\mathfrak a(s) = \sum_{n\leq s}\Bigg(\Bigg\lfloor{n+2  \above 1.5pt
 4}\Bigg\rfloor+\Bigg\lfloor{n+1  \above 1.5pt 4}\Bigg\rfloor\Bigg)$$ 

Note that $\mathfrak a(s)$ is the sequence A054040. 
 [/INST] Note that
$$H_{n,2^{-1}}=\int_1^{n+1}\frac{dt}{\sqrt{\lfloor t\rfloor}}$$
So
$$\int_1^{n+1}\frac{dt}{\sqrt{t}}\le H_{n,2^{-1}}\le1+\int_2^{n+1}\frac{dt}{\sqrt{t-1}}$$
That is,
$$2(\sqrt{n+1}-1)\le H_{n,2^{-1}}\le 1+2(\sqrt{n}-1)$$
For example, $1998 \le H_{999,999,\,2^{-1}} < 1999$. This means that $\mathfrak a(1998)\le 999,999 < \mathfrak a(1999)$.
But
$$\begin{align}\sum_{n\le 1999}&\left(\left\lfloor\frac{n+2}4\right\rfloor+\left\lfloor\frac{n+1}4\right\rfloor\right)\\
&=\sum_{n\le 1999}\frac{2n+3}4\\
&-\left(\frac12+\frac34\right)\cdot499\qquad\text{for $n=4k+1$}\\
&-\frac34\cdot499\qquad\text{for }n=4k+2\\
&-\frac14\cdot499\qquad\text{for }n=4k+3\\
&-\left(\frac12+\frac14\right)\cdot498\qquad\text{for }n=4k\\
&=999,413<999,999
\end{align}$$
</s>