Hello all, this question is a variant (and probably a more difficult one)
of a (promptly answered ) question that I asked here, at [Is it true that all the "irrational power" functions are almost polynomial ?](https://mathoverflow.net/questions/18054/is-it-true-that-all-the-irrational-power-functions-are-almost-polynomial).


For $n\geq 1$, let $f(n)$ denote the "integer part" 
(largest integer below) $n^{\frac{3}{2}}$, and let
$g(n)=f(n+2)-2f(n+1)+f(n)$. 
 Question : Is it true that $g(n)$ is always in $\lbrace -1,0,1\rbrace$ 
(excepting the initial value $g(1)=2$) ? I checked this up to
$n=100000$.


It is not too hard to check that if $t$ is large enough compared to $r$ (say
$t\geq \frac{3r^2+1}{4}$), 
$f(t^2+r)$ is exactly $t^3+\frac{3rt}{2}$ (or $t^3+\frac{3rt-1}{2}$ if $t$
and $r$ are both odd ) and similarly 
$f(t^2-r)$ is exactly $t^3-\frac{3rt}{2}$ (or $t^3-\frac{3rt+1}{2}$ if $t$
and $r$ are both odd ). So we already know that the answer is "yes" for
most of the numbers.



Growth of the "cube of square root" function
Here is a proof that $|g(n)|\le 1$ for all but finitely many $n$. You can extract an explicit bound for $n$ from the argument and check the smaller values by hand.


If $f(n)=n^{3/2}$ without the floor, then $g(n)\sim \frac{3}{4\sqrt n}$, so it is positive and tends to 0. When you replace $n^{3/2}$ by its floor, $g(n)$ changes by at most 2, hence the only chance for failure is to have $g(n)=2$ when the fractional parts of $n^{3/2}$ and $(n+2)^{3/2}$ are very small and the fractional part of $(n+1)^{3/2}$ is very close to 1 (the difference is less than $\frac{const}{\sqrt{n}}$).


Let $a,b,c$ denote the nearest integers to $n^{3/2}$, $(n+1)^{3/2}$ and $(n+2)^{3/2}$. Then $c-2b+a=0$ because it is an integer very close to $(n+2)^{3/2}-2(n+1)^{3/2}+n^{3/2}$. Denote $m=c-b=b-a$. Then $(n+1)^{3/2}-n^{3/2}<m$ and $(n+2)^{3/2}-(n+1)^{3/2}>m$. Observe that
$$
 \frac{3}{2}\sqrt{n}<(n+1)^{3/2} - n^{3/2} < \frac{3}{2}\sqrt{n+1}
$$
(the bounds are just the bounds for the derivative of $x^{3/2}$ on $[n,n+1]$. Therefore
$$
\frac{3}{2}\sqrt{n} < m < \frac{3}{2}\sqrt{n+2}
$$
or, equivalently,
$$
 n < \frac49 m^2 < n+2.
$$
If $m$ is a multiple of 3, this inequality implies that $n+1=\frac49 m^2=(\frac23m)^2$, then $(n+1)^{3/2}=(\frac23m)^3$ is an integer and not slightly smaller than an integer as it should be. If $m$ is not divisible by 3, then
$$
 n+1 = \frac49 m^2 + r
$$
where $r$ is a fraction with denominator 9 and $|r|<1$. From Taylor expansion
$$
 f(x+r) = f(x) +r f'(x) +\frac12 r^2 f''(x+r\_1), \ \ 0<r\_1<r,
$$
for $f(x)=x^{3/2}$, we have
$$
 (n+1)^{3/2} = (\frac49 m^2 + r)^{3/2} = \frac8{27}m^3 + mr + \delta
$$
where $0<\delta<\frac1{4m}$.
This cannot be close to an integer because it is close (from above) to a fraction with denominator 27.


