Growth of the "cube of square root" function 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 ?.
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.

It's not too hard to put a bound on the size of second differences (since without the truncation, they are bounded above by a constant times $n^{-1/2}$), but getting the bound down to one seems delicate.  It looks like it can be done with mindless brute force, though.  I won't write all of the cases, but here is a start.  Write $n = t^2 + r$, for integers $t,r$ satisfying $|r| \leq t$.  The binomial theorem implies $n^{3/2} = t^3 + (3/2)tr + (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)r^4/t^5 - \dots$.
I'll look at the case where $t$ is even.  Then


*

*$f(n) = t^3 + (3/2)tr + \lfloor (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)r^4/t^5 - \dots \rfloor$

*$f(n+1) = t^3 + (3/2)t(r+1) + \lfloor (3/8)(r+1)^2/t - (1/16)(r+1)^3/t^3 + (3/128)(r+1)^4/t^5 - \dots \rfloor$

*$f(n+2) = t^3 + (3/2)t(r+2) + \lfloor (3/8)(r+2)^2/t - (1/16)(r+2)^3/t^3 + (3/128)(r+2)^4/t^5 - \dots \rfloor$.


$g(n)$ then has no contributions from the first two terms of each series, since they cancel.  Therefore:
$g(n) = \lfloor (3/8)r^2/t - (1/16)r^3/t^3 + (3/128)(r+1)^4/t^5 - \dots \rfloor +$
$\qquad + 2\lfloor (3/8)r^2/t + (3/4)r/t - (1/16)(r+1)^3/t^3 + (3/8t) + (3/128)(r+2)^4/t^5 - \dots \rfloor +$
$\qquad + \lfloor (3/8)r^2/t + (3/2)r/t - (1/16)(r+2)^3/t^3 + (3/2t) + (3/128)(r+2)^4/t^5 - \dots \rfloor$.
At this point, you can break the analysis into more cases involving the fractional part of $(3/8)r^2/t$ and the size of $r/t$, and then invoke some estimates about the remaining parts of the sum.
