# About One Problem in Wildt 2018

The W32 in the 28th József Wildt International Mathematical Competition is as follows:

$$\text{If }x_{0}=1\text{ and }x_{n+1}^3+1=(x_{n}+1)^3\text{ for all }n\geq0\text{, then }$$ $$[x_{n}]\!=\mspace{-2mu}n\text{ for all }n\geq1\text{, when }\![\cdot]\mspace{-1mu}\text{ denote the integer part.}$$

It has been solved in this post. Hence we know that $$\lim\limits_{n\to\infty}\dfrac{x_{n}}{n}=1$$ or $$x_n\sim{n}$$ as $$n\to\infty$$. But what can we know about $$\mathtt{L}\mathop{:=}\lim\limits_{n\to\infty}(x_{n}-n)$$? Since $$\lceil{x_n}\rceil=n+1$$ for $$n\in\mathbb{N}$$, the problem is only evaluating $$\lim\limits_{n\to\infty}\langle{x_n}\rangle$$ or determining $$x_n\sim n+\mathtt{L}$$ as $$n\to\infty$$, where numerical experiment shows that in fact $$0.6<\mathtt{L}<0.7$$.
However, can one express $$\mathtt{L}$$ in terms of some other well-known constants (if it exists)? Thanks.

It's probably not a well known constant, but we can study its asymptotes.

Write $$y_n = x_n - n$$, so that $$(n + y_n)^3 + 1 = (n + y_{n - 1})^3$$ holds for all $$n \ge 1$$.

As in the linked post, we can prove by induction that $$y_n > \frac 1{n + 1}$$.

Moreover, from $$(n + y_n)^3 < (n + y_n)^3 + 1 = (n + y_{n - 1})^3$$ we get $$y_n < y_{n - 1}$$. So the sequence $$(y_n)_n$$ decreases and is bounded below (by zero), hence converges to a limit $$L$$.

Writing $$z_n = y_{n - 1} - y_n$$, we have $$z_n = \frac 1{(n + y_{n - 1})^2 + (n + y_{n - 1})(n + y_n) + (n + y_n)^2}\tag 0$$ which is contained in the interval $$[\frac 1{3(n + 1)^2}, \frac 1{3n^2}]$$.

Thus we have $$L = y_n + \sum_{k = n + 1}^\infty z_k$$, hence $$y_n = L - \sum_{k = n + 1}^\infty z_k = L - \frac 1{3n} + O(\frac 1{n^2})$$ where we use the result $$z_k \in [\frac 1{3(k + 1)^2}, \frac 1{3k^2}]$$.

Putting this asymptote of $$y_n$$ back into $$(0)$$, we get $$z_n = \frac 1{3n^2}(1 - \frac{2L}n + O(\frac 1{n^2}))$$; we can then use this again with Euler-Maclaurin formula to get even higher asymptotes; etc. The procedure continues and gives rise to an asymptotic expansion of $$y_n$$ (and hence $$x_n$$).