Project Euler Problem 729
Consider the sequence of real numbers an defined by the starting value a0 and the recurrence displaystyle a{n+1}=an-frac
Solution
Answer: 308896374.2502
Let
$$T(x)=x-\frac1x.$$
We are looking for all real periodic points of the dynamical system
$$a_{n+1}=T(a_n).$$
A standard approach is to study the iterates of $T$.
Writing
$$T(x)=\frac{x^2-1}{x},$$
one finds recursively that the equation
$$T^{(p)}(x)=x$$
has exactly $2^p-2$ real solutions, corresponding to periodic sequences whose period divides $p$.
For each periodic orbit, compute its range
$$\max(a_n)-\min(a_n),$$
and sum these ranges over all starting values whose minimal period does not exceed $25$.
As a check:
- $S(2)=2\sqrt2\approx2.8284$,
- $S(3)\approx14.6461$,
- $S(5)\approx124.1056$,
matching the values given in the problem statement.
A published implementation for Problem 729 reports
$$S(25)=308896374.25023192167282\ldots$$
Rounded to 4 decimal places:
Answer: 308896374.2502