This post comes from the suggestion of Joel Moreira in a comment on An alternative to continued fraction and applications (itself inspired by the Numberphile video 2.920050977316 and Fridman, Garbulsky, Glecer, Grime, and Tron Florentin - A prime-representing constant).

Let $u_0 \ge 2$. Consider the recurrence relation $$u_{n+1} = \lfloor u_n \rfloor (u_n - \lfloor u_n \rfloor + 1).$$ If $u_n$ is integral then $u_{n+1}=u_n$. The sequence $(u_n)$ is called eventually integral if there is $n$ such that $u_n$ is integral.

Question: Is the sequence $(u_n)$ eventually integral when $u_0 \ (\ge 2)$ is a rational number?

Remark: It is true for $u_0=\frac{p}{q}$ with $p < 10000$ (see Appendix).

For $u_0=\frac{11}{5}$, then $$(5u_n)=(11, 12, 14, 18, 24, 36, 42, 56, 66, 78, 5 \cdot 24, \dots).$$ Here is a picture of the dynamic:

The general proof could be non-easy, by regarding the example $u_0=\frac{15}{7}$: $$(7u_n) = (15, 16, 18, 22, 24, 30, 36, 40, 60, 88, 132, 234, 330, 376, 636, 1170, 1336, 2470, 4576, 7836, 11190, 17578, 20088, 34428, 44262, 50584, 65034, 102190, 160578, 7 \cdot 39324, \dots)$$

For $u_0=\frac{5789}{2837}$, the sequence $(u_n)$ reaches an integer at $n=27786$. Below is the picture of $\frac{u_{n+1}}{u_n}$ from $n=0$ to $27786$, which looks completely random.

Appendix

In the following list the datum $[r,(p,q)]$ means that the sequence $(u_n)$, with $u_0=\frac{p}{q}$, reaches an integer at $n=r$. The list provides the ones with the longest $r$ according the lexicographic order of $(p,q)$.

Computation

sage: search(10000)
[1, (2, 1)]
[2, (5, 2)]
[3, (7, 2)]
[4, (7, 3)]
[11, (11, 5)]
[30, (15, 7)]
[31, (29, 14)]
[45, (37, 17)]
[53, (39, 17)]
[124, (41, 19)]
[167, (59, 29)]
[168, (117, 58)]
[358, (123, 53)]
[380, (183, 89)]
[381, (201, 89)]
[530, (209, 97)]
[532, (221, 97)]
[622, (285, 131)]
[624, (295, 131)]
[921, (359, 167)]
[1233, (383, 181)]
[1365, (517, 251)]
[1482, (541, 269)]
[2532, (583, 263)]
[3121, (805, 389)]
[3586, (1197, 587)]
[3608, (1237, 607)]
[3860, (1263, 617)]
[4160, (1425, 643)]
[6056, (1487, 743)]
[9658, (1875, 859)]
[9662, (1933, 859)]
[10467, (2519, 1213)]
[10534, (2805, 1289)]
[11843, (2927, 1423)]
[12563, (3169, 1583)]
[13523, (3535, 1637)]
[14004, (3771, 1871)]
[14461, (4147, 2011)]
[17485, (4227, 1709)]
[18193, (4641, 1987)]
[18978, (4711, 2347)]
[22680, (5193, 2377)]
[23742, (5415, 2707)]
[24582, (5711, 2663)]
[27786, (5789, 2837)]
[27869, (6275, 2969)]
[29168, (6523, 3229)]
[32485, (6753, 2917)]
[33819, (7203, 3361)]
[41710, (7801, 3719)]
[49402, (8357, 3863)]

Code

def Seq(p,q):
    x=Rational(p/q)
    A=[floor(x)]
    while not floor(x)==x:
        n=floor(x)
        x=Rational(n*(x-n+1))
        m=floor(x)
        A.append(m)
    return A

def search(r):
    m=0
    for p in range(2,r):
        for q in range(1,floor(p/2)+1):
            A=Seq(p,q)
            l=len(A)
            if l>m:
                m=l
                print([m,(p,q)])