Function of a sequence of numbers You're given a sequence of number: $1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{64}$.
A function takes two numbers in the sequence, say $a,b$ and replaces them both with $a+b+ab$. For example, we could take the initial sequence and apply the operation to the numbers $1$ and $1/64$ to obtain the new list
$$\frac{1}{2}, \frac{1}{3},\cdots, \frac{1}{63}, \frac{33}{32}, \frac{33}{32}$$

Is there a solution so that after some number of steps, all terms in the sequence are equal? How many unique solutions are there?


Note that this question is different from the question asked here, because the number of terms in that question changes, while in this question, the number of terms remains constant ($64$).
By the reasoning in the linked question, we can observe that the operation $a*b = a + b + ab = (a+1)(b+1)-1$ on $\mathbb{Q}$ is identical to the usual multiplication, under the isomorphism $(\mathbb{Q}, *) \to (\mathbb{Q}, \times)$ given by $x \mapsto x+1$.  
 A: The first value is
$a + b + ab
=(1+a)(1+b)-1
$.
The next value,
if $c$ is combined with this,
 is
$(1+c)(1+(1+a)(1+b)-1)-1
=(1+c)(1+a)(1+b)-1
$.
By induction,
it looks like the result is
$\prod_{a \in A}(1+a) -1
$,
and this is independent
of the order of the
items operated upon.
A: After each operation,
the number of distinct values
is reduced by one.
So,
if there were n values
initially,
after n-1 of the operations,
there will be only
one distinct item left.
Or am I misunderstanding something?
A: I am assuming that the number of terms remains unchanged, so that the terms $a$ and $b$ are replaced by two terms, each equal to $a + b + ab$.
If this is the case, then the starting values, and the function $(a,b) \mapsto a+b+ab$, are irrelevant: you can always make all the terms equal, because $64$ is a power of $2$.
First, merge pairs of neighbouring terms: $x_1$ with $x_2, x_3$ with $x_4,\ldots$
This creates $32$ pairs of equal terms.
Next, merge terms $x_1$ with $x_3, x_2$ with $x_4,x_5$ with $x_7,\ldots$
This creates $16$ blocks of $4$ equal terms.
Next, merge terms $x_1$ with $x_5, x_2$ with $x_6,x_3$ with $x_7,\ldots$
This creates $8$ blocks of $8$ equal terms.
And so on. After six steps, we have a single block of $64$ equal terms.
As for the number of unique solutions $-$ I'll leave that to someone else.
