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 $x \mapsto x+1$.

  • $\begingroup$ Please help me write the question so that it looks more formal and more pleasing to the eye. $\endgroup$ – Mike B. Mar 18 '16 at 23:50
  • $\begingroup$ I think you need to clarify what you mean by "solution". I also don't understand what you mean by "all terms in the sequence are equal"... clearly $1 \neq \frac{1}{64}$. $\endgroup$ – MartianInvader Mar 19 '16 at 0:25
  • $\begingroup$ As marty cohen says, each time you apply the function, the number of terms is reduced by $1$, so in this case after $63$ applications to pairs of terms, there will be only one term left. The usual question is to show that the result is independent of how you select the pair of terms to apply the function to at each step. $\endgroup$ – Ross Millikan Mar 19 '16 at 0:40
  • $\begingroup$ My second answer shows that the result is indeed independent of the order of the items chosen. $\endgroup$ – marty cohen Mar 19 '16 at 0:57
  • 2
    $\begingroup$ Possible duplicate of The final number after $999$ operations. $\endgroup$ – Arnaud D. Aug 1 '18 at 12:50

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.


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?

  • $\begingroup$ Changed the question, sorry for the inaccuracy in the phrasing... $\endgroup$ – Mike B. Mar 19 '16 at 0:49

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.


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