Sequence A112088 and rounds in Josephus' problem

[Rainer Rosenthal]

Max asked:
> I still do not see how A112088 appears here.
> Could you please clarify a little bit further?

there are two methods producing A112088:

   B_SS   Counting leaves in certain
          binary trees (Simon Strandgaard) and

   J_RR   Counting rounds in the Josephus
          game Third-Out (Rainer Rosenthal).

While B_SS is still to be answered, I will again
explain J_RR shortly, so that I can make a comment
in the next days.

We consider the Josephus game where every third
person is eliminated from the circle. We mark the
first person with a hat. The third person will
be eliminated and the next one in the circle gets
the hat. The person with the hat counts to three
again and the third person will be eliminated etc.

In this way the hat makes its rounds. The larger n
the more rounds we have. I will describe some of
these games in order to derive the function rounds(k).
After that it will be clear how the function

      J_RR(n) = smallest k with "rounds(k) = n"

is beginning. Programming J_RR yielded coincidence
with A112088 in all of the 18 first values. I am looking
for a well written description of method J_RR and I hope
that the hat-technique will help.

Not being a native English speaker I ask for a concise
and yet nicely readable desription. The following
examples shall convince the friendly SeqFan that J_RR
isn't too complicated.

For each k we start with the circle (1,2,...,k) and
mark person 1 with a hat:  (1^2 ... k). After each
elimination the smaller circle is shown, the hat being
given to the next person.

==================== k =  2 ================
Round 1: (1^ 2)
Stop:    (2^)
==================== k =  3 ================
Round 1: (1^ 2 3)
Round 2: (1^ 2)
Stop:    (2^)
==================== k =  4 ================
Round 1: (1^ 2 3 4)
         (1 2 4^)
Round 2: (1 4^)
Stop:    (1^)
==================== k =  5 ================
Round 1: (1^ 2 3 4 5)
         (1 2 4^ 5)
Round 2: (2^ 4 5)
Round 3: (2^ 4)
Stop:    (4^)
==================== k =  6 ================
Round 1: (1^ 2 3 4 5 6)
         (1 2 4^ 5 6)
Round 2: (1^ 2 4 5)
         (1 2 5^)
Round 3: (1 5^)
Stop:    (1^)
==================== k =  7 ================
Round 1: (1^ 2 3 4 5 6 7)
         (1 2 4^ 5 6 7)
         (1 2 4 5 7^)
Round 2: (1 4^ 5 7)
Round 3: (1^ 4 5)
Round 4: (1^ 4)
Stop:    (1^)

For k=8 up to k=10 there are only 4 rounds as well. For
k=10 I will demonstrate this, followed by k=11 with
5 rounds.

==================== k = 10 ================
Round 1: (1^ 2 3 4 5 6 7 8 9 10)
         (1 2 4^ 5 6 7 8 9 10)
         (1 2 4 5 7^ 8 9 10)
         (1 2 4 5 7 8 10^)
Round 2: (1 4^ 5 7 8 10)
         (1 4 5 8^ 10)
Round 3: (4^ 5 8 10)
         (4 5 10^)
Round 4: (4 10^)
Stop:    (4^)
==================== k = 11 ================
Round 1: ( 1^ 2 3 4 5 6 7 8 9 10 11)
         ( 1 2 4^ 5 6 7 8 9 10 11)
         ( 1 2 4 5 7^ 8 9 10 11)
         ( 1 2 4 5 7 8 10^ 11)
Round 2: ( 2^ 4 5 7 8 10 11)
         ( 2 4 7^ 8 10 11)
         ( 2 4 7 8 11^)
Round 3: ( 2 7^ 8 11)
Round 4: ( 2^ 7 8)
Round 5: ( 2^ 7)
Stop:    ( 7^)
=============================================

Remark: the situation in round 2 for k=11 is
essentially the same as the one in round 1
for k=7: There are 7 persons and the first one
has the hat. That implies rounds(11)=rounds(7)+1.

I am sorry for my blindness yesterday, when I
thought my J_RR method was incorrect. It is OK.

The first elements of A112088, i.e.
2, 3, 5, 7, 11, 16, 24, 36, 54, 81, 122, 183, ...
will be clear now from the desription above.
The smallest k for rounds(k) = 1,2,3,4,5 are
shown above and they are 2,3,5,7,11.

Thanks for any comments to come as well as for
those already received.

Best regards,
Rainer Rosenthal
r.rosenthal at web.de