Sequence A112088 without example

[Hugo Pfoertner]

Max wrote:
> 
> 
> Ops! I've tried to recompute that sequence myself and I've got
> sequence A005428(n+1) which is different from Rainer's one.
> 
> In particular, for n=7 we have A005428(7+1)=21 while Rainer's value is 24.
> Let see how the executioner proceeds at 21 people.
> At the beginning we have
> 0: * 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
> where * is the executioner's position.
> After first round we left with
> 1: * 1 2 4 5 7 8 10 11 13 14 16 17 19 20
> after second:
> 2: * 2 5 7 10 11 14 16 19 20
> after third:
> 3: * 2 5 10 11 16 19
> after fourth:
> 4: * 2 5 11 16
> after fifth:
> 5: 2 * 16
> the next step of the executioner makes two rounds (sixth and seventh)
> at a time ending with
> 7: * 2
> 
> So 21 people require 7 rounds to be eliminated, then according to
> Hugo's definition the the 7th element of the discussed sequence must
> be <=21.
> 
> What's wrong?
> 
> Thanks,
> Max

Using my definition (e.g. "Survivor announces the number of rounds when
executioner comes to him after having eliminated the final victim) I get
the following sequence (which is A073941)

Round
    People present at start of game
1   1
2   2
3   3
4   4
5   6
6   9
7  14

Example: 7 Rounds are first needed for 14 initial persons:

  1  2  3  4  5  6  7  8  9 10 11 12 13 14

  1  1  x  1  1  x  1  1  x  1  1  x  1  1
  x  2  .  2  x  .  2  2  .  x  2  .  2  x
  .  3  .  3  .  .  x  3  .  .  3  .  x  .
  .  4  .  4  .  .  .  x  .  .  4  .  .  .
  .  5  .  x  .  .  .  .  .  .  5  .  .  .
  .  6  .  .  .  .  .  .  .  .  x  .  .  .
  .  7S

because the executioner needs round 7 to go from the last victim (11)
to  inform the survivor "all done".

For Rainers definition "final victim announces number of round when he
is killed" I get

1   undefined, because there is no victim to announce anything
2   2
3   3
4   5
5   7
6  11

BTW, a comment should be added to A073941 pointing to the "Position of
survivor" formula in the Odlyzko/Wilf article, because it describes the
smallest n, where the exponent of (3/2) first assumes
k=1,2,3,4,5,6,......
 
J_3(n) = 3*n+1-floor(K(3)*(3/2)^ceiling(log_(3/2)((2*n+1)/K(3)))) for
n=0,1,.... 
with the Josephus Constant K(3)=1.6222705028847673....

Benoit Cloitre's formula seems to be derived from this original formula.

Hugo Pfoertner