A080169

[Brian L. Galebach]

75 is what I got as well.  Possibly what might have happened is that the
total of 81 double counts the two-pairs possibilities.  I almost did this
myself when working it out.

Brian Galebach

-----Original Message-----
From: Jens Voss [mailto:jens at voss-ahrensburg.de]
Sent: Tuesday, February 04, 2003 9:17 AM
To: Sequence Fanatics
Subject: A080169



The definition of the new sequence A080169 reads:

> ID Number: A080169
> Sequence:  1,3,13,81,541
> Name:      Ways n competitors can rank in a competition, allowing for the
>               possibility of ties.
> Example:   Three competitors can finish in 13 ways: 1,2,3; 1,3,2; 2,1,3;
2,3,1;
>               3,1,2; 3,2,1; 1,1,3; 2,2,1; 1,3,1; 2,1,2; 3,1,1; 1,2,2;
1,1,1.
> Keywords:  more,nonn,new
> Offset:    1
> Author(s): Robert Lozyniak (nothing at example.com), Feb 04 2003

Either I misunderstood this, or I can't count or there's a mistake:
I am getting only 75 for a(4):

 The  1 way of all four coming in first,
 the  8 ways of having three come in simultaneously and one separately,
 the  6 ways of having two pairs,
 the 36 ways of having one pair and two separate competitors as well as
 the 24 ways of all four coming in separately
-------
sum: 75

Which ones did I miss?

Regards,
Jens