A102687

[David Wasserman]

Dear Vladeta,
    That is very interesting, and rather surprising.  At first I thought a(n, 5) should be nonincreasing, but when I thought about it some more, I saw why it isn't.  
    Did you also compute a(n, 3) and a(n, 4)?  
    I notice that for any given f: m -> m, the sequence f^(n) is eventually periodic.  Therefore the sequence of sets {f^(n) | f: m -> m} is also eventually periodic, its period being the lcm of the periods for each f.  So a(n, m) is also eventually periodic, probably with the same period but possibly smaller.  It looks like a(n, 5) has period 60 after the first 3 terms.
    {f: m -> m} is a symmetric group.  We can ask the same question about any finite group.  For example, with the cyclic group of order k, we get the sequence a(n) = n/gcd(n, k).  The quaternion group gives 8,2,8,1,8,2,8,1...

 - David

>Dear Seqfans,
> 
>There is an interesting sequence of Eric Wegrzynowski:
> 
><http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A102687>http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?Anum=A102687
> 
>I just submitted to the OEIS:
> 
>%I A000001
>%S A000001 3125,1075,985,580,1281,295,1305,580,925,631,1305,220,1305,655,901,580,1305,295,1305,556,925,655,1305,220,1281,655,925,580,1305,271,1305,580,925,655,1281,220,1305,655,925,556,1305,295,1305,580,901,655,1305,220,1305,631,925,580,1305,295,1281,580,925,655,1305,196,1305,655,925,580,1281,295,1305,580,925,631,1305,220,1305,655,901,580,1305,295,1305,556,925,655,1305,220,1281,655,925,580,1305,271,1305,580,925,655,1281,220,1305,655,925,556,1305
>%N A000001 Let a(n,m) = card{f^(n) : f is a mapping from a set of m elements into itself}, where f^(l)(x) = f^(l-1)(f(x)),l>0, f^(0)(x) = x; sequence gives a(n,5).
>%Y A000001 Cf. A102687.
>%O A000001 1
>%K A000001 ,nonn,
>
> 
>Best regards to all,
> 
>Vladeta Jovovic