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A225491 Maximal frequency depth for multisets over an alphabet of n letters.

%I A225491 #5 May 09 2013 12:35:37
%S A225491 0,4,5,6,6,6,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,
%T A225491 8,8,8,8,8,8
%N A225491 Maximal frequency depth for multisets over an alphabet of n letters.
%C A225491 Frequency depth is defined at A225485.  Suppose S is a multiset on an alphabet y(1),..,y(n).  Let f(n) > 0 be the frequency of y(i) in S, so that F(S) (as at A225485) is the multiset {f(1),..,f(m)}, where m is the number of distinct terms in S.  Let {g(1),..,g(k)} be the set of distinct terms of F(S), and let h(i) be the number of occurrences of g(i) in F(S).  Then F(F(S)) is a partition p(m) of m, and D(F(F(S))) = D(p(m)), where D denotes frequency depth.  To maximize D for n>1, put m = n to get a(n) = 2 + A225486(n), for n > 1.
%H A225491 Clark Kimberling, <a href="/A225491/b225491.txt">Table of n, a(n) for n = 1..40</a>
%F A225491 a(1) = 0, a(n) = 2 + A225486(n) for n > 1.
%e A225491 For n = 2, let the alphabet be {u,v}.  Then for some p>=0 and q>=0, S consists of p u's and q v's, so that F(S) = {p,q}.  Assume without loss of generality that p<=q.  If 1 <= p < q, then the depth of 4 is the number of arrows when we write S -> pq -> 11 -> 2 -> 1.  The other possibilities (p = 0, or p=q) for p and q lead to depths < 4, so that a(2) = 4.
%t A225491 c[s_] := c[s] = Select[Table[Count[s, i], {i, 1, Max[s]}], # > 0 &]
%t A225491 f[s_] := f[s] = Drop[FixedPointList[c, s], -2]
%t A225491 t[s_] := t[s] = Length[f[s]]
%t A225491 u[n_] := u[n] = Table[t[Part[IntegerPartitions[n], k]], {k, 1,
%t A225491      Length[IntegerPartitions[n]]}];
%t A225491 v = Table[Max[u[n]], {n, 2, 40}]; (* A225491 *)
%t A225491 Prepend[2 + v, 0]
%Y A225491 Cf. A225485, A225486.
%K A225491 nonn
%O A225491 1,2
%A A225491 _Clark Kimberling_, May 09 2013