A225491 Maximal frequency depth for multisets over an alphabet of n letters.
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, 8, 8, 8, 8, 8, 8
Offset: 1
Keywords
Examples
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.
Links
- Clark Kimberling, Table of n, a(n) for n = 1..40
Programs
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Mathematica
c[s_] := c[s] = Select[Table[Count[s, i], {i, 1, Max[s]}], # > 0 &] f[s_] := f[s] = Drop[FixedPointList[c, s], -2] t[s_] := t[s] = Length[f[s]] u[n_] := u[n] = Table[t[Part[IntegerPartitions[n], k]], {k, 1, Length[IntegerPartitions[n]]}]; v = Table[Max[u[n]], {n, 2, 40}]; (* A225491 *) Prepend[2 + v, 0]
Formula
a(1) = 0, a(n) = 2 + A225486(n) for n > 1.
Comments