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%I #22 Sep 27 2023 13:49:45
%S 0,1,1,1,1,1,1,2,1,1,1,1,2,3,1,3,4,3,1,1,4,8,1,1,3,6,9,3,1,2,8,12,7,1,
%T 3,11,17,10,1,1,11,26,17,1,5,19,25,27,1,1,17,44,38,1,3,25,53,52,1,1,3,
%U 29,63,76,4
%N Number of partitions of n that have frequency depth k, an array read by rows.
%C Let S = {x(1),...,x(k)} be a multiset whose distinct elements are y(1),...,y(h). Let f(i) be the frequency of y(i) in S. Define F(S) = {f(1),..,f(h)}, F(1,S) = F(S), and F(m,S) = F(F(m-1),S) for m>1. Then lim(F(m,S)) = {1} for every S, so that there is a least positive integer i for which F(i,S) = {1}, which we call the frequency depth of S.
%C Equivalently, the frequency depth of an integer partition is the number of times one must take the multiset of multiplicities to reach (1). For example, the partition (32211) has frequency depth 5 because we have: (32211) -> (221) -> (21) -> (11) -> (2) -> (1). - _Gus Wiseman_, Apr 19 2019
%C From _Clark Kimberling_, Sep 26 2023: (Start)
%C Below, m^n abbreviates the sum m+...+m of n terms. In the following list, the numbers p_1,...,p_k are distinct, m >= 1, and k >= 1. The forms of the partitions being counted are as follows:
%C column 1: [n],
%C column 2: [m^k],
%C column 3: [p_1^m,...,p_k^m],
%C column 4: [(p_1^m_1)^m,..., (p_k^m_k)^m], distinct numbers m_i.
%C Column 3 is of special interest. Assume first that m = 1, so that the form of partition being counted is p = [p_1,...,p_k], with conjugate given by [q_1,...,q_m] where q_i is the number of parts of p that are >= i. Since the p_i are distinct, the distinct parts of q are the integers 1,2,...,k. For the general case that m >= 1, the distinct parts of q are the integers m,...,km. Let S(n) denote the set of partitions of n counted by column 3. Then if a and b are in the set S*(n) of conjugates of partitions in S(n), and if a > b, then a - b is also in S*(n). Call this the subtraction property. Conversely, if a partition q has the subtraction property, then q must consist of a set of numbers m,..,km for some m. Thus, column 3 counts the partitions of n that have the subtraction property. (End)
%H Alois P. Heinz, <a href="/A225485/b225485.txt">Rows n = 1..200, flattened</a> (first 40 rows from Clark Kimberling)
%e The first 9 rows:
%e n = 1 .... 0
%e n = 2 .... 1..1
%e n = 3 .... 1..1..1
%e n = 4 .... 1..2..1..1
%e n = 5 .... 1..1..2..3
%e n = 6 .... 1..3..4..3
%e n = 7 .... 1..1..4..8..1
%e n = 8 .... 1..3..6..9..3
%e n = 9 .... 1..2..8.12..7
%e For the 7 partitions of 5, successive frequencies are shown here:
%e 5 -> 1 (depth 1)
%e 41 -> 11 -> 2 -> 1 (depth 3)
%e 32 -> 11 -> 2 -> 1 (depth 3)
%e 311 -> 12 -> 11 -> 2 -> 1 (depth 4)
%e 221 -> 12 -> 11 -> 2 -> 1 (depth 4)
%e 2111 -> 13 -> 11 -> 2 -> 1 (depth 4)
%e 11111 -> 5 -> 1 (depth 2)
%e Summary: 1 partition has depth 1; 1 has depth 2; 2 have 3; and 3 have 4, so that the row for n = 5 is 1..1..2..3 .
%t c[s_] := c[s] = Select[Table[Count[s, i], {i, 1, Max[s]}], # > 0 &]
%t f[s_] := f[s] = Drop[FixedPointList[c, s], -2]
%t t[s_] := t[s] = Length[f[s]]
%t u[n_] := u[n] = Table[t[Part[IntegerPartitions[n], i]],
%t {i, 1, Length[IntegerPartitions[n]]}];
%t Flatten[Table[Count[u[n], k], {n, 2, 25}, {k, 1, Max[u[n]]}]]
%Y Row sums are A000041.
%Y Column k = 2 is A032741.
%Y Column k = 3 is A325245.
%Y a(n!) = A325272(n).
%Y Cf. A181819, A182850, A225486, A323014, A323023, A325238, A325239, A325254.
%Y Integer partition triangles: A008284 (first omega), A116608 (second omega), A325242 (third omega), A325268 (second-to-last omega), A225485 or A325280 (length/frequency depth).
%K nonn,tabf
%O 1,8
%A _Clark Kimberling_, May 08 2013
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