%I #10 Sep 30 2023 21:54:48
%S 1,1,1,2,1,2,2,1,3,4,0,4,6,1,5,9,1,6,11,4,1,8,20,2,0,10,25,7,0,12,37,
%T 6,1,15,47,13,2,18,67,15,1,22,85,25,3,27,122,26,1,32,142,46,10,1,38,
%U 200,53,6,0,46,259,74,6,0,54,330,92,13,1,64,412,136
%N Irregular triangle read by rows: T(n,k) is the number of partitions of n that have depth k.
%C Suppose that P is a partition of n. Let x(1), x(2),...,x(k) be the distinct parts of P, and let m(i) be the multiplicity of x(i) in P. Let f(P) be the partition [m(1)*x(1), m(2)*x(2),...,x(k)*m(k)] of n. Define c(0,P) = P, c(1,P) = f(P), ..., c(n,P) = f(c(n-1,P), and define d(P) = least n such that c(n,P) has no repeated parts; d(P) is as the depth of P, as defined in A237685. Clearly d(P) = 0 if and only if P is a strict partition, as in A000009. Conjecture: if d >= 0, then 2^d is the least n that has a partition of depth d.
%e First 20 rows:
%e 1
%e 1 1
%e 2 1
%e 2 2 1
%e 3 4 0
%e 4 6 1
%e 5 9 1
%e 6 11 4 1
%e 8 20 2 0
%e 10 25 7 0
%e 12 37 6 1
%e 15 47 13 2
%e 18 67 15 1
%e 22 85 25 3
%e 27 122 26 1
%e 32 142 46 10 1
%e 38 200 53 6 0
%e 46 259 74 6 0
%e 54 330 92 13 1
%e 64 412 136 15 0
%t z = 36; c[n_] := c[n] = Map[Length[FixedPointList[Sort[Map[Total, Split[#]], Greater] &, #]] - 2 &, IntegerPartitions[n]]
%t t = Table[Count[c[n], k], {n, 1, z}, {k, 0, Floor[Log[2, n]]}]
%t TableForm[t] (* this sequence as an array *)
%t Flatten[t] (* this sequence *)
%Y Cf. A000009, A000041, A237685 (column 1), A237750 (column 2), A237978 (column 3), A225485 (frequency depth array).
%K nonn,tabf
%O 1,4
%A _Clark Kimberling_, Sep 28 2023