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%I #16 Apr 06 2020 22:15:13
%S 1,2,2,3,2,3,4,3,2,3,4,5,4,3,3,3,4,5,6,5,4,4,3,3,4,3,4,5,6,7,6,5,5,4,
%T 4,4,3,3,4,5,4,5,6,7,8,7,6,6,5,5,5,4,4,4,4,5,3,4,4,5,6,4,5,6,7,8,9,8,
%U 7,7,6,6,6,5,5,5,5,5,4,4,4,4,5,6,3,4
%N Triangular array read by rows: t(n,k) = size of the Ferrers matrix of p(n,k).
%C Suppose that p is a partition of n, and let m = max{greatest part of p, number of parts of p}. Write the Ferrers graph of p with 1's as nodes, and pad the graph with 0's to form an m X m square matrix, which is introduced at A237981 as the Ferrers matrix of p, denoted by f(p). The size of f(p) is m.
%F t(n,k) = max{max(p(n,k)), length(p(n,k)}, where p(n,k) is the k-th partition of n in Mathematica order.
%e First 8 rows:
%e 1
%e 2 2 2
%e 3 2 3
%e 4 3 2 3 4
%e 5 4 3 3 3 4 5
%e 6 5 4 4 3 3 4 3 4 5 6
%e 7 6 5 5 4 4 4 3 3 4 5 4 5 6 7
%e 8 7 6 6 5 5 5 4 4 4 4 5 3 4 4 5 6 4 5 6 7 8
%e For n = 3, the three partitions are [3], [2,1], [1,1,1]. Their respective Ferrers matrices derive from Ferrers graphs as follows:
%e The partition [3] has Ferrers graph 1 1 1, with Ferrers matrix of size 3:
%e 1 1 1
%e 0 0 0
%e 0 0 0
%e The partition [2,1] has Ferrers graph
%e 11
%e 1
%e with Ferrers matrix of size 2:
%e 1 1
%e 1 0
%e The partition [1,1,1] has Ferrers graph
%e 1
%e 1
%e 1
%e with Ferrers matrix of size 3
%e 1 0 0
%e 1 0 0
%e 1 0 0
%e Thus row 3 is (3,2,3).
%t p[n_, k_] := p[n, k] = IntegerPartitions[n][[k]]; a[t_] := Max[Max[t], Length[t]]; t = Table[a[p[n, k]], {n, 1, 10}, {k, 1, PartitionsP[n]}]
%t u = TableForm[t] (* A238943 array *)
%t v = Flatten[t] (* A238943 sequence *)
%Y Cf. A238944, A238945, A237981, A000041.
%K nonn,tabf,easy
%O 1,2
%A _Clark Kimberling_, Mar 07 2014
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