

A238943


Triangular array read by rows: t(n,k) = size of the Ferrers matrix of p(n,k).


3



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, 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, 7, 7, 6, 6, 6, 5, 5, 5, 5, 5, 4, 4, 4, 4, 5, 6, 3, 4
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OFFSET

1,2


COMMENTS

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 1s as nodes, and pad the graph with 0s 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.


LINKS

Table of n, a(n) for n=1..86.


FORMULA

t(n,k) = max{max(p(n,k)), length(p(n,k)}, where p(n,k) is the kth partition of n in Mathematica order.


EXAMPLE

First 8 rows:
1
2 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 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
For n = 3, the three partitions are [3], [2,1], [1,1,1]. Their respective Ferrers matrices derive from Ferrers graphs as follows:
The partition [3] has Ferrers graph 1 1 1, with Ferrers matrix of size 3:
1 1 1
0 0 0
0 0 0
The partition [2,1] has Ferrers graph
11
1
with Ferrers matrix of size 2:
1 1
1 0
The partition [1,1,1] has Ferrers graph
1
1
1
with Ferrers matrix of size 3
1 0 0
1 0 0
1 0 0
Thus row 3 is (3,2,3).


MATHEMATICA

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]}]
u = TableForm[t] (* A238943 array *)
v = Flatten[t] (* A238943 sequence *)


CROSSREFS

Cf. A238944, A238945, A237981, A000041.
Sequence in context: A221999 A222334 A181948 * A070081 A034883 A071647
Adjacent sequences: A238940 A238941 A238942 * A238944 A238945 A238946


KEYWORD

nonn,tabf,easy


AUTHOR

Clark Kimberling, Mar 07 2014


STATUS

approved



