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%I #9 Nov 12 2023 22:00:09
%S 0,2,2,1,2,1,2,2,0,4,1,2,2,2,4,0,1,2,0,4,7,0,0,2,2,2,2,9,2,0,4,1,2,1,
%T 4,11,2,0,4,6,2,2,2,14,6,0,4,9,2,0,0,1,2,0,4,17,6,0,2,19,4,0,0,0,2,2,
%U 4,2,16,10,1,6,17,14,0,0,0,4,1,2,0,4,23
%N Irregular triangular array, read by rows: T(n,k) = number of vertices (partitions) that have degree k in the distance graph of the partitions of n, for k = 1..A366429(n).
%C The distance graph of the partitions of n is defined in A366156.
%e First fourteen rows:
%e 1
%e 2
%e 2 1
%e 2 1 2
%e 2 0 4 1
%e 2 2 2 4 0 1
%e 2 0 4 7 0 0 2
%e 2 2 2 9 2 0 4 1
%e 2 1 4 11 2 0 4 6
%e 2 2 2 14 6 0 4 9 2 0 0 1
%e 2 0 4 17 6 0 2 19 4 0 0 0 2
%e 2 4 2 16 10 1 6 17 14 0 0 0 4 1
%e 2 0 4 23 10 0 2 27 22 1 0 0 4 6
%e 2 2 2 22 18 2 4 27 32 4 0 0 6 12 2
%e Enumerate the 7 partitions (vertices) of 5 as follows:
%e 1: 5
%e 2: 4,1
%e 3: 3,2
%e 4: 3,1,1
%e 5: 2,2,1
%e 6: 2,1,1,1
%e 7: 1,1,1,1,1
%e Call q a neighbor of p if d(p,q)=2, where d is the distance function in A366156.
%e The set of neighbors for vertex k, for k = 1..7, is given by
%e vertex 1: {2}
%e vertex 2: {1,3,4}
%e vertex 3: {2,4,5}
%e vertex 4: {2,3,5,6}
%e vertex 5: {3,4,6}
%e vertex 6: {4,5,7}
%e vertex 7: {6}
%e The number of vertices having degrees 1,2,3,4 are 2,0,4,1, respectively, so that row 5 is 2 0 4 1.
%t c[n_] := PartitionsP[n]; q[n_, k_] := q[n, k] = IntegerPartitions[n][[k]];
%t r[n_, k_] := r[n, k] = Join[q[n, k], ConstantArray[0, n - Length[q[n, k]]]];
%t d[u_, v_] := Total[Abs[u - v]];
%t s[n_, k_] := Select[Range[c[n]], d[r[n, k], r[n, #]] == 2 &];
%t s1[n_] := s1[n] = Table[s[n, k], {k, 1, c[n]}];
%t m[n_] := m[n] = Map[Length, s1[n]];
%t m1[n_] := m1[n] = Max[m[n]]; (* A366429 *)
%t t1 = Join[{1}, Table[Count[m[n], i], {n, 2, 15}, {i, 1, m1[n]}]]
%t Column[t1]
%t Flatten[t1]
%Y Cf. A000041 (row sums), A366429 (row lengths), A366598 (row maxima).
%K nonn,tabf
%O 1,2
%A _Clark Kimberling_, Oct 16 2023