A241815 Irregular triangular array : t(n, k) = number of vertices of degree k in graph S(n) of strict partitions, where two partitions have an edge if and only if their intersection is empty.
2, 2, 3, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 3, 1, 3, 1, 1, 2, 2, 2, 2, 1, 1, 3, 2, 4, 1, 1, 1, 2, 4, 2, 1, 4, 1, 3, 4, 4, 1, 4, 1, 1, 1, 3, 1, 5, 1, 4, 1, 2, 3, 1, 1, 6, 5, 6, 1, 1, 2, 2, 2, 1, 1, 2, 6, 1, 2, 5, 4, 3, 1, 2, 4, 1, 2, 6, 5, 1, 2, 8, 3, 2, 1, 1
Offset: 3
Examples
Rows 3 to 15 (counting the top row as row 3): 2 2 3 1 2 1 1 3 1 1 1 2 1 1 3 1 3 1 1 2 2 2 2 1 1 3 2 4 1 1 1 2 4 2 1 4 1 3 4 4 1 4 1 1 1 3 1 5 1 4 1 2 3 1 1 6 5 6 1 1 2 2 2 1 The graph S(3) is given by 3 -> 21; S(4), by 4 -> 31, S(6), by 6 -> 51, 6 -> 42, 6 -> 321, 51 -> 42, and S(8), by 8 -> 71, 8 -> 62, 8 -> 53, 8 -> 521, 8 -> 431, 71 -> 62, 71 -> 53, 62, -> 53, 62 -> 431. The vertices of S(8) and their degrees d are easily read from the graph: d(521) = 1, d(431) = 2, d(71) = 3, d(53) = 3, d(62) = 4, and d(8) = 5, so that row 8 (counting the top row as row 3) is 1,1,2,1,1.
Links
- Clark Kimberling, Table of n, a(n) for n = 3..3000
Programs
-
Mathematica
z = 20; p[n_] := p[n] = Select[IntegerPartitions[n], DeleteDuplicates[#] == # &]; d[n_] := Table[Map[{p[n][[k]], #} &, DeleteCases[Flatten[Select[Map[{#, Intersection[p[n][[k]], #]} &, p[n]], #[[2]] == {} &], 1], {}]], {k, Length[p[n]]}]; u[n_] := Sort[Split[Sort[Flatten[d[n], 2]]]]; t = Table[Map[Length, u[n]]/2, {n, 1, z}]; Join[{0, 0}, Flatten[t]] (* Peter J. C. Moses, Apr 17 2014 *)
Comments