|
%I #16 Apr 23 2023 07:25:15
%S 2,2,1,3,1,1,2,3,2,4,2,4,1,2,6,5,1,1,4,5,8,3,2,3,8,10,4,5,4,10,13,5,9,
%T 1,2,13,17,8,14,1,1,6,12,22,10,22,3,2,2,19,27,11,32,5,5,4,21,33,15,43,
%U 9,10,4,20,44,21,57,10,19,1,5,28,50,20,77,20
%N Irregular triangle read by rows T(n,k) = number of partitions of degree k in the partition graph G(n), for n >= 2; G(n) is defined in Comments.
%C The partition graph G(n) of n has the partitions of n as nodes, and nodes p and q have an edge if one of them can be obtained from the other by a substitution x -> x-1,1 for some part x. G(n) is nonplanar for n >= 8. Column 1: divisors of n, A000005(n), for n >= 2. A000041(n) = sum of numbers in row n, for n >= 2 (counting the top row as row 2). Number of numbers in row n (i.e., maximal degree in G(n)): A241151(n), n >= 2. Last term in row n (the number of partitions having maximal degree): A241153(n), n >= 2. Maximal number in row n: A241152(n), n >= 2. Let u(n,k) be the array at A029205 (where n >= 0, k=0..n). Then u(n,k) is the number of edges in G(n+2) between partitions of n+2 that having length k+1 and those having length k+2.
%H Clark Kimberling, <a href="/A241150/b241150.txt">Table of n, a(n) for n = 1..500</a>
%e The first 12 rows:
%e 2
%e 2 ... 1
%e 3 ... 1 ... 1
%e 2 ... 3 ... 2
%e 4 ... 2 ... 4 ... 1
%e 2 ... 6 ... 5 ... 1 ... 1
%e 4 ... 5 ... 8 ... 3 ... 2
%e 3 ... 8 ... 10 .. 4 ... 5
%e 4 ... 10 .. 13 .. 5 ... 9 ... 1
%e 2 ... 13 .. 17 .. 8 ... 14 .. 1 ... 1
%e 6 ... 12 .. 22 .. 10 .. 22 .. 3 ... 2
%e 2 ... 19 .. 27 .. 11 .. 32 .. 5 ... 5
%e The graph can be represented by these transformations:
%e 6 -> 51, 51 -> 411, 42 -> 321, 42 -> 411, 411 -> 3111, 33 -> 321, 321 -> 2211, 321 -> 3111, 3111 -> 21111, 222 -> 2211, 2211 -> 21111, 21111 -> 111111. These 4 partitions p have degree 1 (i.e., number of arrows to or from p): 6, 33, 222, 111111; these 2 have degree 2: 51, 42; these 4 have degree 3: 411, 3111, 2211, 21111; the remaining partition, 321, has degree 4. So, row 6 of the array is 4 2 4 1.
%t z = 25; spawn[part_] := Map[Reverse[Sort[Flatten[ReplacePart[part, {# - 1, 1}, Position[part, #, 1, 1][[1]][[1]]]]]] &, DeleteCases[DeleteDuplicates[part], 1]];
%t unspawn[part_] := If[Length[Cases[part, 1]] > 0, Map[ReplacePart[Most[part], Position[Most[part], #, 1, 1][[1]][[1]] -> # + 1] &, DeleteDuplicates[Most[part]]], {}]; m = Map[Last[Transpose[Tally[Map[#[[2]] &, Tally[Flatten[{Map[unspawn, #], Map[spawn, #]}, 2] &[IntegerPartitions[#]]]]]]] &, 1 + Range[z]];
%t Column[m] (* A241150 as an array *)
%t Flatten[m] (* A241150 as a sequence *)
%t Table[Length[m[[n]]], {n, 1, z}] (* A241151 *)
%t Table[Max[m[[n]]], {n, 1, z}] (* A241152 *)
%t Table[Last[m[[n]]], {n, 1, z}] (* A241153 *)
%t (* Next, show the graph G(k) *)
%t k = 8; graph = Flatten[Table[part = IntegerPartitions[k][[n]]; Map[FromDigits[part] -> FromDigits[#] &, spawn[part]], {n, 1, PartitionsP[k]}]]; Graph[graph, VertexLabels -> "Name", ImageSize -> 500, ImagePadding -> 20] (* _Peter J. C. Moses_, Apr 15 2014 *)
%Y Cf. A241151, A241152, A241153, A029205, A000041.
%K nonn,easy,tabf
%O 1,1
%A _Clark Kimberling_ and _Peter J. C. Moses_, Apr 17 2014
|
