A321223 a(n) is the number of recursively self-conjugate partitions of n.

1, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 2, 1, 0, 0, 1, 0, 1, 1, 0, 2, 1, 2, 0, 1, 0, 1, 1, 1, 1, 0, 1, 2, 2, 0, 1, 0, 0, 1, 2, 1, 2, 1, 1, 1, 2, 0, 0, 1, 0, 2, 1, 1, 1, 2, 1, 2, 1, 0, 1, 1, 0, 1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 1, 2, 1, 0, 2, 1, 0, 1, 1, 1, 2, 2, 1, 1, 3, 0, 2

Offset: 1

Author

Michael De Vlieger, Oct 31 2018

Keywords

• nonn
• easy

Comments

A recursively self-conjugate partition L has a conjugate L* = L. Further, elimination of the Durfee square and leg (conjugate with the arm) to leave the arm L_1. L_1 likewise has conjugate L_1* = L_1. We continue taking the arm, eliminating the new Durfee square and leg in this manner until the entire partition is processed and all arms are self-conjugate.

We can define a recursively self-conjugate partition L by placing a series S of squares s_k in position k, whose side-lengths decrease as k increases, in the following manner. We place the first square in the upper left corner, then set 2^(k - 1) squares s_k in all places wherein we have bounds by axis or previous square to the left and top. Thereby we can abbreviate all recursively self-conjugate partitions L by S(L). For example, (5,4,4,4,1) = {4,1}, and (10,9,8,7,6,5,4,3,2,1) = {5,3,1,1}. (See Keith 2011 page 9 Fig. 3.)

A190900 = positions of 0 in a(n).

Observation: the graph of this sequence separates into two distinct bands for n greater than approximately 10,000. Values of a(n) for n mod 3 = 0 or 1 tend to be greater than a(n) for n mod 3 = 2. Even within the upper band, we have the mean a(n) for n mod 3 = 0 distinct from the mean a(n) for n mod 3 = 1. See linked graphs. - Michael De Vlieger, Dec 10 2018

Examples

			a(2) = 0 since neither (2) nor (1,1) is recursively symmetrical.
a(6) = 1 since the partition (3,2,1) of 6 is recursively symmetrical. S(3,2,1) = {2,1}.
a(27) = 2 since both (6,6,6,3,3,3) and (6,5,5,5,5,1) are recursively self-conjugate. S(6,6,6,3,3,3) = {3,3}; S(6,5,5,5,5,1) = {5,1}.
a(103) = 3 since there are 3 recursively self-conjugate partitions of 103: (13,13,13,10,10,10,7,6,6,6,3,3,3), (13,12,12,12,12,8,7,6,5,5,5,5,1), and (13,12,12,10,9,9,9,9,9,4,3,3,1). These can be stated in terms of recursive squares as {7,3,3}, {7,5,1}, and {9,3,1} respectively.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Michael De Vlieger, Diagrams of recursively self-conjugate partitions for 1 <= n <= 150.
• Michael De Vlieger, Graph of A321223(n) for 1 <= n <= 65536.
• Michael De Vlieger, Graph of A321223(n) for 1 <= n <= 65536, color coded mod 3.
• William J. Keith, Recursively Self-Conjugate Partitions, Integers 11A, (2011) Article 12.

Crossrefs

Cf. A190899, A190900.

Programs

Mathematica

f[w_] := Block[{k}, k = Total@ w; Total@ Map[Apply[Function[{s, t}, s Array[Boole[t <= # <= s + t - 1] &, k] ], #] &, Apply[Join, Prepend[Table[Function[{v, c}, Map[{w[[k]], # + 1} &, Map[Total[v #] &, Tuples[{0, 1}, {Length@ v}]]]] @@ {Most@ #, ConstantArray[1, Length@ # - 1]} &@ Take[w, k], {k, 2, Length@ w}], {{w[[1]], 1}}]]] ]; g[n_] := Block[{w = {n}, c}, c[x_] := Apply[Times, Most@ x - Reverse@ Accumulate@ Reverse@ Rest@ x]; Reap[Do[Which[And[Length@ w == 2, SameQ @@ w], Sow[w]; Break[], Length@ w == 1, Sow[w]; AppendTo[w, 1], c[w] > 0, Sow[w]; AppendTo[w, 1], True, Sow[w]; w = MapAt[1 + # &, Drop[w, -1], -1] ], {i, Infinity}] ][[-1, 1]] ]; Block[{n = 12, a}, a = Merge[Map[<| #1 -> #2 |> & @@ # &, #], Identity] &@ TakeWhile[Sort@ Map[{Total@ #2, #1, #2} & @@ {#, f[#]} &, Apply[Join, Array[g, n]] ], First@ # <= n^2 &][[All, 1 ;; 2]]; Array[Length[Lookup[a, #] /. k_ /; MissingQ@ k -> {}] &, Length@ a] ]