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A partition is self-conjugate if it is fixed under conjugation and it is recursively self-conjugate if it is self-conjugate and the portions below and to the right of its Durfee square are recursively self-conjugate. (See the Keith paper for a more detailed description.)

Only a finite number of positive integers do not have a recursively self-conjugate partition. The list is given in A190900.

Integers expressible as a_0^2 + 2*a_1^2 + ... + 2^k*a_k^2 with [a_0, a_1, .., a_k] a non-squashing partition. [See Keith link, p. 6]

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

Algorithm:

Let S be a sequence starting with n. Let k be the index of a term in S, with n at position k = 0. Let S_r be the r-th sequence in row n.

Starting with S_1 = {n}, we either (A) append a 1 to the left of S_r, or (B) we drop the most recently-appended term S_(k) and increment the rightmost term (k - 1).

By default we execute (A) and test according to the following. Consider the reversed accumulation A_(r + 1) = Sum(reverse(S_(k + 1))) = Sum(k_m, k_(m - 1), ..., k_2, k_1). If S_r - A_(r + 1) contains nothing less than 0, then S_(k + 1) is retained, otherwise we execute (B).

We end after k_1 = n, since otherwise we would enter an endless loop that also increments k_0 ad infinitum.

The first sequence S in row n is {n} while the last is {n, n}.

All rows n contain {{n}, {n, 1}, {n, n}}.

Only one repeated term k may appear at the end of any S in row n.

The longest possible sequence S in row n has 2 + floor(log_2(n)) terms = 2 + A113473(n).

The sequence S describes unique integer partitions L that are recursively symmetrical. Example: We can convert S = {4, 2, 1} into the partition (7, 6, 5, 4, 3, 2, 1), a partition of N = 28. We set a 4X Durfee square with its upper-left corner at origin. Then we set 2^k = 2^1 = 2 2X squares, each with its upper-left corner in any coordinate bounded at left and top by either a previously-lain square or an axis. Finally, we set 2^2 = 4 1X squares as above once again. We obtain a Ferrer diagram as below, with the k marked, i.e., the 1st term 4X, the 2nd term 2X, the 3rd term 1X squares:

0 0 0 0 1 1 2

0 0 0 0 1 1

0 0 0 0 2

0 0 0 0

1 1 2

1 1

2

The resulting partition L is recursively self-conjugate; its arms are identical to its legs. We can eliminate the Durfee square and the other appendage and have a symmetrical partition L_1 with Durfee square of k_1 units, etc.

Were we to admit either more than 1 repeated k or a term such that S_k - A_(k + 1) had differences less than 1, we would have overlapping squares in the Ferrer diagram. Such diagrams are generated by larger n and all resulting diagrams are unique given the described algorithm.

The sequences S in row n, converted into integer partitions L, sum to n^2 <= N <= 3 * n^2.

For all n, n^2 <= k <= 3*n^2.

For n > 5, some k may have more than 1 recursively self-conjugate partitions in the same row. For example, k = 90 in row 6 has two recursively self-conjugate partitions (RSCPs) with Durfee square of 6: (12,12,12,9,9,9,6,6,6,3,3,3) and (12,11,11,11,11,7,6,5,5,5,5,1). These RSCPs can be defined by dendritically laying out squares in the series {6,3,3} and {6,5,1} respectively.

Numbers k that set records for the number m of recursively self-conjugate partitions (RCSPs).

1 is the only square in the sequence.

The graph of A321223 suggests there is a finite number of numbers k with a given number m of RSCPs (not all such k appear here). We know that A190900 (positive integers without RSCPs) is finite. For index i <= 2^16, there are 6 squares in A321223, i.e., those of {1, 2, 3, 5, 8}, that have just 1 RSCP; there are 120 nonsquares 3 <= k <= 590 in A321223 that have m = 1 RSCP. In the same range, there are 127 numbers 27 <= k <= 830 in A321223 that have m = 2 RSCPs, and 142 numbers 103 <= k <= 1280 in A321223 that have m = 3 RSCPs. This sequence includes many of the first terms k of these finite sequences, all k having m RSCPs.

Examining the smallest 381 terms (i.e., all k < 2^16) and the plot of A321223, we observe the following:

1. a(3) = 103 and a(23) = 2011 are the only primes.

2. a(2) = 27 = 3^3 and a(64) = 6561 = 3^8 are the only prime powers.

3. Numbers k such that k mod 3 = 2 are never in this sequence.

4. Only k in {1, 103, 175, 310, 838, 880, 2011, 2416, 4531, 4720, 5872, 11248, 11632, 12400, 15136, 16081, 19696, 20464, 29296, 40816, 51568, 52336} are congruent to 1 (mod 3); this of course includes both primes 103 and 2011. It appears that there are yet more k congruent to 1 (mod 3) greater than 2^16.