cp's OEIS Frontend

Permutes the 4 axes of the 4D-Ferrers plot of the solid partitions as 2143.

The hooklength list of a plane partition is the sorted list of 3D hooklengths of its 3D Ferrers plot, analogous to the classic 2D case.

Conjecture: equal to 3*(2*A000219 -A000990 -2*A000041 +1) tested up to n=20.

Appears to equal 3*A097391(n-3).

A solid (or 3D) partition of n describes a piling of boxes in a corner with heights nonincreasing away from the corner, and containing integers, similarly nonincreasing, that sum to n.

The shape of a solid partitions is defined as the plane partition containing the heights of the piling, irrespective of the numerical content of the boxes.

Row sums equal A000293, T(n,n) = T(n+1,n) equals A000219;

Equals number of solid partitions with total by layer equal to partitions of n with largest part m.

Partitions are generated in reverse lexicographic order.

Remark that A262030 uses Abramowitz-Stegun (A-St) order.

Sum of row r equals r^r for r > 0 (Robinson-Schensted correspondence).

Triangular table with alternating signed sum equal to 0 for n>0,

1

1,-1

1,-2,1

1,-4,5,-2

1,-6,13,-12,4

..

and with alternating signed weighted sum (first moment) also equal to 0 for n>1,

0

0,-1

0,-2,2

0,-4,10,-6

0,-6,26,-36,16

..

also when shifting the weights to start at 1,

1

1,-2

1,-4,3

1,-8,15,-8

1,-12,39,-48,20

Length of row n is A370164(n).

No Markov number is divisible by any prime congruent to 3 (mod 4). Proof: Let m be a Markov number and let p be an odd prime dividing m. Then there exist positive integers b and c such that m^2 + b^2 + c^2 = 3 * m * b * c (Markov's equation). Furthermore it is known that b and c must be coprime to m. Evaluating Markov's equation (mod p) gives b^2 = -c^2 (mod p), which implies -1 is a square (mod p). This is only possible if p = 1 (mod 4).

Given a prime p congruent to 3 (mod 4) and greater than 3, no Markov number is congruent to 1/3 * (p - 2 * sigma(p)) or 2/3 * (p + sigma(p)) (mod p), where sigma(p) = -1 if p is congruent to 7 (mod 12) and sigma(p) = 1 if p is congruent to 11 (mod 12). Proof: Let m be a Markov number congruent to one of the forbidden residues (mod p). Then evaluating Markov's equation (mod p) using the forms of the forbidden residues above and doing a few simplifications implies that -4 is a square (mod p). This contradicts that p = 3 (mod 4).

The first comment above implies that an odd Markov number is congruent to 1 (mod 4). It has also been shown that any positive even integer satisfying all of the constraints of the first two comments above is congruent to 2 (mod 32).

It is conjectured that, modulo n, all residues not forbidden by the constraints in the three comments above are actually realized by some Markov number. Specifically, mod n the forbidden residues include the following: (1) if p is a prime congruent to 3 (mod 4) that divides n, then any residue congruent to 0 (mod p) or congruent to either of the forbidden residues listed in the second comment (mod p), (2) if 4 divides n, then any odd residue congruent to 3 (mod 4), (3) if 2^r is the highest power of 2 dividing n and 2 <= r <= 5, then any even residue not congruent to 2 (mod 2^r), (4) if 2^r is the highest power of 2 dividing n and r > 5, then any even residue not congruent to 2 (mod 32). It is conjectured that all other residues occur. This has been verified for all n <= 38000.

If the conjecture in the previous comment is correct, then it follows from the Chinese remainder theorem that a(n) may be found by writing n = 2^s * 3^t * u * p_1^r_1 * p_2^r_2 * ... * p_k^r_k, where p_1, ..., p_k are distinct primes greater than 3 and congruent to 3 (mod 4) and r_1, ..., r_k are positive and where the prime divisors of u are all congruent to 1 (mod 4). Then a(n) = u * C_s * D_t * Product_{j=1..k} (p_j - 3) * p_j^(r_j - 1), where C_s = 2^s if s < 2, 1 + 2^(s-2) if 2 <= s <= 5, and 2^(s - 5) + 2^(s - 2) if s > 5, and where D_t = 1 if t = 0 and 2 * 3^(t-1) if t > 0. If the conjecture holds then a(n) is a multiplicative function.

Both following comments are empirical observations:

1) also accumulant of A119907;

2) the characters of exactly these partitions do not occur in the decomposition of the count of parts 1<=k<=n into the characters of the symmetric group of n (Elders' Theorem).

3) the complement (partitions with no more than 2 parts >1) is counted by A033638.