A356802 A refinement of the Mahonian numbers (non-canonical ordering).

1, 1, 1, 1, 2, 2, 1, 1, 3, 5, 3, 3, 5, 3, 1, 1, 4, 9, 6, 9, 16, 4, 11, 11, 4, 16, 9, 6, 9, 4, 1, 1, 5, 14, 10, 19, 35, 14, 26, 40, 5, 10, 61, 19, 35, 26, 40, 40, 26, 35, 19, 61, 10, 5, 40, 26, 14, 35, 19, 10, 14, 5, 1

Offset

1,5

Comments

Let T(N,d) be a Mahonian number.

(1) Find all partitions k(1) >= k(2) >= ... >= k(N) = 0 of the number d with at most N-1 parts, such that k(i) - k(i+1) <= 1 and k(N) = 0.

(2) For each such partition, draw a ribbon Young diagram with N boxes at matrix (row-column) coordinates (k(i), k(i) + i - 1), i = 1, ..., N.

(3) For each ribbon diagram, count all standard Young skew tableaux.

The numbers under (3) will add up to T(N,d), as proven in the cited reference. These refinements appear consecutively as subsequences in the sequence, ordered by increasing N and increasing d from 0 to N(N-1)/2 for each N. The ordering within each subsequence is reverse-lexicographic by the partitions (1), i.e., from the largest k(1) down.

This ordering corresponds to the ordering of generating polynomials from the lowest power in d up, and from the highest power in k(1) down. Because of certain symmetries in the sequence (cf. the example), it is the same as decreasing d from N(N-1)/2 to 0 and ordering the partitions lexicographically. For the converse (canonical) convention and a program implementation, cf. A357611.

The sums of rows (3) are A008302. Because the latter is itself a triangle, arranged in the (N, d) plane, the present sequence is actually three-dimensional: each number in the triangle A008302 can be replaced by the row numbers (3), each of which is thus coordinatized by (N, d, m), the last coordinate being its position in the row.

The range of the m coordinate is the number of partitions satisfying the constraint (1). It is proven in the cited reference to be equal to the coefficient of q^d in the q-binomial theorem: coeff[q^d] Product_{k=1..N-1} (1 + q^k).

Links

Table of n, a(n) for n=1..63.

D. K. Sunko, Evaluation and spanning sets of confluent Vandermonde forms, arXiv:2209.02523 [math-ph], 2022.

D. K. Sunko, Evaluation and spanning sets of confluent Vandermonde forms, J. Math. Phys. 63, 082101 (2022).

Example

The first nontrivial terms in the sequence are a(11) = a(12) = 3, corresponding to the refinement T(4, 3) = 6 = 3 + 3. The terms from a(1) to a(10) are the Mahonian numbers themselves, because the refinement is trivial for them (there is only one partition satisfying the given constraints).
The data in triangular form are:
N, d
1, 0    1
2, 0    1
   1    1
3, 0    1
   1    2
   2    2
   3    1
4, 0    1
   1    3
   2    5
   3    3, 3
   4    5
   5    3
   6    1
5, 0    1
   1    4
   2    9
   3    6, 9
   4    16, 4
   5    11, 11
   6    4, 16
   7    9, 6
   8    9
   9    4
  10    1
6, 0    1
   1    5
   2    14
   3    10, 19
   4    35, 14
   5    26, 40, 5
   6    10, 61, 19
   7    35, 26, 40
   8    40, 26, 35
   9    19, 61, 10
  10    5, 40, 26
  11    14, 35
  12    19, 10
  13    14
  14    5
  15    1
One can check the generating function for the number of terms in a row, e.g., for N = 4: (1 + q)(1 + q^2)(1 + q^3) = q^6 + q^5 + q^4 + 2q^3 + q^2 + q + 1.

Crossrefs

Cf. A008302, A357611.

Sequence in context: A157103 A135966 A292741 * A060351 A335845 A357611

Adjacent sequences: A356799 A356800 A356801 * A356803 A356804 A356805

Keyword

nonn,tabf

Author

Denis K. Sunko, Aug 28 2022

Status

approved