

A226651


Multidimensional Young numbers: Given a ddimensional partition of n, this is the number of ways to fill the associated ddimensional Young diagram with the integers 1 to n such that the entries are increasing in each positive (orthogonal) direction.


1



1, 1, 1, 2, 1, 3, 2, 6, 1, 4, 5, 6, 12, 8, 24, 1, 5, 9, 10, 5, 16, 20, 25, 30, 20, 16, 60, 40, 120, 1, 6, 14, 15, 14, 25, 20, 21, 30, 54, 60, 30, 96, 40, 66, 61, 75, 90, 48, 120, 150, 180, 120, 96, 80, 360, 240, 720, 1, 7, 20, 21, 28, 64, 35, 14, 70, 56, 90, 42, 42, 98, 105, 98, 245, 140, 147
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OFFSET

1,4


COMMENTS

Generalization of the number of standard Young tableaux on a given Young diagram to arbitrary dimension.
The multidimensional Young numbers of partitions which are conjugate are equal. Therefore, the multidimensional Young numbers listed above are indexed with respect to an ordering of the "conjugacy classes" of partitions. This ordering is defined in the attached pdf file.
The number of entries between the mth and (m+1)th appearance of 1 (including the mth appearance, but excluding the (m+1)th) is the number of infinite dimensional partitions of m up to conjugacy, i.e., sequence A119268.
Let f(m) give the number of 2dimensional partitions of m up to conjugacy (sequence A005987). Then the first f(m) entries following (and including) the mth appearance of 1 are standard Young tableaux numbers on 2dimensional partitions of m, and can be found in sequence A117506.


LINKS

Table of n, a(n) for n=1..76.
Graham H. Hawkes, MATLAB Program
Graham H. Hawkes, Multidimensional Young Numbers


FORMULA

Let p be a partition of n. Let Q be the set of partitions of n1 such that for all q in Q, p covers q. Then the Young number of p is given by Y(p) = Sum_{q in Q} Y(q).


EXAMPLE

The ordering of "conjugacy classes" of partitions begins:
(1), (2), (3), (2+1), (4), (3+1), (2+2), ((2+1)+(1)), (5), (4+1), (3+2), (3+1+1), ((3+1)+(1)), ((2+2)+(1)),
(((2+1)+(1))+((1))), ...
The 14th partition, ((2+2)+(1)), is associated to the Young diagram with cubes centered at p_1=(0,0,0), p_2=(1,0,0), p_3=(0,1,0), p_4=(1,1,0), and p_5=(0,0,1). The possible ways to fill the cubes centered on these points so that the numbers are increasing in all directions are;
(For each i=1:5, the ith integer in a sequence below is placed on p_i.)
12345
13245
12354
13254
12453
14253
13452
14352
Hence the 14th term is 8.
The 48th partition, ((2+2)+(2+2)), can be represented as a cube divided into octants. The integers 1 and 8 must lie in opposite octants. Of the three octants adjacent to the one which contains 1, one must contain 2 and one must contain 3. This gives 6 possibilities. For each of these possibilities there are 4 numbers (4, 5, 6, and 7) to choose from for the number placed in the remaining cube in the plane that contains 1, 2, and 3. Regardless of this choice, there are 2 ways to fill in the remaining three octants. Thus there are 6*4*2=48 ways to fill the octants all togetherthat is, the 48th multidimensional Young number is 48.
Example of recursion:
The partition: p6=((3+2)+(1)) covers the following partitions of 5:
q_15=(3+2)
q_25=((3+1)+(1))
q_35=((2+2)+(1)) Thus Y(p)=Y(q_1)+Y(q_2)+Y(q_3)=5+12+8=25


PROG

See MATLAB function in Links.


CROSSREFS

Sequence in context: A075257 A260618 A306286 * A073711 A071690 A319179
Adjacent sequences: A226648 A226649 A226650 * A226652 A226653 A226654


KEYWORD

nonn,tabf


AUTHOR

Graham H. Hawkes, Jul 30 2013


STATUS

approved



