A195822 - OEIS

A195822

Triangle read by rows in which row n lists the Dyson's ranks of all partitions of n that do not contain 1 as a part, in nonincreasing order.

0, 1, 2, 3, 0, 4, 1, 5, 2, 1, -1, 6, 3, 2, 0, 7, 4, 3, 2, 1, 0, -2, 8, 5, 4, 3, 2, 1, 0, -1, 9, 6, 5, 4, 3, 3, 2, 1, 1, 0, -1, -3, 10, 7, 6, 5, 4, 4, 3, 2, 2, 1, 1, 0, -1, -2, 11, 8, 7, 6, 5, 5, 4, 4, 3, 3, 2, 2, 2, 1, 1, 0, 0, -1, -1, -2, -4

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OFFSET

1,3

COMMENTS

The sum of row n is equal to A000041(n-1), if n >= 2. Proof: Dyson defined the rank of a partition as the largest part minus the number of parts. On the other hand the total number of parts in all partitions of n equals the sum of largest parts of all partitions of n (Cf. A006128), hence the sum of the ranks of all partitions of n is equal to zero. Let p(n) be the number of partitions of n. If we now add an part equal to 1 in each partition of n we obtain the partitions of n+1 that contain 1 as a part. The sum of the ranks of these partitions is p(n)*(-1) because the largest parts are the same but now there is an additional part in each partition. On the other hand the sum of the ranks of all partitions of n+1 is equal to zero, hence the sum of the ranks of all partitions of n+1 that do not contain 1 as a part is equal to p(n).

LINKS

Table of n, a(n) for n=1..77.

A. O. L. Atkins and F. G. Garvan, Relations between the ranks and cranks of partitions, arXiv:math/0208050 [math.NT], 2002.

EXAMPLE

Triangle begins:

3, 0;

4, 1;

5, 2, 1, -1;

6, 3, 2, 0;

7, 4, 3, 2, 1, 0, -2;

8, 5, 4, 3, 2, 1, 0, -1;

9, 6, 5, 4, 3, 3, 2, 1, 1, 0, -1, -3;