A331553 Irregular triangle T(n,k) = A115722(n,k)^2 - n.

0, 0, 1, 1, 2, 2, 2, 3, 3, 0, 3, 3, 4, 4, 1, 4, 1, 4, 4, 5, 5, 2, 5, 2, 2, 5, 2, 2, 5, 5, 6, 6, 3, 6, 3, 3, 6, 3, 3, 3, 6, 3, 3, 6, 6, 7, 7, 4, 7, 4, 4, 7, 4, 4, 4, 4, 7, 4, 4, 4, 4, 7, 4, 4, 4, 7, 7, 8, 8, 5, 8, 5, 5, 8, 5, 5, 5, 5, 8, 5, 5, 5, 5, 5, 8, 0, 5

Offset: 0

Author

Michael De Vlieger, Jan 20 2020

Keywords

• nonn
• tabf

Comments

Let P be an integer partition of n, and let D be the Durfee square of P with side length s, thus area s^2. We borrow the term "square excess" from A053186(n), which is simply the difference n - floor(sqrt(n)). This sequence lists the "Durfee square excess" of P = n - s^2 for all partitions P of n in reverse lexicographic order.

Zero appears in row n for n that are perfect squares. Let r = sqrt(n). For perfect square n, there exists a partition of n that consists of a run of r parts that are each r themselves; e.g., for n = 4, we have {2, 2}, for n = 9, we have {3, 3, 3}. It is clear through the Ferrers diagram of these partitions that they are equivalent to their Durfee square, thus n - s^2 = 0.

Since the partitions of any n contain Durfee squares in the range of 1 <= s <= floor(sqrt(n)) (with perfect square n also including k = 0), the distinct Durfee square excesses must be the differences n - s^2 for 1 <= s <= floor(sqrt(n)).

Examples

			Table begins:
0: 0;
1: 0;
2: 1, 1;
3: 2, 2, 2;
4: 3, 3, 0, 3, 3;
5: 4, 4, 1, 4, 1, 4, 4;
6: 5, 5, 2, 5, 2, 2, 5, 2, 2, 5, 5;
7: 6, 6, 3, 6, 3, 3, 6, 3, 3, 3, 6, 3, 3, 6, 6;
...
Table of distinct terms:
1:  0;
2:  1;
3:  2;
4:  0,  3;
5:  1,  4;
6:  2,  5;
7:  3,  6;
8:  4,  7;
9:  0,  5,  8;
...
For n = 4, the partitions are {4}, {3, 1}, {2, 2}, {2, 1, 1}, {1, 1, 1, 1}. The partition {2, 2} has Durfee square s = 2; for all partitions except {2, 2}, we have Durfee square with s = 1.
		

Links

• Eric Weisstein's World of Mathematics, Durfee Square.

Crossrefs

Cf. A115722, A116365.

Programs

Mathematica

{0}~Join~Array[Map[Total@ # - Block[{k = Length@ #}, While[Nand[k > 0, AllTrue[Take[#, k], # >= k &]], k--]; k]^2 &, IntegerPartitions[#]] &, 12] // Flatten

Formula

T(n,k) = A115722(n,k)^2 - n.

2 * A116365(n) = sum of row n.