

A331478


Irregular triangle T(n,k) = n  (s  k + 1)^2 for 1 <= k <= s, with s = floor(sqrt(n)).


1



0, 1, 2, 0, 3, 1, 4, 2, 5, 3, 6, 4, 7, 0, 5, 8, 1, 6, 9, 2, 7, 10, 3, 8, 11, 4, 9, 12, 5, 10, 13, 6, 11, 14, 0, 7, 12, 15, 1, 8, 13, 16, 2, 9, 14, 17, 3, 10, 15, 18, 4, 11, 16, 19, 5, 12, 17, 20, 6, 13, 18, 21, 7, 14, 19, 22, 8, 15, 20, 23, 0, 9, 16, 21, 24, 1
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OFFSET

1,3


COMMENTS

Row n begins with n  floor(sqrt(n)).
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)).
We borrow the term "square excess" from A053186(n), which is simply the difference n  floor(sqrt(n)).
Row n of this sequence contains distinct Durfee square excesses among all integer partitions of n (see example below).


LINKS

Michael De Vlieger, Table of n, a(n) for n = 1..10125 (rows 1 <= n <= 625, flattened)
Eric Weisstein's World of Mathematics, Durfee Square.


FORMULA

Let s = floor(sqrt(n));
T(n,1) = A053186(n) = n  s;
T(n,k) = T(n,1) + partial sums of 2(s  k + 1) + 1 for 2 <= k <= s + 1.
A000196(n) = Length of row n.
A022554(n) = Sum of row n.
Last term in row n = T(n, A000196(n)) = n  1.


EXAMPLE

Table begins:
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;
10: 1, 6, 9;
11: 2, 7, 10;
12: 3, 8, 11;
13: 4, 9, 12;
14: 5, 10, 13;
15: 6, 11, 14;
16: 0, 7, 12, 15;
...
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. Therefore we have two unique solutions to n  s^2 for n = 4, i.e., {0, 3}, so row 4 contains these values.


MATHEMATICA

Array[#  Reverse@ Range[Sqrt@ #]^2 &, 625] // Flatten


CROSSREFS

Cf. A000196, A022554, A053186, A117522.
Sequence in context: A025636 A025637 A195826 * A097065 A084964 A267182
Adjacent sequences: A331475 A331476 A331477 * A331479 A331480 A331481


KEYWORD

nonn,easy,tabf


AUTHOR

Michael De Vlieger, Jan 17 2020


STATUS

approved



