A270060 Number of incomplete rectangles of area n.

0, 0, 1, 1, 3, 3, 6, 7, 9, 11, 14, 15, 19, 22, 23, 28, 30, 34, 36, 41, 42, 51, 49, 57, 55, 68, 64, 75, 71, 84, 79, 95, 89, 106, 92, 116, 104, 127, 116, 134, 121, 150, 130, 160, 143, 172, 148, 188, 156, 193, 177, 209, 177, 226, 185, 231, 210, 246, 207, 269, 218, 272, 239, 287, 238, 312, 250, 317, 279, 320, 271, 359, 283, 355, 316

Offset: 1

Author

Stanislav Mikusek, Mar 09 2016

Keywords

• nonn

Comments

An incomplete rectangle is a six-sided figure obtained when two rectangles with different widths are coupled together so that two of the edges form a straight line.

In other words, this shape is a rectangle from which a smaller rectangle has been removed from one corner.

Incomplete rectangles which differ by a rotation and/or reflection are not counted as different.

Also the number of integer partitions of n into parts of 2 distinct sizes, where any integer partition and its conjugate are considered equivalent. For example a(8)=7 counts (7,1), (6,2), (6,1,1), (5,3), (5,1,1,1), (4,2,2), and (3,3,2).

The unit squares composing the incomplete rectangle can be viewed as the boxes of a Ferrers diagram of an integer partition of n with 2 different sizes of rows. A002133(n) counts all Ferrers diagrams with 2 different sizes of rows. A100073(n) counts all self-conjugate Ferrers diagrams with 2 different sizes of rows since these Ferrers diagrams look like a square with a smaller square removed from the corner. Thus a(n)=(A002133(n)+A100073(n))/2. Lara Pudwell, Apr 03 2016

Examples

			n = 3
.___.
| ._|
|_|
.
n = 4
._____.
| .___|
|_|
.
n = 5
._______. ._____. ._____.
| ._____| |   ._| | .___|
|_|       |___|   | |
                  |_|
.
The three solutions for n = 6:
XXXXX
X
.....
XXXX
XX
.....
XXXX
X
X
.....
		

Crossrefs

Cf. A038548 (number of complete rectangles of area n), A002133, A100073, A067627.

Programs

Maple

# see A067627(n,k=2).

Pseudocode

/* rectangle : LL = long side, SS = short side
removed corner : L = long side, S = short side */
{
int a[100];
int LL,SS,L,S,area;
for(area:=1;area<=100;area++){
a[area]:=0;
};
for(LL:=1;LL<=100;LL++){
for(SS:=1;SS<=LL;SS++){
  for(L:=1;L<=LL;L++){
    for(S:=1;S<=LL;S++){
      area=LL*SS-L*S;
      if( area>=1 && area<=100 ){
        if( L>=S || LSS ){
          a[area]++;
        };
      };
    };
  };
};
};
for(area:=1;area<=100;area++){
  print a[area];
};
}

Formula

a(n)=(A002133(n)+A100073(n))/2. See the integer partition comment above. Lara Pudwell, Apr 03 2016

G.f.: sum(sum(x^(i+j)/(2*(1-x^i)*(1-x^j))+x^(i^2-j^2)/2,j=1..i-1),i=1..infinity). See the integer partition comment above. Lara Pudwell, Apr 03 2016