

A256535


The largest number of Ttetrominos that fit within an n X n square.


0



0, 0, 1, 4, 5, 8, 11, 16, 19, 24, 29, 36, 41, 48, 55, 64, 71, 80, 89, 100, 109, 120, 131, 144, 155, 168, 181, 196, 209, 224, 239, 256, 271, 288, 305, 324, 341, 360, 379, 400, 419, 440, 461, 484, 505, 528, 551, 576, 599, 624, 649, 676, 701, 728, 755, 784, 811
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OFFSET

1,4


COMMENTS

No Ttetromino fits in a 1 X 1 or 2 X 2 square: a(1)=a(2)=0. A single Ttetromino can be placed in a 3 X 3 square, and must occupy the center square. Four Ttetrominos fit within a 4 X 4 square with no spaces left over, in a rotationally symmetric tiling: a(4)=4.
For n = 4m, it is obvious that a(n) = n^2/4, by repeating the construction for n=4. For n = 4m + 2, it can be shown, using a chessboard coloring, that a(n) < n^2/4. By tiling an Lshaped strip of width 2 in a manner that can be indefinitely extended, one can show that a(n) = n^2/4  1.
Shuxin Zhan proved that it is not possible to tile a square of side 4m+1 or 4m+3 with Ttetrominos and a single monomino. Thus there must be at least 5 empty squares in any partial tiling by Ttetrominos. This bound is achieved for tilings in 5 X 5, 7 X 7, 9 X 9 and 11 X 11 squares. Robert Hochberg proved that for n > 11, there must be either 5 or 9 empty squares. He conjectured that 5 is always enough.
Jack W Grahl proved that, for squares, 5 monominos are always sufficient. This means that the sequence is given by n^2/4, (n^21)/41, n^2/41, (n^21)/41, for n = 4m, n = 4m+1, n = 4m+2 and n = 4m+3, respectively (which the exception of a(1) = 0), and generating function x^3*(12*x+2*x^22*x^3+x^4) ) / ( (1+x)*(x^2+1)*(x1)^3 ).  Jack W Grahl, Jul 25 2018


LINKS

Table of n, a(n) for n=1..57.
Jack W Grahl, Every square can be tiled with Ttetrominos and no more than 5 monominos, arXiv:1807.09201 [math.CO], 2018.
R. Hochberg, The gap number of the Ttetromino arxiv:1403.6730, [math.CO], June 2014.
S. Zhan, Tiling a deficient rectangle with ttetrominoes, August 2012.


FORMULA

From Jack W Grahl, Jul 25 2018: (Start)
a(4m) = 4m^2;
a(4m+1) = 4m^2 + 2m  1;
a(4m+2) = 4m^2 + 4m;
a(4m+3) = 4m^2 + 6m + 1. (End)


EXAMPLE

The optimal tiling for a 4 X 4 square is:
AAAB
DABB
DDCB
DCCC
This forms the building block of a solution for all n a multiple of four.
For n=7 a solution is given by:
ABBBCCC
AABD/CE
A/DDDEE
F/E
FFG
FGG
//G
with 5 empty squares, and the 4 X 4 square in the lower left filled in as above.
For n=6, a tiling of the excess after removing a 4 X 4 square shows us how optimal solutions can be generated for any even number that is not a multiple of 4:
//ABBB
/AAABC
CC
DC
DD
D/
The pairs A&B and C&D can be extended in the manner of a frieze. A nice solution for 9 X 9 does not include tilings of smaller even squares:
ABBBCDDDE
AABFCCDEE
AGFFCHHHE
GGGFI/HJ/
KKKIIIJJJ
/KL//MNNN
OLLLPMMNQ
OORPPMSQQ
ORRRPSSSQ


MATHEMATICA

Delete[Flatten[ Table[{4n^2, 4n^2 + 2n  1, 4n^2 + 4n, 4n^2 + 6n + 1}, {n, 0, 14}]], 2] (* or *)
CoefficientList[ Series[1 + (x^4  2x^3  2x + 1)/((x  1)^3 (x^3 + x^2 + x + 1)), {x, 0, 58}], x] (* Robert G. Wilson v, Jul 25 2018 *)


CROSSREFS

Sequence in context: A293790 A190778 A117573 * A249669 A144062 A066233
Adjacent sequences: A256532 A256533 A256534 * A256536 A256537 A256538


KEYWORD

nonn,changed


AUTHOR

Jack W Grahl, Sep 15 2015


STATUS

approved



