OFFSET
1,2
COMMENTS
Each integer k can be mapped onto a black-and-white checkerboard pattern if we read the digits of its binary representation into the diagonals of a quadrant form, the least-significant digit b(0) into the corner, taking 1, then 2, then 3 etc. digits to fill consecutive diagonals of the quadrant:
b(0) b(2) b(5) b(9) ..
b(1) b(4) b(8) ..
b(3) b(7) ..
b(6) ..
b(10)
This will leave one last diagonal partially filled unless the number of binary digits in k is a triangular number.
Replace the "1" bits by black unit squares and the "0" or unset bits by white squares.
If the black squares define a singly connected polyomino considering position and up-down-left-right connectivity, and if the same polyomino cannot be created by a number smaller than k, we add k to the sequence. Polyominoes are considered the same if they can be matched by translations, rotations or flips.
For k >= 1, the number of terms with k 1's in their binary expansions equals A000105(k). - Pontus von Brömssen, Mar 02 2026
LINKS
Pontus von Brömssen, Table of n, a(n) for n = 1..12604 (all terms below 2^18)
EXAMPLE
(i) The triangular representations of 3 = 11, 5 = 101, and 10 = 1010 are
1
1
and
11
0
and
00
1
1
The 1's define the same 2-omino in all 3 cases, so only the smallest representative, the 3, enters the sequence.
(ii) For k = 181 = 10110101 the quadrant is filled with
111
01
01
0
No smaller number leads with this method to this T-shaped 5-omino, so 181 enters the sequence.
(iii) the representations of 6 = 110, 9 = 1001, and 29 = 11101 are
01
1
and
10
0
1
and
11
01
1
In all of these 3 cases, the 1's are not singly connected and do not represent polyominoes, so neither 6 nor 9 nor 29 can make it into the sequence.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Leonhard Kreissig, Sep 06 2009
EXTENSIONS
Explanation expanded by R. J. Mathar, Sep 22 2009
Term 220 removed (since it corresponds to the same polyomino as 94) by Joonas Jürgen Kisel, Mar 02 2026
STATUS
approved