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A165197
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Smallest integers whose binary digits "1" define a polyomino if sorted into a quadrant shape.
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2
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1, 3, 7, 11, 15, 23, 27, 30, 31, 47, 62, 63, 75, 79, 91, 94, 95, 111, 126, 127, 143, 159, 175, 181, 182, 183, 188, 189, 190, 191, 207, 219, 220, 221, 222, 223, 239, 252, 253, 254, 255, 347, 350, 351, 367, 378, 379, 382, 383, 406, 407, 413, 415, 431, 443, 446
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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EXAMPLE
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(i) The triangular representations of 3= 11, 5 = 101, 10=1010 and 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.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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