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A278388
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Lexicographically earliest sequence such that (i*2^a(i)) AND (j*2^a(j)) = 0 for any distinct i and j (AND stands for the bitwise AND operator).
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1
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0, 0, 2, 2, 5, 7, 10, 3, 13, 14, 18, 20, 24, 27, 31, 10, 35, 36, 41, 34, 44, 48, 53, 55, 60, 64, 69, 72, 77, 81, 86, 15, 51, 42, 61, 89, 93, 95, 101, 102, 108, 109, 115, 119, 123, 128, 134, 136, 138, 143, 145, 149, 155, 160, 166, 169, 175, 180, 186, 190, 196
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history;
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OFFSET
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1,3
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COMMENTS
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By analogy with A275152, this sequence can be obtained by the following algorithm:
- we start with a half-open line of empty squares with coordinates X=0, X=1, X=2, etc.,
- for n=1, 2, 3, ...: we choose the least k such that the polyomino corresponding to n, shifted by k squares to the right, does not overlap one of the previous polyominoes.
a(2*k+1) > a(2*k) for any k>0.
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LINKS
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EXAMPLE
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The following table depicts the first terms, alongside the corresponding polyominoes ("X" denotes a filled square, "_" denotes an empty square):
n n in binary a(n) n as a polyomino shifted by a(n) to the right
-- ----------- ---- ---------------------------------------------
1 1 0 X
2 10 0 _X
3 11 2 XX
4 100 2 __X
5 101 5 X_X
6 110 7 _XX
7 111 10 XXX
8 1000 3 ___X
9 1001 13 X__X
10 1010 14 _X_X
11 1011 18 XX_X
12 1100 20 __XX
13 1101 24 X_XX
14 1110 27 _XXX
15 1111 31 XXXX
16 10000 10 ____X
17 10001 35 X___X
18 10010 36 _X__X
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PROG
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(PARI) sumn2a = 0; for (n=1, 1 000, a=0; while (bitand(sumn2a, n<<a), a++); print1 (a ", "); sumn2a += n<<a)
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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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STATUS
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approved
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