A278388 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).

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

Offset

1,3

Comments

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.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Example

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

Prog

(PARI) sumn2a = 0; for (n=1, 1 000, a=0; while (bitand(sumn2a, n<<a), a++); print1 (a ", "); sumn2a += n<<a)

Crossrefs

Cf. A275152.

Sequence in context: A256358 A308842 A241761 * A239737 A262883 A308908

Adjacent sequences: A278385 A278386 A278387 * A278389 A278390 A278391

Keyword

nonn,base

Author

Rémy Sigrist, Nov 20 2016

Status

approved