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
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, Nov 20 2016
STATUS
approved