

A279125


Lexicographically earliest sequence such that, for any distinct i and j, a(i)=a(j) implies (i AND j)=0 (where AND stands for the bitwise AND operator).


5



0, 0, 1, 0, 2, 3, 4, 0, 3, 2, 5, 1, 6, 7, 8, 0, 7, 6, 9, 5, 10, 11, 12, 4, 13, 14, 15, 16, 17, 18, 19, 0, 11, 10, 16, 9, 14, 13, 20, 12, 21, 22, 23, 24, 25, 26, 27, 1, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 0, 18, 17, 24, 15, 22, 21, 35, 9
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OFFSET

1,5


COMMENTS

This sequence is similar to A279119 in the sense that here we check for common ones in binary representation and there we check for common prime factors.
By analogy with A275152, this sequence can be seen as a way to tile the first quadrant with fixed disconnected 2dimensional polyominoes: the (vertical) polyomino corresponding to n is shifted to the right as little as possible so as not to overlap a previous polyomino, and a(n) gives the corresponding number of steps to the right (see illustration in Links section).


LINKS



FORMULA



MAPLE

with(Bits):
n:= 100:
l:= []:
g:=[seq(0, i = 0..n1)]:
for i from 1 to n by 1
do
a:= 0;
while (And(g[a + 1], i)) > 0
do
a++;
end do:
g[a + 1] += i;
l:= [op(l), a];
end do:


MATHEMATICA

n = 100;
l = {};
g = ConstantArray[0, n];
For[i = 0, i < n, i++; a = 0; While[BitAnd[g[[a + 1]], i] > 0, a++];
g[[a + 1]] += i;
l = Append[l, a]];


PROG

(PARI) g = vector(72); for (n=1, #g, a = 0; while (bitand(g[a+1], n)>0, a++); g[a+1] += n; print1 (a", "))
(Python)
n = 100
g = n * [0]
for i in range(1, n + 1):
a = 0
while g[a] & i:
a += 1
g[a] += i


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



