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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

%I #32 Jan 27 2022 21:27:41

%S 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,

%T 19,0,11,10,16,9,14,13,20,12,21,22,23,24,25,26,27,1,28,29,30,31,32,33,

%U 34,35,36,37,38,39,40,41,42,0,18,17,24,15,22,21,35,9

%N 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).

%C 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.

%C By analogy with A275152, this sequence can be seen as a way to tile the first quadrant with fixed disconnected 2-dimensional 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).

%H Rémy Sigrist, <a href="/A279125/b279125.txt">Table of n, a(n) for n = 1..10000</a>

%H Rémy Sigrist, <a href="/A279125/a279125.jpg">Illustration of the first terms (by way of polyominos)</a>

%H N. J. A. Sloane and Brady Haran, <a href="https://www.youtube.com/watch?v=j0o-pMIR8uk">Amazing Graphs III</a>, Numberphile video (2019).

%F a(n)=0 iff n belongs to A000079.

%F a(n)=1 iff n belongs to A164346.

%p with(Bits):

%p n:= 100:

%p l:= []:

%p g:=[seq(0, i = 0..n-1)]:

%p for i from 1 to n by 1

%p do

%p a:= 0;

%p while (And(g[a + 1], i)) > 0

%p do

%p a++;

%p end do:

%p g[a + 1] += i;

%p l:= [op(l), a];

%p end do:

%p print(l); # _Reza K Ghazi_, Dec 29 2021

%t n = 100;

%t l = {};

%t g = ConstantArray[0, n];

%t For[i = 0, i < n, i++; a = 0; While[BitAnd[g[[a + 1]], i] > 0, a++];

%t g[[a + 1]] += i;

%t l = Append[l, a]];

%t l (* _Reza K Ghazi_, Dec 29 2021 *)

%o (PARI) g = vector(72); for (n=1, #g, a = 0; while (bitand(g[a+1],n)>0, a++); g[a+1] += n; print1 (a", "))

%o (Python)

%o n = 100

%o g = n * [0]

%o for i in range(1, n + 1):

%o a = 0

%o while g[a] & i:

%o a += 1

%o g[a] += i

%o print(a, end=', ') # _Reza K Ghazi_, Dec 29 2021

%Y Cf. A000079, A164346, A275152, A279119.

%K nonn,base,look

%O 1,5

%A _Rémy Sigrist_, Dec 06 2016

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Last modified August 13 02:28 EDT 2024. Contains 375113 sequences. (Running on oeis4.)