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a(n) = (1 + A151548(n))/2.
1

%I #7 Feb 17 2015 00:13:27

%S 1,2,3,4,3,6,9,8,3,6,9,10,11,20,25,16,3,6,9,10,11,20,25,18,11,20,27,

%T 30,41,64,65,32,3,6,9,10,11,20,25,18,11,20,27,30,41,64,65,34,11,20,27,

%U 30,41,64,67,46,41,66,83,100,145,192,161,64,3,6,9,10,11,20

%N a(n) = (1 + A151548(n))/2.

%C It appears that when A255045 is regarded as a triangle with rows of lengths 1, 2, 4, 8, 16, ..., this is what the rows converge to.

%H <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>

%e Written as an irregular triangle in which the row lengths are the terms of A011782:

%e 1;

%e 2;

%e 3,4;

%e 3,6,9,8;

%e 3,6,9,10,11,20,25,16;

%e 3,6,9,10,11,20,25,18,11,20,27,30,41,64,65,32;

%e 3,6,9,10,11,20,25,18,11,20,27,30,41,64,65,34,11,20,27,30,41,64,67,46,41,66,83,100,145,192,161,64;

%e ...

%e Right border gives A000079.

%Y Cf. A000079, A011782, A139251, A151548, A160552, A255045.

%K nonn,tabf

%O 0,2

%A _Omar E. Pol_, Feb 15 2015