cp's OEIS Frontend

A292681 Rule 6: (000, ..., 111) -> (0, 1, 1, 0, 0, 0, 0, 0), without extending to the right of input bit 0.

%I A292681 #16 Feb 16 2025 08:33:51
%S A292681 0,3,6,4,12,13,8,8,24,27,26,24,16,17,16,16,48,51,54,52,52,53,48,48,32,
%T A292681 35,34,32,32,33,32,32,96,99,102,100,108,109,104,104,104,107,106,104,
%U A292681 96,97,96,96,64,67,70,68,68,69,64,64,64,67,66,64,64,65,64,64,192,195,198,196,204,205,200,200,216
%N A292681 Rule 6: (000, ..., 111) -> (0, 1, 1, 0, 0, 0, 0, 0), without extending to the right of input bit 0.
%C A292681 This is A292680 / 2; the orbit of 1 under A292680 is A266180.
%C A292681 Rule 6 is the smallest rule which is even (otherwise infinitely many bits would be switched on at step 1, for any finite starting value) and nontrivial (i.e., does not lead to extinction nor simple reproduction, possibly shifted left or right, of a single-bit initial state).
%C A292681 In the context of elementary cellular automata, one would usually consider one further bit of output (i.e., the cell which has bit 0 of the input as left neighbor). This leads to A292680 which has all terms even, so it is natural to consider this variant.
%H A292681 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/ElementaryCellularAutomaton.html">Elementary Cellular Automaton</a>
%H A292681 <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>
%H A292681 <a href="https://oeis.org/wiki/Index_to_Elementary_Cellular_Automata">Index to Elementary Cellular Automata</a>
%e A292681      n        |         a(n)
%e A292681    0 =   0[2] |      0[2] =  0
%e A292681    1 =   1[2] |     11[2] =  3 (bits below 001 and 01(0) are on)
%e A292681    2 =  10[2] |    110[2] =  6 (1 below 001 and 010, 0 below 10(0))
%e A292681    3 =  11[2] |    100[2] =  4 (1 below 001, 0 below 011 and 11(0))
%e A292681    4 = 100[2] |   1100[2] = 12 (as n = 1 and n = 2, shifted right once more)
%e A292681    5 = 101[2] |   1101[2] = 13 (1 below 001 and 010 (twice), 0 below 101)
%e A292681    6 = 110[2] |   1000[2] =  8 (as n = 3, shifted right once)
%e A292681    7 = 111[2] |   1000[2] =  8 (1 below 001, 0 below 011, 111 and 11(0)).
%o A292681 (PARI) apply( A292681(n,r=6)=sum(i=0,logint(!n+n<<=1,2)+1,bittest(r,(n>>i)%8)<<i),[0..60])
%Y A292681 Cf. A292680, A266178, A266179, A266180, A019590.
%K A292681 nonn,easy
%O A292681 0,2
%A A292681 _M. F. Hasler_, Oct 09 2017