%I #7 Oct 09 2017 23:32:57

%S 0,3,6,5,12,15,10,11,24,27,30,29,20,23,22,23,48,51,54,53,60,63,58,59,

%T 40,43,46,45,44,47,46,47,96,99,102,101,108,111,106,107,120,123,126,

%U 125,116,119,118,119,80,83,86,85,92,95,90,91,88,91,94,93,92,95,94,95,192,195,198,197,204,207

%N Rule 230: (000, ..., 111) -> (0, 1, 1, 0, 0, 1, 1, 1), without extending to the right of input bit 0.

%C The orbit of 1 under this rule is A006977.

%C The substitution rules 000 -> 0 and 100 -> 0 ensure that no (even or odd) input value can ever extend / "propagate" to the right, therefore it is not required to consider the additional digit to the right of input bit 0 (i.e., the cell which would have this bit 0 as left neighbor), as one would usually do in the context of elementary cellular automata (cf., e.g., A292680 vs. A292681).

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ElementaryCellularAutomaton.html">Elementary Cellular Automaton</a>

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

%H <a href="https://oeis.org/wiki/Index_to_Elementary_Cellular_Automata">Index to Elementary Cellular Automata</a>

%e n | a(n)

%e 0 = 0[2] | 0[2] = 0

%e 1 = 1[2] | 11[2] = 3 (bits below 001 and 01(0) are on)

%e 2 = 10[2] | 110[2] = 6 (1 below 001 and 010, 0 below 10(0))

%e 3 = 11[2] | 101[2] = 5 (1 below 001 and 11(0), 0 below 011.)

%e 4 = 100[2] | 1100[2] = 12 (as n = 1 and n = 2, shifted left once more)

%e 5 = 101[2] | 1111[2] = 15 (1 below 001, 010 (twice) and 101)

%e 6 = 110[2] | 1010[2] = 10 (as n = 3, shifted left once)

%e 7 = 111[2] | 1011[2] = 11 (1 below 001, 111 and 11(0), 0 below 011).

%o (PARI) apply( A292682(n,r=230)=sum(i=0,logint(!n+n<<=1,2)+1,bittest(r,bitand(n>>i,7))<<i),[0..60])

%Y Cf. A006977, A292680, A292681, A266178, A266179, A266180, A019590.

%K nonn,easy

%O 0,2

%A _M. F. Hasler_, Oct 09 2017