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A292682
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Rule 230: (000, ..., 111) -> (0, 1, 1, 0, 0, 1, 1, 1), without extending to the right of input bit 0.
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2
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0, 3, 6, 5, 12, 15, 10, 11, 24, 27, 30, 29, 20, 23, 22, 23, 48, 51, 54, 53, 60, 63, 58, 59, 40, 43, 46, 45, 44, 47, 46, 47, 96, 99, 102, 101, 108, 111, 106, 107, 120, 123, 126, 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
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OFFSET
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0,2
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COMMENTS
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The orbit of 1 under this rule is A006977.
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).
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LINKS
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EXAMPLE
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n | a(n)
0 = 0[2] | 0[2] = 0
1 = 1[2] | 11[2] = 3 (bits below 001 and 01(0) are on)
2 = 10[2] | 110[2] = 6 (1 below 001 and 010, 0 below 10(0))
3 = 11[2] | 101[2] = 5 (1 below 001 and 11(0), 0 below 011.)
4 = 100[2] | 1100[2] = 12 (as n = 1 and n = 2, shifted left once more)
5 = 101[2] | 1111[2] = 15 (1 below 001, 010 (twice) and 101)
6 = 110[2] | 1010[2] = 10 (as n = 3, shifted left once)
7 = 111[2] | 1011[2] = 11 (1 below 001, 111 and 11(0), 0 below 011).
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PROG
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(PARI) apply( A292682(n, r=230)=sum(i=0, logint(!n+n<<=1, 2)+1, bittest(r, bitand(n>>i, 7))<<i), [0..60])
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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