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A180001 Eventual period of a single cell in rule 110 cellular automaton in a cyclic universe of width n. 24
1, 1, 1, 2, 1, 9, 14, 16, 7, 25, 110, 9, 351, 91, 295, 32, 7, 27, 285, 30, 630, 44, 1058, 36, 250, 7, 405, 1652, 1044, 60, 7, 64, 495, 51, 1050, 72, 4403, 76, 390, 60, 7, 630, 1548, 88, 7, 7, 705, 96, 1470, 100, 765, 195, 8109, 7, 825, 7, 2052, 116, 7, 19560, 915 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,4
COMMENTS
The first 21 terms match the most frequent possible outcome (see comment in A332717) with the exception of a(14) which is the second-most frequent. - Hans Havermann, Jun 11 2020
LINKS
Pontus von Brömssen, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Rule 110
EXAMPLE
For n=4, the evolution of a single cell is:
0001
0011
0111 <--= period starts
1101
0111 <--= again start of period
etc, so a(4)=2.
MATHEMATICA
a[n_] := -Subtract @@
Flatten[Map[Position[#, #[[-1]]] &,
NestWhileList[CellularAutomaton[110],
Prepend[Table[0, {n - 1}], 1], Unequal, All], {0}]]
PROG
(Sage)
def A180001(n):
def bit(x, i): return (x >> i) & 1
rulemap = dict((tuple(bit(i, k) for k in reversed(range(3))), bit(110, i)) for i in range(8))
def neighbours(d, i): return tuple(d[k % n] for k in [i-1..i+1])
v = [0]*n; v[-1] = 1;
history = [v]
while True:
v2 = [rulemap[neighbours(history[-1], i)] for i in range(n)]
if v2 in history: return len(history)-history.index(v2)
history.append(v2) # D. S. McNeil, Jan 15 2011
CROSSREFS
Sequence in context: A085488 A072265 A192352 * A204371 A199887 A058876
KEYWORD
nonn
AUTHOR
Ben Branman, Jan 13 2011
EXTENSIONS
More terms from Alois P. Heinz, Jan 14 2011
STATUS
approved

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Last modified April 25 03:15 EDT 2024. Contains 371964 sequences. (Running on oeis4.)