

A106109


Let S_0 = {1}; let S_n be the image of S_{n1} under the morphism 1>{3}, 2>{3, 4}, 3>{6, 5, 6}, 4>{6, 6, 6}, 5>{1}, 6>{1, 2}; sequence gives the concatenation S_0, S_1, S_2, ...


0



1, 3, 6, 5, 6, 1, 2, 1, 1, 2, 3, 3, 4, 3, 3, 3, 4, 6, 5, 6, 6, 5, 6, 6, 6, 6, 6, 5, 6, 6, 5, 6, 6, 5, 6, 6, 6, 6, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 4, 3, 3, 3, 4, 3
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OFFSET

0,2


COMMENTS

This simulates a threelevel twostate neural net on six symbols: FibonacciCantorFibonacci.
From Michel Dekking, Oct 13 2020: (Start)
This sequence is a morphic sequence, i.e., the lettertoletter image of a fixed point of a morphism. Let
alpha: 1>3, 2>34, 3>656, 4>666, 5>1, 6>12
be the defining morphism for this sequence.
Define the morphism beta on {1,2,3,4,5,6,7} as follows:
beta(j) = alpha(j) for j<7, beta(7) = 73.
Let y be the fixed point of beta starting with 7.
Define the lettertoletter map lambda by lambda(j) = j for j<7, and lambda(7) = 1. Then we have for all n:
lambda(beta^n(7)) = 1 alpha(1) ... alpha^n(1) = S_0 S_1 ... S_n.
This is easily proved by induction, using that
lambda(beta^n(3)) = alpha^n(3) = alpha^{n+1}(1).
Letting n tend to infinity we find that
lambda(y) = (a(n)).
(End)


LINKS

Table of n, a(n) for n=0..105.


FORMULA

1>{3}, 2>{3, 4}, 3>{6, 5, 6}, 4>{6, 6, 6}, 5>{1}, 6>{1, 2}.


MATHEMATICA

s[1] = {3}; s[2] = {3, 4}; s[3] = {6, 5, 6}; s[4] = {6, 6, 6}; s[5] = {1}; s[6] = {1, 2}; t[a_] := Flatten[s /@ a]; p[0] = {1}; p[1] = t[p[0]]; p[n_] := t[p[n  1]] aa = Flatten[Table[p[i], {i, 1, 8}]]


CROSSREFS

Sequence in context: A291050 A268981 A245652 * A275925 A282581 A247581
Adjacent sequences: A106106 A106107 A106108 * A106110 A106111 A106112


KEYWORD

nonn,tabf


AUTHOR

Roger L. Bagula, May 07 2005


EXTENSIONS

Edited by N. J. A. Sloane, Aug 23 2007


STATUS

approved



