A106109 Let S_0 = {1}; let S_n be the image of S_{n-1} 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, ...

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

Offset

0,2

Comments

This simulates a three-level two-state neural net on six symbols: Fibonacci-Cantor-Fibonacci.

From Michel Dekking, Oct 13 2020: (Start)

This sequence is a morphic sequence, i.e., the letter-to-letter 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 letter-to-letter 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: A268981 A353909 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