A133162 Trajectory of 1 under the morphism 1 -> {1,1,2,1}, 2 -> {2}.

1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1

Offset: 1

Author

N. J. A. Sloane, Oct 09 2007, Oct 10 2007

Keywords

• nonn
• easy

Comments

It can be shown that this is lim_{t -> oo} S_t, where S_0 = 1, S_{t+1} = S_t S_t 2 S_t.

Suggested by A131989: a(n) = length of n-th run of 1's in A131989.

For a proof of this see the Comments of A131989. - Michel Dekking, Oct 19 2019

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• Index entries for sequences that are fixed points of mappings

Crossrefs

Cf. A049320, A131989, A317962.

Programs

Mathematica

Nest[Function[l, {Flatten[(l /. {1 -> {1,1,2,1}, 2 -> {2} })] }], {1}, 5] (* Georg Fischer, Jul 19 2019 *)

Formula

Denote the sequence by a(1), a(2), ...

Block t, that is, S_t, extends from n=1 through n=(3^(t+1)-1)/2.

Given n, to find a(n): first find t from

p = (3^t-1)/2 < n <= (3^(t+1)-1)/2.

Then if n=3^t, a(n) = 2. Otherwise, a(n) = a(n'), where

n' = n-p if n<3^t, otherwise n' = n-2p-1.