A306211 Concatenation of the current sequence with the lengths of the runs in the sequence, with a(1) = 1.

1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 3, 1, 4, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 3, 1, 4, 1, 2, 1, 1, 1, 1, 3

Offset: 1

Author

A. D. Skovgaard, Jan 29 2019

Keywords

• nonn
• nice

Comments

Conjecture: All terms are less than or equal to 5. - Peter Kagey, Jan 29 2019

Conjecture: Every number appears! (Based on the analogy with the somewhat similar sequence A090822, where the first 5 appeared at around 10^(10^23) steps). - N. J. A. Sloane, Jan 29 2019

An alternative definition: Start with 1, extend the sequence by appending its RUNS transform, recompute the RUNS transform, append it, repeat. - N. J. A. Sloane, Jan 29 2019

The first time we see 1, 2, 3, 4, 5 is at n=1, 3, 37, 60, 255 (A323829). After 65 generations (10228800161220 terms) the largest term is 5. The relative frequencies of 1..5 are roughly 0.71, 6.7e-9, 0.23, 1.6e-8, 0.061. 2s and 4s appear to get rarer as n increases. - Benjamin Chaffin, Feb 07 2019

If we record the successive RUNS transforms and concatenate them, we get 1; 2; 2, 1; 2, 2, 1; 2, 2, 1, 2, 1; ..., which is this sequence without the initial 1. - A. D. Skovgaard, Jan 30 2019 (Rephrased by N. J. A. Sloane, Jan 30 2019)

Examples

			a(2) = 1, since there is a run of length 1 at a(1).
a(3) = 2, since there is a run of length 2 at a(1..2).
a(4..5) = 2, 1, since the runs are as follows:
  1, 1, 2  a(1..3)
  \__/  |
  2,    1  a(4..5)
a(37) = 3, since a(20..22) = 1, 1, 1.
Steps in construction:
  [1]  initial sequence
  [1]  its run length
 .
  [1, 1] concatenation of above is new sequence
  [2]  its run length
 .
  [1, 1, 2] concatenation of above is new sequence
  [2, 1]  its run lengths
 .
  [1, 1, 2, 2, 1]
  [2, 2, 1]
 .
  [1, 1, 2, 2, 1, 2, 2, 1]
  [2, 2, 1, 2, 1]
 .
  [1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1]
  [2, 2, 1, 2, 1, 2, 1, 1, 1]
 .
  [1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1]
  [2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3]
 .
  [1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3]
From _N. J. A. Sloane_, Jan 31 2019: (Start)
The first 9 generations, in compressed notation (see A323477) are:
  1
  11
  112
  11221
  11221221
  1122122122121
  1122122122121221212111
  1122122122121221212111221212111211113
  1122122122121221212111221212111211113221212111211113211113141
  ... (End)
		

Links

• Peter Kagey, Table of n, a(n) for n = 1..10029 (first 20 generations)
• N. J. A. Sloane, Table of n, a(n) for n = 1..236878 (first 27 generations)
• N. J. A. Sloane, Notes on A306211, Feb 01 2019

Crossrefs

Cf. A000002, A107946, A306215, A090822.

Positions of 3's, 4's, 5's: A323476, A306222, A306223.

Successive generations: A323477, A323478, A306215, A323475, A306333.

See also A323479, A323480, A323481, A323826 (RUNS transform), A323827, A323829 (where n first appears).

Programs

Haskell

group [] = []
group (x:xs)= (x:ys):group zs where (ys,zs) = span (==x) xs
a306211_next_gen xs = xs ++ (map length $ group xs)
a306211_gen 1 = [1]
a306211_gen n = a306211_next_gen $ a306211_gen (n-1)
a306211 n = a306211_gen n !! (n-1)
-- Jean-François Antoniotti, Jan 31 2021

Mathematica

seq[n_] := seq[n] = If[n==1, {1}, Join[seq[n-1], Length /@ Split[seq[n-1]]]];
seq[10] (* Jean-François Alcover, Jul 19 2022 *)