|
%I #96 Feb 16 2024 01:31:30
%S 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,
%T 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,
%U 1,1,2,1,1,1,1,3,2,1,1,1,1,3,1,4,1,2,1,1,1,1,3
%N Concatenation of the current sequence with the lengths of the runs in the sequence, with a(1) = 1.
%C Conjecture: All terms are less than or equal to 5. - _Peter Kagey_, Jan 29 2019
%C 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
%C 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
%C The first time we see 1, 2, 3, 4, 5 is at n=1, 3, 37, 60, 255. 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
%C 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)
%H Peter Kagey, <a href="/A306211/b306211.txt">Table of n, a(n) for n = 1..10029</a> (first 20 generations)
%H N. J. A. Sloane, <a href="/A306211/a306211_1.txt">Table of n, a(n) for n = 1..236878</a> (first 27 generations)
%H N. J. A. Sloane, <a href="/A306211/a306211.txt">Notes on A306211</a>, Feb 01 2019
%e a(2) = 1, since there is a run of length 1 at a(1).
%e a(3) = 2, since there is a run of length 2 at a(1..2).
%e a(4..5) = 2, 1, since the runs are as follows:
%e 1, 1, 2 a(1..3)
%e \__/ |
%e 2, 1 a(4..5)
%e a(37) = 3, since a(20..22) = 1, 1, 1.
%e Steps in construction:
%e [1] initial sequence
%e [1] its run length
%e .
%e [1, 1] concatenation of above is new sequence
%e [2] its run length
%e .
%e [1, 1, 2] concatenation of above is new sequence
%e [2, 1] its run lengths
%e .
%e [1, 1, 2, 2, 1]
%e [2, 2, 1]
%e .
%e [1, 1, 2, 2, 1, 2, 2, 1]
%e [2, 2, 1, 2, 1]
%e .
%e [1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1]
%e [2, 2, 1, 2, 1, 2, 1, 1, 1]
%e .
%e [1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1]
%e [2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3]
%e .
%e [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]
%e From _N. J. A. Sloane_, Jan 31 2019: (Start)
%e The first 9 generations, in compressed notation (see A323477) are:
%e 1
%e 11
%e 112
%e 11221
%e 11221221
%e 1122122122121
%e 1122122122121221212111
%e 1122122122121221212111221212111211113
%e 1122122122121221212111221212111211113221212111211113211113141
%e ... (End)
%t seq[n_] := seq[n] = If[n==1, {1}, Join[seq[n-1], Length /@ Split[seq[n-1]]]];
%t seq[10] (* _Jean-François Alcover_, Jul 19 2022 *)
%o (Haskell)
%o group [] = []
%o group (x:xs)= (x:ys):group zs where (ys,zs) = span (==x) xs
%o a306211_next_gen xs = xs ++ (map length $ group xs)
%o a306211_gen 1 = [1]
%o a306211_gen n = a306211_next_gen $ a306211_gen (n-1)
%o a306211 n = a306211_gen n !! (n-1)
%o -- _Jean-François Antoniotti_, Jan 31 2021
%Y Cf. A000002, A107946, A306215, A090822.
%Y Positions of 3's, 4's, 5's: A323476, A306222, A306223.
%Y Successive generations: A323477, A323478, A306215, A323475, A306333.
%Y See also A323479, A323480, A323481, A323826 (RUNS transform), A323827, A323829 (where n first appears).
%K nonn,nice
%O 1,3
%A _A. D. Skovgaard_, Jan 29 2019
|
