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Cumulative counting sequence: method A (adjective-before-noun)-pairs with first term 0.
9

%I #29 Apr 26 2023 05:36:36

%S 0,1,0,2,0,1,1,3,0,3,1,1,2,4,0,5,1,2,2,2,3,5,0,6,1,5,2,3,3,1,4,1,5,6,

%T 0,9,1,6,2,5,3,2,4,4,5,1,6,7,0,11,1,8,2,6,3,4,4,6,5,4,6,0,7,0,8,1,9,

%U 10,0,13,1,9,2,7,3,7,4,7,5,7,6,2,7,2,8,2,9,0,10,1,11,12,0,15,1,13,2,8,3,8,4

%N Cumulative counting sequence: method A (adjective-before-noun)-pairs with first term 0.

%C Start with 0; at n-th step, write down what is in the sequence so far.

%C "Look and Say" how many times each integer (in increasing order), <= max {existing terms} appears in the sequence. Then concatenate. Sequence's graph looks roughly like that of A080096.

%C For the original version, where "increasing order..." is "order of 1st appearance", see A217760. The conjecture formerly placed here applies to A217760. - _Clark Kimberling_, Mar 24 2013

%H Zak Seidov, <a href="/A055186/b055186.txt">Table of n, a(n) for n = 1..1019</a> (the first 22 steps)

%F Conjectures: a(n) < 2*sqrt(n); limit as n goes to infinity Max( a(k) : 1<=k<=n)/sqrt(n) exist = 2. - _Benoit Cloitre_, Jan 28 2003

%e Write 0, thus having 1 0, thus having 2 0's and 1 1, thus having 3 0's and 3 1's and 1 2, etc. 0; 1,0; 2,0,1,1; 3,0,3,1,1,2; ...

%t s={0};Do[ta=Table[{Count[s, # ], # }&/@Range[0,Max[s]]]; s=Flatten[{s,ta}],{22}];s (* _Zak Seidov_, Oct 23 2009 *)

%Y Cf. A005150. For other versions see A051120, A079668, A079686.

%Y Cf. A055168-A055185 (method B) and A055187-A055191 (method A).

%Y Cf. A217760.

%K nonn,base

%O 1,4

%A _Clark Kimberling_, Apr 27 2000

%E Edited by _N. J. A. Sloane_, Jan 17 2009 at the suggestion of _R. J. Mathar_

%E Removed a conjecture. - _Clark Kimberling_, Oct 24 2009

%E Entries changed to match b-file. - _N. J. A. Sloane_, Oct 04 2010