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A055186
Cumulative counting sequence: method A (adjective-before-noun)-pairs with first term 0.
9
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, 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, 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
OFFSET
1,4
COMMENTS
Start with 0; at n-th step, write down what is in the sequence so far.
"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.
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
LINKS
Zak Seidov, Table of n, a(n) for n = 1..1019 (the first 22 steps)
FORMULA
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
EXAMPLE
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; ...
MATHEMATICA
s={0}; Do[ta=Table[{Count[s, # ], # }&/@Range[0, Max[s]]]; s=Flatten[{s, ta}], {22}]; s (* Zak Seidov, Oct 23 2009 *)
CROSSREFS
Cf. A005150. For other versions see A051120, A079668, A079686.
Cf. A055168-A055185 (method B) and A055187-A055191 (method A).
Cf. A217760.
Sequence in context: A064922 A303337 A340502 * A217760 A339218 A263412
KEYWORD
nonn,base
AUTHOR
Clark Kimberling, Apr 27 2000
EXTENSIONS
Edited by N. J. A. Sloane, Jan 17 2009 at the suggestion of R. J. Mathar
Removed a conjecture. - Clark Kimberling, Oct 24 2009
Entries changed to match b-file. - N. J. A. Sloane, Oct 04 2010
STATUS
approved