

A055186


Cumulative counting sequence: method A (adjectivebeforenoun)pairs with first term 0.


8



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
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OFFSET

1,4


COMMENTS

Start with 0; at nth 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. A055168A055185 (method B) and A055187A055191 (method A).
Cf. A217760.
Sequence in context: A029353 A064922 A303337 * A217760 A263412 A321258
Adjacent sequences: A055183 A055184 A055185 * A055187 A055188 A055189


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 bfile.  N. J. A. Sloane, Oct 04 2010


STATUS

approved



