|
Search: id:A055186
|
|
|
| A055186 |
|
Cumulative counting sequence: method A (adjective,noun)-pairs with first term 0. |
|
+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, 1, 9, 8, 0, 13, 1, 9, 2, 7, 3, 7, 4, 7, 5, 7, 6, 2, 9, 1, 7, 1, 11, 1, 8, 9, 0, 17, 1
(list; graph; listen)
|
|
|
OFFSET
|
1,4
|
|
|
COMMENT
|
Start with 0; at n-th step, write down what is in the sequence so far.
"Look and Say" how many time each integer (in increasing order), <= max {existing terms} appear in the sequence. Then concatenate. Start with 0.
Sequence's graph looks roughly like that of A080096
|
|
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 (benoit7848c(AT)orange.fr), 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).
Sequence in context: A135830 A029353 A064922 this_sequence A124035 A157897 A140129
Adjacent sequences: A055183 A055184 A055185 this_sequence A055187 A055188 A055189
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Clark Kimberling (ck6(AT)evansville.edu), Apr 27 2000
|
|
EXTENSIONS
|
Edited by N. J. A. Sloane (njas(AT)research.att.com), Jan 17 2009 at the suggestion of R. J. Mathar
Removed a conjecture. - Clark Kimberling (ck6(AT)evansville.edu), Oct 24 2009
|
|
|
Search completed in 0.002 seconds
|
