A055186 Cumulative counting sequence: method A (adjective-before-noun)-pairs with first term 0.

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

Author

Clark Kimberling, Apr 27 2000

Keywords

• nonn
• base

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

Examples

			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; ...
		

Links

• Zak Seidov, Table of n, a(n) for n = 1..1019 (the first 22 steps)

Crossrefs

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

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

Cf. A217760.

Programs

Mathematica

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

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

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