%I
%S 3,3,1,3,2,1,1,3,3,1,3,2,1,3,6,1,5,2,2,3,7,1,6,2,4,6,1,5,1,3,8,1,9,2,
%T 5,6,3,5,2,7,1,4,1,3,10,1,12,2,7,6,4,5,4,7,2,4,2,8,1,9,1,3,11,1,15,2,
%U 10,6,5,5,5,7,4,4,5,8,2,9,2,10,1,12,1,3,12,1
%N Cumulative counting sequence: method B (noun,adjective)pairs with first term 3.
%C Segments (generated as at A055168): 3; 3,1; 3,2,1,1; 3,3,1,3,2,1; ...
%C Conjecture: every positive integer occurs.
%e Write 3, thus having 3 once, thus having 3 twice and 1 once, thus having 3 3 times and 1 3 times and 2 once, etc.
%t s = {3}; Do[s = Flatten[{s, {#,Count[s, #]} & /@ DeleteDuplicates[s]}], {24}]; s (* _Peter J. C. Moses_, Mar 21 2013 *)
%Y Cf. A055168.
%K nonn
%O 1,1
%A _Clark Kimberling_, Apr 27 2000
