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%I #16 Sep 06 2018 17:26:09
%S 1,3,5,7,9,11,14,17,20,23,26,29,32,35,38,41,44,47,50,53,56,59,62,65,
%T 68,71,74,77,80,83,86,89,92,95,98,101,105,109,113,117,121,125,129,133,
%U 137,141,145,149,153,157,161,165,169,173,177,181,185,189,193,197,201,205,209,213
%N Define b(0)=0, b(1)[1]=1, b(1)[2]=1; and for n>=2, b(n)[1] = total number of digits in b(n-1), and b(n)[2] = total number of digits in b(0),...,b(n-1); a(n) = b(n)[2].
%D Eric Angelini, Posting to Math Fun Mailing List, Aug 20 2007.
%e The initial values of b(0), b(1), ... are:
%e b(0) = 0
%e b(1) = [1, 1]
%e b(2) = [2, 3]
%e b(3) = [2, 5]
%e b(4) = [2, 7]
%e b(5) = [2, 9]
%e b(6) = [2, 11]
%e b(7) = [3, 14]
%e b(8) = [3, 17]
%e b(9) = [3, 20]
%e b(10) = [3, 23]
%e b(11) = [3, 26]
%e b(12) = [3, 29]
%e ...
%e The second terms give the present sequence.
%e The sequence of values of the first terms b(i)[1] for i >= 1 consists of 1 (once), 2 (5 times), 3 (30 times), 4 (225 times), 5 (1800 times), 6 (15000 times), ... (see A318920).
%p A055642 := proc(n) max(1, ilog10(n)+1) ; end proc:
%p read transforms;
%p M:=1000;
%p b[1]:=[1,1];
%p for n from 2 to M do
%p b[n][1]:=A055642(b[n-1][1]) + A055642(b[n-1][2]);
%p b[n][2]:=b[n-1][2]+b[n][1];
%p od:
%p s2:=[seq(b[n][2],n=1..M)]; # A318919
%p s1:=[seq(b[n][1],n=1..M)]: RUNS(s1); # A318920
%Y For b(n)[1] see A318920.
%K nonn,base
%O 1,2
%A _N. J. A. Sloane_, Sep 06 2018
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