%I
%S 1,12,121,1011,1121,10111,11121,109911,111311,111211,1101111,1112211,
%T 1111211,11011111,11192111,11111211,11112111,111011111,111113111,
%U 111122111,111112111,1110111111,1111122111,1111921111,1111112111
%N Smallest number appearing exactly n times in the concatenation of all integers from 1 to itself.
%C For m>1, is the number of mdigit terms in the sequence always Int(m/2)?
%C For 4<=m<=10, the last mdigit term consists of m1 1's and a single 2 located at the first digit position to the right of the middle, i.e., 1121, 11121, 111211, 1111211, 11112111, 111112111, 1111121111. Does this pattern hold for all m>3?
%C Is there an easy way to extend the sequence indefinitely?
%e Let s(k) be the string of digits obtained by concatenating all integers from 1 to k. Then a(3)=121 because the substring 121 appears exactly 3 times in s(121)=123..1213..112113..119120121, and there is no smaller number having this property.
%K base,nonn
%O 1,2
%A _Jon E. Schoenfield_, Apr 25 2010
