

A176779


Smallest number appearing exactly n times in the concatenation of all integers from 1 to itself.


0



1, 12, 121, 1011, 1121, 10111, 11121, 109911, 111311, 111211, 1101111, 1112211, 1111211, 11011111, 11192111, 11111211, 11112111, 111011111, 111113111, 111122111, 111112111, 1110111111, 1111122111, 1111921111, 1111112111
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OFFSET

1,2


COMMENTS

For m>1, is the number of mdigit terms in the sequence always Int(m/2)?
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?
Is there an easy way to extend the sequence indefinitely?


LINKS



EXAMPLE

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.


CROSSREFS



KEYWORD

base,nonn


AUTHOR



STATUS

approved



