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I read about the Champernowne constant on Wikipedia a couple of days ago, and I got curious about something similar: is there some "Champernowne-like" number; that is, a concatenation of all numbers up to some $n \ge 2$ (like $1234567891011$), that is a perfect square? I've done a computer search up to $n=2000$, but I haven't found any.

Are there any? If not, how may we prove this? Any thoughts on this are welcome!

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Well, $n=1$ clearly fits the bill. You should probably explicitly mention $n \geq 2$ in your post. – Sasha Dec 7 '11 at 15:26
@Sasha: fixed! Thanks. – Carolus Dec 7 '11 at 15:29
This says that the proper name for your sequence is the "Smarandache consecutive sequence". See Smarandache's book for instance. – J. M. Dec 14 '11 at 16:16
None to $10^4.$ It should be possible to test this more efficiently than testing individual members (find the first digits of a square root, then square and see if the result is close to an integer) but I haven't attempted anything along those lines. – Charles Dec 14 '11 at 16:27
@J.M.: Thanks, I will have a look. – Carolus Dec 14 '11 at 17:12

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