Champernowne-like squares, are there any?

Asked Dec 07 '11 at 15:21

Active Dec 14 '11 at 16:29

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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!

edited Dec 14 '11 at 16:29

J. M. ain't a mathematician

75,051

asked Dec 07 '11 at 15:21

Carolus

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This says that the proper name for your sequence is the "Smarandache consecutive sequence". See Smarandache's book for instance. – J. M. ain't a mathematician Dec 14 '11 at 16:16
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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
@J.M.: I'm pretty sure Smarandache was not the first to study this sequence! – Charles Dec 14 '11 at 19:32
@Charles: I'm sure you're right, but it was his name that got used... :) – J. M. ain't a mathematician Dec 14 '11 at 23:37
Does this answer your question? Is there any perfect square in the sequence $12,123,1234,12345,...$? – Thirdy Yabata Mar 21 '24 at 08:07

Champernowne-like squares, are there any?

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