zeroless squares

[David Wilson]

A naive estimate:

There are approximately s(d) = (10^d)^(1/2) - (10^(d-1))^(1/2) d-digit 
squares.
A random d-digit number has the probability p(d) = (9/10)^(d-1) of being
zeroless (exponent d-1 as opposed to d because the first digit is not zero).
So we expect p(d)s(d) zeroless d-digit squares.  For d = 1 through 12, we
get (truncating):

    1, 5, 15, 44, 127, 363, 1034, 2943, 8377, 23841, 67854, 193117

This is not bad agreement with your actual quoted counts.  The estimates
tend to be a little small, possibly due to inaccuracy in the s(d) estimate
and/or nonrandomness in the digits of square numbers.  The elements grow 
approximately geometrically with limit ratio (9/10)*10^(1/2) = 2.846+.

The same naive estimate can easily be generalize to kth powers, giving the
estimate s(d) = (10^d)^(1/k) - (10^(d-1))^(1/k) for d-digit kth powers.
p(d) remains the same.  The resulting estimates have ratio (9/10)*10^(1/k).
We should expect an infinite number of zeroless kth powers when this ratio
is >= 1, which it is for k <= 21.  For k >= 22, the ratio is < 1 and we
should expect a finite number of zeroless kth powers.

This argument can be extended to other bases, but I don't want to do it.

----- Original Message ----- 
From: "Ron" <ron at ronknott.com>
To: <ham>; <seqfan at ext.jussieu.fr>
Sent: Saturday, February 26, 2005 7:37 PM
Subject: zeroless squares


> Thank you Wouter for passing on the correction and I'm hoping this gets
> to the list this time!
>
> For the record, a brute-force search with the Maple code of that message
> gives
> "A(n) = the number of n-digit numbers whose square has no zero digits"
> as:
>        3, 6, 19, 44, 136, 376, 1061, 2985, 8431, 24009, 67983, 193359
> index from n=1 to 12
>
> Ron Knott 



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