# An application for zeroless squares

Matthijs Coster matthijs at coster.demon.nl
Fri Apr 29 07:56:38 CEST 2005

Hello Seqfans,

A month ago there was a discussion about zeroless squares.
If you are interested read the (recreational) application below.

Suppose you have to solve a crossword-puzzle of n x n.
You have to fill it by the digits 0,...,9 and all the descriptions are
equal, namely square of a number. How would you solve this
crossword-puzzle, let's say crosssquare-puzzle?

Important for solving this puzzle is to omit zeros at the first position.

Well the 1 x 1 crosssquare-puzzle is easy. Only 1, 4, 9 are the solutions.

If you are interested:
number of solutions of the 2x2 puzzle is 4,
3x3 : 13
4x4 : 14
5x5 : 76
6x6 : 40
7x7 : 459

So therefore I introduce the sequence of number of solutions:
3, 4, 13, 14, 76, 40, 459, ...

What will happen when n grows? I don't know. Important is the last position
where only 0,1,4,5,6,9 can be placed. Therefore you have to find squares
with only these digits. How do the number of squares behave?

David Wilson <davidwwilson at comcast.net> writes at 26 Feb 2005

>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+.

Now the same calculation has to be made for numbers only containing
0,1,4,5,6,9. We get limit ratio (3/5)*10^(1/2), which is OK.

More important is the remark of Ron Knott <ron at ronknott.com> at 27 Feb 2005.
He says that

>A slightly simpler infinite series of zeroless squares is
>((10^n-1)/3+1)^2 which, for n=1..8 gives
>16, 1156, 111556, 11115556, 1111155556, 111111555556,
>11111115555556, 1111111155555556

Fortunately there are squares which contain only 1,5 and 6.

The question for the math-recreationals under us is what happens with the
number of crosssquare-puzzles when n grows? I don't know.

I wrote a program to find the crosssquare-puzzles 10 years ago. You can
find it at my site (http://www.coster.demon.nl/sequences/index.html#new),
as well as the solutions for 2 up to 7.

--Matthijs
My HOMEPAGE --> http://www.coster.demon.nl/index.html
email: mailto://m@coster.demon.nl

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