Hollow Square Laminae I

We shall define a square lamina to be a square outline with a square “hole” so that the shape possesses vertical and horizontal symmetry. For example, using exactly thirty-two square tiles we can form two different square laminae:

With one-hundred tiles, and not necessarily using all of the tiles at one time, it is possible to form forty-one different square laminae.
Using up to one million tiles how many different square laminae can be formed?

The problem of finding how many different square laminae can be formed with up to one million tiles involves understanding the mathematical structure of a square lamina.

The square lamina is built from concentric squares. The outermost square has side length n (for some positive integer n), the next square in has side length (n-2), the next has side length (n-4), and so on until reaching either a square of side length 1 (for odd n) or a single tile (for even n).

Such a lamina with outer side length n is built from a square of side length n, minus a square of side length (n-2), plus a square of side length (n-4), and so on. The number of tiles it’s built from is thus n^2 + (n-2)^2 + (n-4)^2 + …

Let S be the sum of squares for n, n-2, n-4, …, down to either 1 or 2. Clearly, if n is odd, then we have S = n² + (n-2)² + ….+ 1², and if n is even then S = n² + (n-2)² + ….+ 2².

The formula for the sum of squares of the first m natural numbers is m(m + 1)(2m + 1) / 6

For even n we have S = 0² + 2² + …. + n² = 4[1² + 2² + …. + (n/2)²] = 4 * (n/2) * (n/2 + 1) * (n + 2) / 6;
For odd n we have S = 1² + 3² + …. + n² = 1 + 4[1² + 2² + …. + ((n-1)/2)²] = 1 + 4 * ((n-1)/2) * ((n-1)/2 + 1) * (n + 1) / 6.

The number of laminae for a given number of tiles is the number of S that are less than or equal the number of tiles.

To get the answer, run a count for n from 1, incrementing by 1 each time, until S exceeds one million.

This is essentially an application of summation of squares in mathematics, and can be solved programmatically with integer math given that the limit of one million is quite reasonable.

Please note this involves rather advanced mathematics like combinatorics and sum of squares. The ‘count’ approach is computational but simplifies this given the constraints. In general, there is no straightforward formula to solve the problem directly.

More Answers:
Pandigital Concatenating Products
Square Sum of the Digital Squares
Few Repeated Digits

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