A365236 a(n) is the least number of integer-sided squares that can be packed together with the n squares 1 X 1, 2 X 2, ..., n X n to fill out a rectangle.
0, 1, 1, 3, 2, 4, 3, 3, 4
Offset: 1
Examples
Compositions of rectangles that satisfy the minimal number of augmenting squares for n. Where more than one minimal composition exists for a given n, the table shows a single example. In the table body, the numbers include both the specific mandatory and augmenting squares. a(n) is the total number of squares in the rectangle minus n. | 1^2 2^2 3^2 4^2 5^2 6^2 7^2 8^2 9^2 10^2 | Total ---------------------------------------------------------------------------- a(1) = 0 | 1 | 1 a(2) = 1 | 2 1 | 3 a(3) = 1 | 2 1 1 | 4 a(4) = 3 | 2 1 2 2 | 7 a(5) = 2 | 2 1 1 2 1 | 7 a(6) = 4 | 2 1 3 2 1 1 | 10 a(7) = 3 | 1 1 1 3 1 2 1 | 10 a(8) = 3 | 3 2 1 1 1 1 1 1 | 11 a(9) = 4 | 2 2 2 2 1 1 1 1 1 | 13
Links
- Tamas Sandor Nagy, Examples for a(1) - a(4).
- Tamas Sandor Nagy, Example for a(5).
- Tamas Sandor Nagy, Example for a(6).
- Tamas Sandor Nagy, Original upper bound examples for a(7) with 5 augmenting squares and a(8) with 6 augmenting squares.
- Tamas Sandor Nagy, Example of a conjectured solution for a(10) with 5 augmenting squares, found by Peter Munn.
- Thomas Scheuerle, Example for a(6) with smallest possible area.
- Thomas Scheuerle, Example for a(7).
- Thomas Scheuerle, Example for a(8).
- Thomas Scheuerle, Example for a(9).
- Thomas Scheuerle, Original upper bound example for a(10) with 6 augmenting squares.
- Thomas Scheuerle, Example for a(11) = 4 this is at the same time also a conjectured solution for a(10) = 5.
Formula
a(n) <= 1 + Sum_{k = 1 .. ceiling((n - 1)/2)} (n + (1 - k)*floor(n/k) - 2). This upper bound corresponds to placing the squares with length n up to n - floor((n - 1)/2) all in one row. The remaining mandatory squares will then fit naturally into the rectangle n X (1/2)*(2*n - ceiling((n - 1)/2))*(ceiling((n - 1)/2) + 1).
a(n) > a(n - 1) - 2.
Extensions
Edited by Peter Munn, Oct 04 2023
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