A005842 -id:A005842 - cp's OEIS frontend

A092137 Lower bound for A005842(n).

1, 3, 4, 6, 8, 10, 12, 15, 17, 20, 23, 26, 29, 32, 36, 39, 43, 46, 50, 54, 58, 62, 66, 70, 75, 79, 84, 88, 93, 98, 103, 107, 112, 117, 123, 128, 133, 138, 144, 149, 155, 160, 166, 172, 178, 184, 189, 195, 202, 208

Offset: 1

Rob Pratt, Mar 30 2004

Area of square must be large enough to contain all n squares without overlap.

Cf. A005842.

Table[Ceiling[Sqrt[Sum[k^2, {k, 1, n}]]], {n, 1, 50}]

a(n) = ceiling(sqrt(Sum_{k=1..n} k^2)).

A005670 Mrs. Perkins's quilt: smallest coprime dissection of n X n square.

Original entry on oeis.org

1, 4, 6, 7, 8, 9, 9, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17

Offset: 1

N. J. A. Sloane

The problem is to dissect an n X n square into smaller integer squares, the GCD of whose sides is 1, using the smallest number of squares. The GCD condition excludes dissecting a 6 X 6 into four 3 X 3 squares.
The name "Mrs Perkins's Quilt" comes from a problem in one of Dudeney's books, wherein he gives the answer for n = 13. I gave the answers for low n and an upper bound of order n^(1/3) for general n, which Trustrum improved to order log(n). There's an obvious logarithmic lower bound. - J. H. Conway, Oct 11 2003
All entries shown are known to be correct - see Wynn, 2013. - N. J. A. Sloane, Nov 29 2013

			Illustrating a(7) = 9: a dissection of a 7 X 7 square into 9 pieces, courtesy of _Ed Pegg Jr_:
.___.___.___.___.___.___.___
|...........|.......|.......|
|...........|.......|.......|
|...........|.......|.......|
|...........|___.___|___.___|
|...........|...|...|.......|
|___.___.___|___|___|.......|
|...............|...|.......|
|...............|___|___.___|
|...............|...........|
|...............|...........|
|...............|...........|
|...............|...........|
|...............|...........|
|___.___.___.___|___.___.___|
The Duijvestijn code for this is {{3,2,2},{1,1,2},{4,1},{3}}
Solutions for n = 1..10: 1 {{1}}
2 {{1, 1}, {1, 1}}
3 {{2, 1}, {1}, {1, 1, 1}}
4 {{2, 2}, {2, 1, 1}, {1, 1}}
5 {{3, 2}, {1, 1}, {2, 1, 2}, {1}}
6 {{3, 3}, {3, 2, 1}, {1}, {1, 1, 1}}
7 {{4, 3}, {1, 2}, {3, 1, 1}, {2, 2}}
8 {{4, 4}, {4, 2, 2}, {2, 1, 1}, {1, 1}}
9 {{5, 4}, {1, 1, 2}, {4, 2, 1}, {3}, {2}}
10 {{5, 5}, {5, 3, 2}, {1, 1}, {2, 1, 2}, {1}}

H. T. Croft, K. J. Falconer and R. K. Guy, Unsolved Problems in Geometry, C3.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Ed Wynn, Table of n, a(n) for n = 1..120
J. H. Conway, Mrs. Perkins's quilt, Proc. Camb. Phil. Soc., 60 (1964), 363-368.
A. J. W. Duijvestijn, Table I
A. J. W. Duijvestijn, Table II
R. K. Guy, Letter to N. J. A. Sloane, 1987
Ed Pegg, Jr., Mrs Perkins's Quilts (best known values to 40000)
G. B. Trustrum, Mrs Perkins's quilt, Proc. Cambridge Philos. Soc., 61 1965 7-11.
Eric Weisstein's World of Mathematics, Mrs. Perkins's Quilt
Ed Wynn, Exhaustive generation of 'Mrs Perkins's quilt' square dissections for low orders, arXiv:1308.5420 [math.CO], 2013-2014.
Ed Wynn, Exhaustive generation of 'Mrs. Perkins's quilt' square dissections for low orders, Discrete Math. 334 (2014), 38--47. MR3240464

Cf. A005842, A089046, A089047.

b-file from Wynn 2013, added by N. J. A. Sloane, Nov 29 2013

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.

Original entry on oeis.org

0, 1, 1, 3, 2, 4, 3, 3, 4

Offset: 1

Tamas Sandor Nagy and Thomas Scheuerle, Sep 25 2023

Warning: several terms are provisional as their intended verification effectively assumed the augmenting squares were not larger than n X n. - Peter Munn, Oct 02 2023
The definition does not exclude squares larger than n X n.
Terms for n < 10 were verified by the use of a program.
a(10) <= 5.

			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

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.

Cf. A000290, A005670, A005842, A038666, A081287.

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.

Edited by Peter Munn, Oct 04 2023

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A092137 Lower bound for A005842(n).

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A005670 Mrs. Perkins's quilt: smallest coprime dissection of n X n square.

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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.

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