cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A051602 a(n) is the maximal number of squares that can be formed from n points in the plane.

Original entry on oeis.org

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Sascha Kurz has proved that one can assume that the points belong to the square grid Z X Z.
So we obtain the same values if we replace the definition by: a(n) is the maximal number of squares that can be formed from n points chosen from the infinite square grid.
In other words, take the infinite square grid. Pick a set S of n grid points, and let c(S) be the number of subsets of four points of S that form a square of any (nonzero) size. Then a(n) = maximum of c(S) over all choices of S.
The general problem of estimating the maximal number of similar figures within point-configurations was studied in [Elekes--Erdos]. References can be found in [Matousek, p. 47] and [AFKK]. Note that the present problem concerning squares shares several properties with the case of right isosceles triangles treated in [AFR].
The following remarks are out of date, and will be revised soon to reflect progress made in October 2021 by a number of members of the Sequence Fans Mailing List.
A more detailed account will be found in the report by Sascha Kurz et al. which is nearing completion.
[Comments revised to this point by N. J. A. Sloane, Nov 02 2021]
Values for n <= 9 [now 16] are exact and are the same for a and b (see proofs below by Hugo van der Sanden and Benoît Jubin for n <= 7 and Sascha Kurz for n <= 9, which furthermore classify all optimal configurations for these values). Values for n > 9 are conjectural since they were obtained by exhaustive search for grid points within a square of side ceil(sqrt(n)), which is a reasonable assumption. A proof that optimal configurations (with gcd of side lengths equal to 1) have a diameter at most f(n) with f of moderate growth would permit exact computation of values from exhaustive searches.
Asymptotic behavior:
One has a(n), b(n) = Theta(n^2):
Upper bound: Since two vertices determine squares, one has b(n) = O(n^2). More explicitly: a pair of points uniquely determines the square that has it as diagonal, and a square has two diagonals, so b(n) <= n(n-1)/4 ~ n^2/4.
Lower bound: When n = m^2, the set of all grid points in [0, m-1]^2 yields S = n(n-1)/12 squares. Indeed, for a in [0, m-1] and b in [1, m], the square formed on (a,0) and (0,b) (as its "lower-left side") has other vertices (a+b,b) and (a,a+b), so there are (m-(a+b))^2 translates of that square. Therefore, S = Sum_{a=0..m-1} Sum_{b=1..m} (m - (a + b))^2. By change of summation indices ((a,c) := (a,a+b)), expanding, using the sum of the first m integers, squares, cubes, and refactoring, one obtains S = n(n-1)/12. Since a is nondecreasing, one has a(n) >= (n-1)(n-2)/12 ~ n^2/12.
We actually have better lower and upper bounds:
0.09... = (1-2/Pi)/4 <= liminf a(n)/n^2 <= limsup b(n)/n^2 <= 1/6 = 0.16...
The upper bound is given in [AFR] (which counts isosceles right triangles, so their value has to be divided by four, the number of isosceles right triangles formed on a square). This gives b(n) <= (2/3*(n - 1)^2 - 5/3)/4.
Lower bound (due to Peter Munn, see SeqFan post in the links): For r >= 0, denote by D(r) the disc centered at the origin with radius r. If A is a point on the boundary of D(r), then the set of points B such that the square with diagonal AB is included in D(r) is a lens-shaped region of area (Pi-2)r^2 (as a proportion of the disc area, this is twice A258146). Therefore, the number S of grid-squares included in D(R) can be estimated as follows: since the set of squares with at least two vertices equidistant from the origin is negligible, we can assume that every square has a unique vertex furthest from the origin, say at distance r, which corresponds to the A above. Then the opposite vertex B is in the region computed above, and the L^1 (aka rectilinear, or Manhattan) distance between A and B is even (being opposite vertices, they are two sides apart), so we have to divide that area by 2. There are approximately 2*Pi*r grid points at a distance approximately r from the origin, so at first order, S ~= Integral_{r=0..R} Pi*r(Pi - 2)r^2 dr = Pi(Pi-2)/4 R^4. Since the disc D(R) contains approximately Pi*R^2 points, one obtains S ~= (1-2/Pi)/4 n^2.
Conjecture: the asymptotic density of the numbers n such that there is no maximal arrangement formed by all the grid points within a suitably chosen circle, is 0. - Peter Munn, Sep 30 2021
For the known values of a(n) (n <= 17), there is a maximal arrangement formed using circles as specified by the conjecture above. For n <= 100 no arrangement has yet been found to contain more squares than the best attained using a circle as specified by A348469. - Peter Munn, Nov 12 2021

Examples

			Lower bounds:
Computer searches, using glutton algorithms starting with all grid points in a convex polygon and adding successive points, have given the following current record configurations, which thus yield lower bounds for a(n).  They are due mainly for n <= 36 to Sean A. Irvine and for 37 <= n <= 50 to Sascha Kurz.  The representations below are given for ranges of n, and the integers indicate at what stage a given node is added (the letters A--Z encode the integers 10--36).
For instance, the first representation encodes the sequence of configurations
                  X
XX ; XXX ; XXX ; XXX ; etc.
XX ; XX  ; XXX ; XXX
----------
n = 4--11
435
0016
0027
----------
n = 12--19
..7
4003
00005
00006
1002
----------
n = 20--36
.GA78B
.93100C
E500000
F400000
.600000
.D2000
----------
n = 37--47
.60007
5000008
0000000
00000009
0000000A
4000003
.20001
----------
n = 48--50
..0000
.000000
20000000
00000000
00000000
.0000000
.000000
...001
----------
In particular, a(25)>=51, a(36)>=109, a(37)>=117, a(48)>=198, a(49)>=207, a(50)>=216.
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Another optimal configuration for a(8) = 4 due to Sascha Kurz:
.XX
XXXX
.XX
----------
Configurations and values for larger values of n can be found in the links below.
		

References

  • Elekes, Erdos, Similar configurations and pseudo grids, Coll. Math. Soc. Janos Bolyai 63 Intuitive Geometry, Budapest (Hungary), 1994.
  • J. Matousek, Lectures on Discrete Geometry, GTM 212, Springer, 2002.

Crossrefs

A348768 gives the number of inequivalent solutions.

Formula

If n = m^2, then a(n) >= m^2*(m^2-1)/12 (see A002415). If n = m^2-1, then a(n) >= (m-1)*(m-2)*(m^2+3*m+6)/12. - N. J. A. Sloane, Sep 28 2021

Extensions

a(13)-a(16) from Sean A. Irvine, Sep 23 2021
a(13)-a(16) confirmed and a(17) from Sascha Kurz, Oct 30 2021
Revised following a rich discussion on the seqfan mailing list, with contributions by the persons cited in the text, Allan Wechsler, Alex Meiburg, and Benoit Jubin. - Benoit Jubin, Oct 07 2021
Partially revised by N. J. A. Sloane, Nov 02 2021

A186705 The Erdős unit distance problem: the maximum number of occurrences of the same distance among n points in the plane.

Original entry on oeis.org

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Michael Somos, Feb 25 2011

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An upper bound is floor(k*n^(4/3)), A129011, if k is close enough to 1. Also a(27)=81 (Hamming 3,3 graph). - Ed Pegg Jr, Feb 02 2018

Examples

			a(4) = 5 because there is a unit distance graph with 4 vertices of an equilateral rhombus such that all but one of the six pairs of vertices are unit distance apart.
Comment from _Allan C. Wechsler_, Sep 17 2018: (Start)
Construction for a(9)=18: Take a convex, equilateral hexagon ABCDEF. Make the angles vary a bit, though, to avoid the hexagon being regular. Now, on each of the six sides, construct an equilateral triangle pointing into the hexagon. In general, the triangles will overlap here and there; this is OK because we aren't going to care about edges crossing each other. So we have triangles ABU, BCV, CDW, DEX, EFY, and FAZ: a total of twelve points with 18 unit distances among them.
Now adjust the hexagon to make some pairs of the internal points coincide. We want to make U=X, V=Y, and W=Z. The resulting linkage still has one degree of freedom, so we can arrange it so that none of the edges coincide (they can and must cross, though). The adjusted hexagon will only have two different angles: ABC = CDE = EFA, and BCD = DEF = FAB. The whole thing will have triangular (D_6) symmetry. It will have nine vertices (after merging three pairs from the original 12) but it will still have 18 unit edges. (End)
		

References

  • P. Brass, W. O. J. Moser, J. Pach, Research Problems in Discrete Geometry, Springer (2005), p. 183

Crossrefs

Cf. A385657 (number of nonisomorphic maximally dense unit-distance graphs).

Extensions

Extended to a(21) using values from Version 2 of the Alexeev et al. arXiv manuscript. - N. J. A. Sloane, Jun 24 2025
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