A014529 -id:A014529 - cp's OEIS frontend

A289944 Sizes of triangles in the tilings of a convex polygon by n equilateral triangles described in A014529 written as triangle T(n,m), n >= 1, 1 <= m <= n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 1, 1, 1, 2, 2, 3, 4, 2, 2, 2, 3, 3, 3, 4, 4, 1, 3, 3, 3, 4, 4, 5, 5, 6, 1, 1, 4, 4, 4, 5, 6, 6, 7, 8, 2, 2, 3, 5, 5, 7, 7, 8, 8, 9, 11, 1, 3, 3, 4, 7, 7, 7, 10, 11, 12, 12, 13

Offset: 1

Hugo Pfoertner, Jul 16 2017

Sizes are counted as multiples of the side length of a smallest triangle with assumed side length 1.
If more than one tiling resulting in a polygon of maximum area exists for a given number of triangles, the lexicographically smallest list of triangles is given.
The sequence is based on an illustration given in the German translation of Ian Stewart's article in Math. Recreations, Scientific American, Jul 15 1997, p. 96.
A continuation of the sequence would need the details of John W. Layman's extension of A014529.
There is a tiling with 12 equilateral triangles and John W. Layman's calculated area of 860 that has triangles of sizes 1, 3, 3, 4, 7, 7, 7, 10, 11, 12, 12, 13. - Peter Munn, Aug 23 2017
There is a tiling with 13 equilateral triangles and John W. Layman's calculated area of 1559 that has 13 triangles of sizes 2, 2, 3, 5, 7, 7, 7, 12, 14, 15, 15, 16, 18. - Peter Munn, Aug 24 2017
From Peter Munn, Jan 01 2018: (Start)
There are tilings matching the other areas calculated by John W. Layman as follows:
A hexagon of area 2831 tiled with 14 equilateral triangles of sizes 1, 4, 5, 5, 6, 6, 11, 11, 16, 17, 20, 20, 23, 24 with the smallest 5 arranged as in the A014529 tiling with 9 triangles, and the size 24 triangle placed at the concave vertex of the region tiled by the smallest 5.
A hexagon of area 5114 tiled with 15 equilateral triangles of sizes 1, 1, 6, 6, 7, 8, 8, 15, 15, 21, 23, 27, 27, 31, 32 with the smallest 6 arranged as in the A014529 tiling with 10 triangles, and the size 32 triangle placed at the concave vertex of the region tiled by the smallest 6.
(End)

			Triangle starts:
  1;
  1, 1;
  1, 1, 1;
  1, 1, 1, 2;
  1, 1, 1, 2, 2;
  1, 1, 1, 2, 2, 3;
  1, 1, 1, 2, 2, 3, 4;
  2, 2, 2, 3, 3, 3, 4, 4;
  1, 3, 3, 3, 4, 4, 5, 5, 6;
  1, 1, 4, 4, 4, 5, 6, 6, 7, 8;
...
For corresponding illustrations see A014529.

Ian Stewart, Die unscheinbare Schwester der goldenen Zahl, Spektrum der Wissenschaft, Dossier 02/2003: Mathematische Unterhaltungen II, 55-57.

Cf. A014529.

Definition and data changed to resolve problem described in discussion dated Jul 19 2017 by Hugo Pfoertner, Aug 23 2017

Terms corresponding to A014529(12) from Peter Munn, Aug 23 2017

A089047 Edge length of largest square dissectable into up to n squares in Mrs. Perkins's quilt problem.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 13, 17, 23, 29, 41, 53, 70, 91, 126, 158, 216, 276, 386, 488, 675, 866, 1179, 1544, 2136, 2739, 3755, 4988, 6443

Offset: 1

R. K. Guy, Dec 03 2003

An inverse to A005670.
More precisely, a(n) = greatest k such that A005670(k) <= n. - Peter Munn, Mar 13 2018
It is not clear which terms have been proved to be correct and which are just conjectures. - Geoffrey H. Morley, Sep 07 2012; N. J. A. Sloane, Jul 06 2017
Terms up to and including a(18) have been proved correct by Ed Wynn (2013). - Stuart E Anderson, Sep 16 2013
A089046 and A089047 are almost certainly correct up to 5000. - Ed Pegg Jr, Jul 06 2017
Deleted terms above 5000. - N. J. A. Sloane, Jul 06 2017
Further best known terms are 8568, 11357, 14877, 19594, 26697, 34632. - Ed Pegg Jr, Jul 06 2017
A290821 is the equivalent sequence for equilateral triangles. - Peter Munn, Mar 06 2018

Ed Pegg, Jr., Mrs. Perkins's Quilts
Ed Pegg Jr. and Richard K. Guy, Mrs. Perkins's Quilts (Wolfram Demonstrations Project)
Ed Wynn, Exhaustive generation of Mrs Perkins's quilt square dissections for low orders, arXiv:1308.5420 [math.CO], 2013-2014.

Cf. A005670, A014529, A018835, A089046, A211302, A290821.

More terms from Ed Pegg Jr, Dec 03 2003
Corrected and extended by Ed Pegg Jr, Apr 18 2010
Duplicate a(6) deleted and a(22)-a(26) revised (from Ed Pegg Jr, Jun 15 2010) by Geoffrey H. Morley, Sep 07 2012
Conjectured terms have been extended up to a(44), based on simple squared square enumeration, by Duijvestijn, Skinner, Anderson, Pegg, Johnson, Milla and Williams. - Stuart E Anderson, Sep 16 2013
a(33) and further terms added by Ed Pegg Jr, Jul 06 2017
Name edited by Peter Munn, Mar 14 2018

A290821 Side length of largest equilateral triangle that can be made from n or fewer equilateral triangles with integer sides s_k, subject to gcd(s_1,s_2,...,s_n) = 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 39, 49

Offset: 1

Hugo Pfoertner, Aug 11 2017

No construction from 2, 3 or 5 equilateral triangles exists. The first difference from the Padovan numbers occurs for a(15)=39, where the corresponding term A000931(19)=37. a(16)=A000931(20)=49. a(n) >= A000931(n+3). From the growth behavior of A290697 it is conjectured that a(k) > A000931(k+3) for all k > 20.
a(19) is at least 130. This compares with A000931(23) = 114. It hints of growth behavior similar to sqrt(A014529) or sqrt(A001590). Ceiling(sqrt(A001590(n))) matches a(n) to n=14, then runs 38, 52, 70, 95, 128, ... . - Peter Munn, Mar 10 2018
From Peter Munn, Mar 14 2018 re monotonicity: (Start)
For n >= 6, a(n+1) > a(n).
Sketch of proof (inductive step) expressed in terms of tiling:
Given a triangle of side a(n) tiled with n equilateral triangular tiles. Let X, Y and Z be the tiles incident on its vertices, with X being not smaller than Y or Z.
Case 1: Y and Z have no vertices coincident. Remove Y and Z, thereby reducing the tiled area to a pentagon that has edges A and C that were previously internal to the area, and an edge B between A and C. Fit a new tile T against edge B, thereby extending edges A and C. Make the tiled area triangular by fitting a new tile against each of the extended edges.
Case 2: X, Y and Z have pairwise coincident vertices. It follows that these tiles are the same size. Remove Y and Z, thereby reducing the tiled area to a rhombus. Remove the tile at the rhombus vertex opposite X. The remaining area is a pentagon, since n >= 6. Extend the area by resiting Y against X, and Z against Y so that X and Z have external edges aligned. Make the area trapezoidal by fitting a new tile against the area's edge that includes an edge of Y. Fit another tile T against the smaller of the trapezoid's parallel edges.
In each case, we now have n+1 tiles, tiling an equilateral triangle with side length a(n) plus the side of T. As the sides of new and removed tiles can be calculated by adding sides of tiles that stayed in place, the GCD of the sides is unchanged.
(End)

			a(12) = 16:
                                  *
                                 / \
                                +   +
                               /     \
                              +       +
                             /         \
                            +           +
                           /             \
                          +               +
                         /                 \
                        +                   +
                       /                     \
                      +                       +
                     /                         \
                    +                           +
                   /                             \
                  +                               +
                 /                                 \
                *---+---*---+---+---+---+---+---+---*
               / \     / \                         / \
              +   +   +   +                       +   +
             /     \ /     \                     /     \
            *---*---*       +                   +       +
           / \ / \ /         \                 /         \
          +   *---*---+---+---*               +           +
         /     \             / \             /             \
        +       +           +   +           +               +
       /         \         /     \         /                 \
      +           +       +       +       +                   +
     /             \     /         \     /                     \
    +               +   +           +   +                       +
   /                 \ /             \ /                         \
  *---+---+---+---+---*---+---+---+---*---+---+---+---+---+---+---*

Stuart Anderson, An Introduction to Triangled Equilateral Triangles
Ales Drapal and Carlo Hamalainen, An enumeration of equilateral triangle dissections, arXiv:0910.5199 [math.CO], 2009-2010.
Hugo Pfoertner, Illustration of a(15)=39.

Cf. A000931, A001590, A167123, A290653, A290697, A290820.
A014529 gives greatest area of any convex polygon constructable from such triangles.
A089047 is this sequence's equivalent for squares.

Definition modified and 5 terms prepended by Peter Munn, Mar 14 2018

A096004 Number of convex triangular polyominoes containing n cells.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 3, 2, 2, 2, 3, 2, 3, 3, 5, 2, 3, 3, 4, 2, 4, 4, 6, 3, 3, 4, 5, 2, 5, 5, 7, 3, 4, 5, 6, 3, 5, 5, 8, 3, 4, 5, 6, 4, 7, 7, 9, 4, 5, 5, 7, 3, 7, 8, 9, 3, 5, 7, 8, 4, 8, 8, 11, 4, 5, 7, 8, 4, 9, 9, 11, 5, 5, 8, 9, 4, 9, 9, 13, 5, 7, 9, 8, 5, 8, 9, 12

Offset: 1

Paul Boddington, Jul 27 2004

The main sequence on triangular polyominoes is A000577. The convexity condition makes enumeration easy as a convex triangular polyomino has at most 6 sides. It is simple to prove that a(n) is also the number of 4-tuples (p,b,c,d) of nonnegative integers satisfying b<=c<=d, b+c+d<=p, n=p^2-b^2-c^2-d^2.

For n = A014529(k) there are a(n) many polygons. At least one of them can be tiled with k equilateral triangles. - Rainer Rosenthal, Sep 20 2017

			a(8)=3 because there are 3 ways to compose a convex polygon of 8 equilateral triangles with side 1:
.
    *---*---*---*---*
   / \ / \ / \ / \ /
  *---*---*---*---*
      *---*---*
     / \ / \ /
    *---*---*
   / \ / \ /
  *---*---*
      *---*
     / \ / \
    *---*---*
   / \ / \ / \
  *---*---*---*

Rainer Rosenthal, Table of n, a(n) for n = 1..5200
Peter Kagey, Examples for a(1)-a(30).
Walter Trump, Number of Convex Polygons of a Given Perimeter or Area on the Triangular Lattice, 2025.

Cf. A000577, A014529.

a:=proc(n) local x,p,d,c,b; x:=0; for p from 0 to ceil((n+1)/2) do; for d from 0 to p do; for c from 0 to min(d,p-d) do; for b from 0 to min(c,p-c-d) do; if p^2-b^2-c^2-d^2=n then x:=x+1 fi; od; od; od; od; x; end; # corrected by Rainer Rosenthal, Sep 20 2017

a(n) >= sqrt(n)/3. - Baohua Tian, Apr 21 2020

a(83) and a(84) corrected by Rainer Rosenthal, Sep 20 2017

A338861 a(n) is the largest area of a rectangle which can be dissected into n squares with integer sides s_i, i = 1 .. n, and gcd(s_1,...,s_n) = 1.

Original entry on oeis.org

1, 2, 6, 15, 42, 143, 399, 1190, 4209, 10920, 37245, 109886, 339745, 1037186, 3205734, 9784263, 29837784, 93313919, 289627536

Offset: 1

Rainer Rosenthal, Nov 12 2020

A219158 gives the minimum number of squares to tile an i x j rectangle. a(n) is found by checking all rectangles (i,j) for which A219158 has a dissection into n squares.

Due to the potential counterexamples to the minimal squaring conjecture (see MathOverflow link), terms after a(19) have to be considered only as lower bounds: a(20) >= 876696755, a(21) >= 2735106696. - Hugo Pfoertner, Nov 17 2020, Apr 02 2021

			a(6) = 11*13 = 143.
Dissection of the 11 X 13 rectangle into 6 squares:
.
          +-----------+-------------+
          |           |             |
          |           |             |
          |   6 X 6   |    7 X 7    |
          |           |             |
          |           |             |
          +---------+-+             |
          |         +-+-----+-------+
          |  5 X 5  |       |       |
          |         | 4 X 4 | 4 X 4 |
          |         |       |       |
          +---------+-------+-------+
.
a(19) = 16976*17061 = 289627536.
Dissection of the 16976 X 17061 rectangle into 19 squares:
.
       +----------------+-------------+
       |                |             |
       |                |             |
       |                |     7849    |
       |      9212      |             |
       |                |             |
       |                |             |
       |                |------+------|
       |________________|      |      |
       |             |   see   | 4109 |
       |             |Rosenthal|      |
       |             |  link +-+------+
       |     7764    |-------|        |
       |             |       |  5018  |
       |             | 4279  |        |
       |             |       |        |
       +-------------+-------+--------+
.

Stuart Anderson, Catalogues of Simple Perfect Squared Rectangles.
Stuart Anderson, Simple Imperfect Squared Rectangles, orders 9 to 24.
Bertram Felgenhauer, Filling rectangles with integer-sided squares.
MathOverflow, tiling a rectangle with the smallest number of squares, answer by Ed Pegg Jr, Jul 09 2017.
Rainer Rosenthal, Rectangle tiled by 19 squares with maximum area a(19)

Cf. A219158, A340726, A340920.

This sequence and A089047 are effectively analogs for dissecting (or tiling) rectangles and squares respectively. Analogs using equilateral triangular tiles are A014529 and A290821 respectively.

a(11)-a(17) from Hugo Pfoertner based on data from squaring.net website, Nov 17 2020
a(18) from Hugo Pfoertner, Feb 18 2021
a(19) from Hugo Pfoertner, Apr 02 2021

A133044 Area of the spiral of equilateral triangles with side lengths which follow the Padovan sequence, divided by the area of the initial triangle.

Original entry on oeis.org

1, 2, 3, 7, 11, 20, 36, 61, 110, 191, 335, 591, 1032, 1816, 3185, 5586, 9811, 17207, 30203, 53004, 93004, 163229, 286430, 502655, 882111, 1547967, 2716528, 4767152, 8365761, 14680930, 25763171, 45211271, 79340235, 139232356, 244335860, 428779421, 752455502, 1320467391

Offset: 1

Omar E. Pol, Nov 04 2007

nonn

First differs from A014529 at a(8).

G. C. Greubel, Table of n, a(n) for n = 1..2000
Index entries for linear recurrences with constant coefficients, signature (1,1,1,-1,1,-1).

Cf. A000931, A014529, A133043.

RecurrenceTable[{a[n + 6] == a[n + 5] + a[n + 4] + a[n + 3] - a[n + 2] + a[n + 1] - a[n], a[1] == 1, a[2] == 2, a[3] == 3, a[4] == 7, a[5] == 11, a[6] == 20}, a, {n, 1, 2000}] (* G. C. Greubel, Dec 17 2015 *)
Rest@ CoefficientList[Series[x (x^3 + x + 1)/((x^3 - x^2 + 2 x - 1) (x^3 - x - 1)), {x, 0, 38}], x] (* Michael De Vlieger, Feb 21 2018 *)

Vec((x^3+x+1)/((x^3-x^2+2*x-1)*(x^3-x-1)) + O(x^40)) \\ Andrew Howroyd, Feb 21 2018

From Colin Barker, Sep 18 2013: (Start)
Conjecture: a(n) = a(n-1) + a(n-2) + a(n-3) - a(n-4) + a(n-5) - a(n-6).
G.f.: x*(x^3+x+1) / ((x^3-x^2+2*x-1)*(x^3-x-1)).
(End)
From Félix Breton, Dec 17 2015: (Start)
a(n) = 2*p(n+4)*p(n+5) - p(n+2)^2 where p is the Padovan sequence (A000931). This establishes Colin Barker's conjecture, because
a(n) = a(n-1) + p(n+4)^2
= a(n-1) + (p(n+1) + p(n+2))^2
= a(n-1) + p(n+1)^2 + p(n+2)^2 + 2*p(n+1)*p(n+2) - p(n-1)^2 + p(n-1)^2
= a(n-1) + (a(n-3)-a(n-4)) + (a(n-2)-a(n-3)) + a(n-3) + (a(n-5)-a(n-6))
= a(n-1) + a(n-2) + a(n-3) - a(n-4) + a(n-5) - a(n-6). (End)

a(27) and beyond taken from G. C. Greubel's table. - Omar E. Pol, Dec 18 2015

a(589) in b-file corrected by Andrew Howroyd, Feb 21 2018

cp's OEIS Frontend

A289944 Sizes of triangles in the tilings of a convex polygon by n equilateral triangles described in A014529 written as triangle T(n,m), n >= 1, 1 <= m <= n.

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A089047 Edge length of largest square dissectable into up to n squares in Mrs. Perkins's quilt problem.

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A290821 Side length of largest equilateral triangle that can be made from n or fewer equilateral triangles with integer sides s_k, subject to gcd(s_1,s_2,...,s_n) = 1.

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A096004 Number of convex triangular polyominoes containing n cells.

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A338861 a(n) is the largest area of a rectangle which can be dissected into n squares with integer sides s_i, i = 1 .. n, and gcd(s_1,...,s_n) = 1.

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A133044 Area of the spiral of equilateral triangles with side lengths which follow the Padovan sequence, divided by the area of the initial triangle.

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