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.

Showing 1-6 of 6 results.

A273464 The number of tilings of an equilateral triangle of side length n with k lozenges and n^2 - 2*k unit triangles. Triangle T(n, k) with n >= 1 and 0 <= k <= n*(n + 1)/2, read by rows.

Original entry on oeis.org

1, 1, 3, 1, 9, 24, 18, 1, 18, 126, 434, 762, 630, 187, 1, 30, 387, 2814, 12699, 36894, 69242, 81936, 57672, 21432, 3135, 1, 45, 915, 11127, 90270, 515970, 2139120, 6523428, 14683401, 24256853, 28975770, 24383838, 13860321, 4966929, 989970, 81462, 1, 63
Offset: 1

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Author

R. J. Mathar, May 23 2016

Keywords

Examples

			Triangle T(n,k) (with rows n >= 1 and columns k >= 0) begins as follows:
  1;
  1,  3;
  1,  9,  24,   18;
  1, 18, 126,  434,   762,   630,   187;
  1, 30, 387, 2814, 12699, 36894, 69242, 81936, 57672, 21432, 3135;
  ...
		

Crossrefs

Cf. A045943 (column k=1), A011555, A011556, A011781, A122722, A326367 (k=2), A326368 (k=3), A326369 (k=4), A000124 (row lengths).

Formula

T(n,2) = 3*(n-1)*(n-2)*(3*n^2+3*n-4)/8 . - R. J. Mathar, May 24 2016
T(n,3) = (n-2)*(9*n^5-9*n^4-81*n^3+81*n^2+160*n-192)/16. - Greg Dresden, Jul 03 2019
Conjecture: T(n,4) = 3*(n-2)*(n-3)*(9*n^6+9*n^5-135*n^4-81*n^3+670*n^2+104*n-1216)/128. - Greg Dresden, Jul 03 2019
Conjecture: T(n,5) = 3*(n-3)*(n+3)* (27*n^8 -135*n^7 -387*n^6 +2835*n^5 -168*n^4 -18732*n^3 +19568*n^2 +36992*n -56320)/1280. - R. J. Mathar, Jul 07 2019
From Petros Hadjicostas, Sep 13 2019: (Start)
Conjecture for rightmost terms: A122722(n) = n! * T(n, n*(n+1)/2) for n >= 1.
Conjectures for column k >= 0: Sum_{0 <= s <= 2*k + 1} (-1)^s * binomial(2*k+1, s) * T(n-s, k) = 0 for n >= 2*k+2.
Sum_{0 <= s <= 2*k} (-1)^s * binomial(2*k, s) * T(n-s, k) = A011781(k) for n >= 2*k+1. (End)

A326367 Number of tilings of an equilateral triangle of side length n with unit triangles (of side length 1) and exactly two unit "lozenges" or "diamonds" (also of side length 1).

Original entry on oeis.org

0, 0, 24, 126, 387, 915, 1845, 3339, 5586, 8802, 13230, 19140, 26829, 36621, 48867, 63945, 82260, 104244, 130356, 161082, 196935, 238455, 286209, 340791, 402822, 472950, 551850, 640224, 738801, 848337, 969615, 1103445, 1250664, 1412136, 1588752, 1781430, 1991115
Offset: 1

Views

Author

Greg Dresden, Jul 01 2019

Keywords

Examples

			We can represent a unit triangle this way:
       o
      / \
     o - o
and a unit "lozenge" or "diamond" has these three orientations:
     o
    / \          o - o            o - o
   o   o  and   /   /   and also   \   \
    \ /        o - o                o - o
     o
and for n=3, here is one of the 24 different tiling of the triangle of side length 3 with exactly two lozenges:
          o
         / \
        o   o
       / \ / \
      o - o - o
     /   / \ / \
    o - o - o - o
		

Crossrefs

Programs

  • Mathematica
    Rest@ CoefficientList[Series[3 x^3*(4 - x) (2 + x)/(1 - x)^5, {x, 0, 37}], x] (* Michael De Vlieger, Jul 04 2019 *)
  • PARI
    concat([0,0], Vec(3*x^3*(4 - x)*(2 + x) / (1 - x)^5 + O(x^40))) \\ Colin Barker, Jul 01 2019

Formula

a(n) = (3/8)*(n-2)*(n-1)*(3*n^2 + 3*n - 4) (conjectured by R. J. Mathar, proved by Greg Dresden and E. Sijaric).
From Colin Barker, Jul 01 2019: (Start)
G.f.: 3*x^3*(4 - x)*(2 + x) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.
(End)
E.g.f.: (3/8)*exp(x)*x^2*(32 + 24*x + 3*x^2). - Stefano Spezia, Jul 01 2019

A326368 Number of tilings of an equilateral triangle of side length n with unit triangles (of side length 1) and exactly three unit "lozenges" or "diamonds" (also of side length 1).

Original entry on oeis.org

0, 0, 18, 434, 2814, 11127, 33365, 83568, 184254, 369254, 686952, 1203930, 2009018, 3217749, 4977219, 7471352, 10926570, 15617868, 21875294, 30090834, 40725702, 54318035, 71490993, 92961264, 119547974, 152182002, 191915700, 239933018, 297560034, 366275889
Offset: 1

Views

Author

Greg Dresden, Jul 01 2019

Keywords

Examples

			We can represent a unit triangle this way:
       o
      / \
     o - o
and a unit "lozenge" or "diamond" has these three orientations:
     o
    / \          o - o            o - o
   o   o  and   /   /   and also   \   \
    \ /        o - o                o - o
     o
and for n=3, here is one of the 18 different tiling of the triangle of side length 3 with exactly three lozenges:
          o
         / \
        o   o
       / \ / \
      o - o   o
     /   / \ / \
    o - o - o - o
		

Crossrefs

Programs

  • Mathematica
    Rest@ CoefficientList[Series[x^3*(18 + 308 x + 154 x^2 - 87 x^3 + 10 x^4 + 2 x^5)/(1 - x)^7, {x, 0, 30}], x] (* Michael De Vlieger, Jul 07 2019 *)
  • PARI
    concat([0,0], Vec(x^3*(18 + 308*x + 154*x^2 - 87*x^3 + 10*x^4 + 2*x^5) / (1 - x)^7 + O(x^40))) \\ Colin Barker, Jul 02 2019

Formula

a(n) = (1/16)*(n-2)*(9*n^5 - 9*n^4 - 81*n^3 + 81*n^2 + 160*n - 192) for n >= 2 (proved by Greg Dresden and E. Sijaric).
From Colin Barker, Jul 02 2019: (Start)
G.f.: x^3*(18 + 308*x + 154*x^2 - 87*x^3 + 10*x^4 + 2*x^5) / (1 - x)^7.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) for n>8.
(End)

A122722 Number of triangulations of Delta^2 x Delta^(k-1).

Original entry on oeis.org

1, 6, 108, 4488, 376200, 58652640, 16119956160, 7519632382080, 5788821019685760, 7197150396467808000, 14206044114169232371200, 43903287397136367836697600, 210012592354755890839147008000, 1540026232221309103088828327116800, 17170286302440610680613970557956096000, 289015112280462271460535463614055526400000
Offset: 1

Views

Author

Jonathan Vos Post, Oct 22 2006

Keywords

Comments

The number of triangulations of Delta^2 x Delta^(k) is between alpha^(k^2) and beta*(k^2) where alpha = (27/16)^(1/4) ~ 1.13975 and beta = 6^(1/6) ~ 1.34800 [p. 10 of Santos's handwritten notes about "The Cayley trick"].
There are arithmetic errors in Santos's lecture notes "The Cayley trick". The same table gives lozenge tilings of k*Delta^2.
From Petros Hadjicostas, Sep 13 2019: (Start)
The first column (indexed by k) of the table on p. 9 in Santos' handwritten notes "The Cayley trick" is actually the sequence (A273464(k, k*(k-1)/2 + 1): k >= 1).
In later published papers, Santos (2004, 2005) mentions that the number of triangulations of Delta^2 x Delta^k grows as exp(A244996*k^2/2 + o(k^2)) as k -> infinity. Notice that exp(A244996 * k^2/2) = A242710^(k^2/2). [See Theorem 1 and Theorem 4.9. Probably Theorem 1, part (2), in Santos (2004) has a typo.]
Note that alpha = (27/16)^(1/4) ~ 1.13975 < A242710^(k^2/2) ~ 1.175311 < beta = 6^(1/6) ~ 1.34800 (where alpha and beta are given on the first paragraph of these comments).
The reason the name of the sequence has "Delta^2 x Delta^(k-1)" rather than "Delta^2 x Delta^k" is because (according to Santos) the number of triangulations of Delta^2 x Delta^(k-1) equals k! times the number of lozenge tilings of k*Delta^2. (End)

Examples

			a(1) = 1 * 1! = 1.
a(2) = 3 * 2! = 6.
a(3) = 18 * 3! = 108.
a(4) = "187 * 4! = 2244" [sic]; actually 187 * 4! = 4488.
a(5) = "3135 * 5! = 188100" [sic]; actually 3135 * 5! = 376200.
		

Crossrefs

Formula

Conjectures: a(n) = n! * A273464(n, n*(n+1)/2) for n >= 1; a(n) = A011555(n-1) for n >= 2. [A273464(n,k) is defined for n >= 1 and 0 <= k <= n*(n+1)/2.] - Petros Hadjicostas, Sep 12 2019

Extensions

More terms (using the references) from Petros Hadjicostas, Sep 12 2019

A011555 Number of vertices of secondary polytope for triangle X n-simplex.

Original entry on oeis.org

6, 108, 4488, 376200
Offset: 1

Views

Author

Keywords

References

  • Gelfand, Kapranov and Zelevinsky, Discriminants, Resultants and Multidimensional Determinants, Birkhäuser, 1994, pp. 243-250.

Crossrefs

A011556 Number of regular triangulations of triangle X n-simplex.

Original entry on oeis.org

1, 5, 35, 530
Offset: 1

Views

Author

Keywords

References

  • Gelfand, Kapranov and Zelevinsky, Discriminants, Resultants and Multidimensional Determinants, Birkhäuser, 1994, pp. 243-250.

Crossrefs

Showing 1-6 of 6 results.