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 1-4 of 4 results.

A058212 a(n) = 1 + floor(n*(n-3)/6).

Original entry on oeis.org

1, 0, 0, 1, 1, 2, 4, 5, 7, 10, 12, 15, 19, 22, 26, 31, 35, 40, 46, 51, 57, 64, 70, 77, 85, 92, 100, 109, 117, 126, 136, 145, 155, 166, 176, 187, 199, 210, 222, 235, 247, 260, 274, 287, 301, 316, 330, 345, 361, 376, 392, 409, 425, 442, 460, 477, 495, 514, 532, 551
Offset: 0

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Author

N. J. A. Sloane, Nov 30 2000

Keywords

Comments

For n >= 3, number of solutions to x+y+z == 0 (mod n) with 0 <= x < y < z < n. E.g., for n=3 there is a unique solution, x=0, y=1, z=2.

Crossrefs

Cf. A003035.
Apart from initial term, same as A007997.
The third diagonal of A061857.

Programs

  • Haskell
    a058212 n = 1 + n * (n - 3) `div` 6  -- Reinhard Zumkeller, May 08 2012
    
  • Mathematica
    Table[Floor[(n(n-3))/6]+1,{n,0,70}] (* or *) LinearRecurrence[{2,-1,1,-2,1},{1,0,0,1,1},70] (* Harvey P. Dale, Jun 21 2021 *)
  • PARI
    a(n)=n*(n-3)\6 + 1 \\ Charles R Greathouse IV, Jun 11 2015
  • Sage
    [ceil(binomial(n,2)/3) for n in range(-1, 55)] # Zerinvary Lajos, Dec 03 2009
    

Formula

From Paul Barry, Mar 18 2004: (Start)
G.f.: (1 - 2x + x^2 + x^4)/((1 - x)^2(1 - x^3)).
a(n) = 4*cos(2*Pi*n/3)/9 + (3*n^2 - 9*n + 10)/18. (End)
E.g.f.: (exp(x)*(10 - 6*x + 3*x^2) + 8*exp(-x/2)*cos(sqrt(3)*x/2))/18. - Stefano Spezia, May 03 2023
Sum_{n>=3} 1/a(n) = 6 - (2*Pi/sqrt(3))*(1 - tanh(sqrt(5/3)*Pi/2)/sqrt(5)). - Amiram Eldar, May 06 2023

Extensions

Zerinvary Lajos, Dec 07 2009

A006065 Maximal number of 4-tree rows in n-tree orchard problem.

Original entry on oeis.org

0, 0, 0, 1, 1, 1, 2, 2, 3, 5, 6, 7, 9, 10, 12, 15, 16, 18, 20, 23
Offset: 1

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Author

Keywords

Comments

Maximum number of rows with exactly 4 trees in each row if there are n trees in the orchard.
For further references and links see A003035.

References

  • M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, Chap. 22.
  • F. Levi, Geometrische Konfigurationen, Hirzel, Leipzig, 1929.
  • Xianzu Lin, A new result about orchard-planting problem, Preprint, 2005. [Shows a(20) >= 23.]
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • For further references and links see A003035.

Crossrefs

Cf. A172992 (the same problem, but with integer-valued tree coordinates).

Formula

a(n) >= A172992(n).

Extensions

a(13)-a(15) from Zhao Hui Du, Aug 24 2008
a(17) from Zhao Hui Du, Nov 11 2008
a(18) from Zhao Hui Du, Nov 25 2008
a(19) from Zhao Hui Du, Dec 17 2009
a(20) from Zhao Hui Du, Feb 01 2010

A008997 Orchard problem with 5 trees in a row (may not have all been proved optimal).

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 6, 6, 7, 9, 10, 11
Offset: 1

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Author

Keywords

Comments

Berloquin shows that a(21) >= 12.

References

  • S. A. Burr, B. Grünbaum and N. J. A. Sloane, The Orchard Problem, Geometriae Dedicata, 2 (1974), 397-424.
  • M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, Chap. 22.

Crossrefs

A337747 Maximal number of 4-point circles passing through n points on a plane.

Original entry on oeis.org

0, 0, 0, 1, 1, 3, 6, 12, 14, 22, 30, 45
Offset: 1

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Author

Dmitry Kamenetsky, Sep 17 2020

Keywords

Comments

This is a variant of the orchard-planting problem that uses circles instead of straight lines.
The maximal number of 3-point circles passing through n points on a plane is binomial(n,3). Given an arrangement of n points in general position, any choice of three points defines a circle. - Peter Kagey, Oct 05 2020
Paul Panzer provides upper and lower bounds:
a(n) <= floor(n*(n-1)*(n-2)/24).
a(n) >= 2 + n*((n-2)*(n-2) + 4)/32 for n == 0 (mod 4) and n >= 8.
a(n) >= 2 + (n-1)*((n-1)*(n-5) + 16)/32 for n == 1 (mod 4) and n >= 9.
a(n) >= 2 + n*(n-2)*(n-2)/32 for n == 2 (mod 4) and n >= 10.
a(n) >= 2 + (n-1)*((n-3)*(n-3) + 16)/32 for n == 3 (mod 4) and n >= 11.
It seems that a(n) = n*((n-2)*(n-2) + 4)/32 + 2*A008610(n/2-4) if n == 0 (mod 4) and n >= 8. - Zhao Hui Du, Dec 14 2022
The number of 4-point circles passing through n points (2*cos(t_k), sin(t_k)) where t_k = (2k-1)*Pi/n, k=1,2,...,n is A008610(n-4), so A337747(n) >= A008610(n-4), so A337747(n) ~ n^3/24 for sufficiently large n. - Zhao Hui Du, Dec 15 2022

Examples

			See examples in links.
		

Crossrefs

Cf. A003035 (the original orchard problem), A006065.

Extensions

a(11) from Zhao Hui Du, Nov 22 2022
a(12) from Zhao Hui Du, Dec 01 2022
Showing 1-4 of 4 results.