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 1-7 of 7 results.

A171972 Greatest integer k such that k/n^2 < sqrt(3).

Original entry on oeis.org

0, 1, 6, 15, 27, 43, 62, 84, 110, 140, 173, 209, 249, 292, 339, 389, 443, 500, 561, 625, 692, 763, 838, 916, 997, 1082, 1170, 1262, 1357, 1456, 1558, 1664, 1773, 1886, 2002, 2121, 2244, 2371, 2501, 2634, 2771, 2911, 3055, 3202, 3353, 3507, 3665, 3826, 3990, 4158
Offset: 0

Views

Author

Reinhard Zumkeller, Jan 20 2010

Keywords

Comments

Integer part of the surface area of a regular tetrahedron with edge length n.
A171970(n)*A005843(n) <= a(n);
a(n) <= 4*A171971(n); 0 <= a(n) - 4*A171971(n) < 4.

Crossrefs

Programs

  • Haskell
    a171972 = floor . (* sqrt 3) . fromInteger . a000290
    -- Reinhard Zumkeller, Dec 15 2012
  • Mathematica
    z = 120; r = Sqrt[3];
    Table[Floor[r*n^2], {n, 0, z}]; (* A171972 *)
    Table[Ceiling[r*n^2], {n, 0, z}]; (* A293410 *)
    Table[Round[r*n^2], {n, 0, z}]; (* A070169. -  Clark Kimberling, Oct 11 2017 *)

Formula

a(n) = floor(n^2 * sqrt(3)).
a(n) = A022838(n^2);
a(n) = A293410(n) - 1 for n > 0.

A171973 Integer part of the volume of a regular tetrahedron with edge length n.

Original entry on oeis.org

0, 0, 3, 7, 14, 25, 40, 60, 85, 117, 156, 203, 258, 323, 397, 482, 579, 687, 808, 942, 1091, 1254, 1433, 1629, 1841, 2071, 2319, 2587, 2874, 3181, 3510, 3861, 4235, 4632, 5052, 5498, 5969, 6466, 6990, 7542, 8122, 8731, 9369, 10039, 10739, 11471, 12235
Offset: 1

Views

Author

Reinhard Zumkeller, Jan 20 2010

Keywords

Comments

Lim{n->oo} a(n)/A000292(n) = sqrt(2)/2;
floor(A171971(n)*A171974(n)/3) <= a(n).

Crossrefs

Programs

  • Haskell
    a171973 = floor . (/ 12) . (* sqrt 2) . fromInteger . a000578
    -- Reinhard Zumkeller, Dec 15 2012

Formula

a(n) = floor(n^3 * sqrt(2) / 12).

A171970 Integer part of the height of an equilateral triangle with side length n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 6, 7, 8, 9, 10, 11, 12, 12, 13, 14, 15, 16, 17, 18, 19, 19, 20, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 30, 31, 32, 32, 33, 34, 35, 36, 37, 38, 38, 39, 40, 41, 42, 43, 44, 45, 45, 46, 47, 48, 49, 50, 51, 51, 52, 53, 54, 55, 56, 57, 58, 58, 59, 60, 61, 62, 63
Offset: 1

Views

Author

Reinhard Zumkeller, Jan 20 2010

Keywords

Crossrefs

Beatty sequence of A010527.

Programs

Formula

a(n) = floor(n*sqrt(3)/2).
a(n) = floor(A022838(n)/2).
a(n)*A004526(n) <= A171971(n)
a(n)*A005843(n) <= A171972(n).

A171974 Integer part of the height of a regular tetrahedron with edge length n.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 5, 6, 7, 8, 8, 9, 10, 11, 12, 13, 13, 14, 15, 16, 17, 17, 18, 19, 20, 21, 22, 22, 23, 24, 25, 26, 26, 27, 28, 29, 30, 31, 31, 32, 33, 34, 35, 35, 36, 37, 38, 39, 40, 40, 41, 42, 43, 44, 44, 45, 46, 47, 48, 48, 49, 50, 51, 52, 53, 53, 54, 55, 56, 57, 57, 58, 59
Offset: 1

Views

Author

Reinhard Zumkeller, Jan 20 2010

Keywords

Comments

-3 <= 4*A171975(n) - 3*a(n) < 3;
a(n)*A171975(n) <= A007590(n);
floor(a(n)*A171971(n)/3) <= A171973(n).

Crossrefs

Cf. A171972, A022840. Beatty sequence of A157697.

Programs

  • Haskell
    a171974 = floor . (/ 3) . (* sqrt 6) . fromInteger
    -- Reinhard Zumkeller, Dec 15 2012

Formula

a(n) = floor(n*sqrt(6)/3).

A255604 Table read by antidiagonals, T(n,k) is the integer part of the area of a regular k-gon with side length n.

Original entry on oeis.org

0, 1, 1, 3, 4, 1, 6, 9, 6, 2, 10, 16, 15, 10, 3, 15, 25, 27, 23, 14, 4, 21, 36, 43, 41, 32, 19, 6, 27, 49, 61, 64, 58, 43, 24, 7, 35, 64, 84, 93, 90, 77, 55, 30, 9, 43, 81, 110, 127, 130, 120, 98, 69, 37, 11, 52, 100, 139, 166, 178, 173, 154, 123, 84, 44, 13, 62, 121, 172
Offset: 1

Views

Author

Kival Ngaokrajang, Feb 27 2015

Keywords

Comments

See illustration in the links.

Examples

			See table in the links.
		

Crossrefs

Row 1: A011865.
Columns 1,2,3,4,10: A171971, A000290, A255605, A255606, A172526.

Programs

  • Mathematica
    t[n_, k_] := Floor[k*n^2/(4 Tan[Pi/k])]; Table[t[n - k + 1, k], {n, 3, 14}, {k, 3 , n}] // Flatten
    (* to view table: Table[t[n, k], {k, 3, 6}, {n, 6}] // TableForm *) (* Robert G. Wilson v, Feb 28 2015 *)
  • PARI
    {for(i=1,20,for(n=3,i-1,a=floor(n*(i-n)^2/(4*tan(Pi/n)));print1(a,", ")))}

Formula

T(n,k) = floor(k*n^2/(4*tan(Pi/k))), n >=1, k >=3.

A308358 Beatty sequence for sqrt(3)/4.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 2, 3, 3, 3, 4, 4, 5, 5, 6, 6, 6, 7, 7, 8, 8, 9, 9, 9, 10, 10, 11, 11, 12, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 16, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 21, 22, 22, 22, 23, 23, 24, 24, 25, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 29, 30, 30, 31, 31, 32, 32, 32, 33, 33, 34, 34
Offset: 0

Views

Author

R. J. Mathar, May 22 2019

Keywords

Comments

Differs from A057357 first at n=37.

Crossrefs

Cf. A120011.

Programs

  • Mathematica
    Floor[Sqrt[3] Range[0, 100]/4] (* Wesley Ivan Hurt, Dec 26 2023 *)

Formula

a(n) = floor(n*A120011).
A171971(n) = a(n^2).

A194082 Sum{floor(sqrt(3)*k/2) : 1<=k<=n}.

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 27, 34, 42, 51, 61, 72, 84, 96, 109, 123, 138, 154, 171, 189, 208, 227, 247, 268, 290, 313, 337, 362, 387, 413, 440, 468, 497, 527, 558, 590, 622, 655, 689, 724, 760, 797, 835, 873, 912, 952, 993, 1035, 1078, 1122, 1167, 1212
Offset: 1

Views

Author

Clark Kimberling, Aug 17 2011

Keywords

Comments

Partial sums of A171970.
Comment from R. J. Mathar, Dec 02 2012 (Start):
a(n-1) is the number of unit squares regularly packed into the isosceles triangle of edge length n.
The triangle may be aligned with the Cartesian axes by putting its bottom edge on the horizontal axis, so its vertices are at (x,y) = (0,0), (n,0) and (n/2,sqrt(3)*n/2), see A010527.
The area inside the triangle is sqrt(3)*n^2/4 = A120011*n^2. There is an obvious upper limit of floor(sqrt(3)*n^2/4) = A171971(n) to the count of non-overlapping unit squares inside this triangle.
Regular packing: We place the first row of unit squares so they touch the bottom edge of the triangle. Their number is limited by the length of the horizontal section of the line y=1 inside the triangle, n-2*y/sqrt(3), which touches all of these first-row squares at their top.
The number of unit squares in the next row, between y=1 and y=2, is limited by the length of the horizontal section of the line y=2 inside the triangle, n-2*y/sqrt(3). Continuing, in row y=1, 2, ... we insert floor(n-2*y/sqrt(3)) unit squares, all with the same orientation.
The total number of squares is sum_{ y=1, 2, ..., floor(n*sqrt(3)/2) } floor( n-2*y/sqrt(3) ), and resummation yields, up to an index shift, this sequence here.
(End)

Crossrefs

Cf. A171970.

Programs

  • Mathematica
    r = Sqrt[3]/2;
    c[k_] := Sum[Floor[j*r], {j, 1, k}];
    Table[c[k], {k, 1, 90}]
  • PARI
    a(n)=sum(k=1,n,sqrtint(3*k^2\4)) \\ Charles R Greathouse IV, Jan 06 2013
Showing 1-7 of 7 results.