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 1-3 of 3 results.

A231348 Number of triangles after n-th stage in a cellular automaton based in isosceles triangles of two sizes (see Comments lines for precise definition).

Original entry on oeis.org

0, 1, 3, 7, 11, 15, 23, 33, 41, 45, 53, 65, 81, 91, 111, 133, 149, 153, 161, 173, 189, 201, 225, 253, 285, 295, 315, 343, 383, 405, 449, 495, 527, 531, 539, 551, 567, 579, 603, 631, 663, 675, 699, 731, 779, 807, 863, 923, 987, 997, 1017, 1045, 1085, 1113, 1169, 1233, 1313, 1335, 1379, 1439, 1527, 1573, 1665, 1759, 1823
Offset: 0

Views

Author

Omar E. Pol, Dec 15 2013

Keywords

Comments

On the semi-infinite square grid the structure of this C.A. contains "black" triangles and "gray" triangles (see the Links section). Both types of triangles have two sides of length 5^(1/2). Every black triangle has a base of length 2 hence its height is 2 and its area is 2. Every gray triangle has a base of length 2^(1/2) hence its height is 3/(2^(1/2)) and its area is 3/2. Both types of triangles are arranged in the same way as the triangles of Sierpinski gasket (see A047999 and A006046). The black triangles are arranged in vertical direction. On the other hand the gray triangles are arranged in diagonal direction in the holes of the structure formed by the black triangles. Note that the vertices of all triangles coincide with the grid points.
The sequence gives the total number of triangles (black and gray) in the structure after n-th stage. A231349 (the first differences) gives the number of triangles added at n-th stage.
For a more complex structure see A233780.

Examples

			We start at stage 0 with no triangles, so a(0) = 0.
At stage 1 we add a black triangle, so a(1) = 1.
At stage 2 we add two black triangles, so a(2) = 1+2 = 3.
At stage 3 we add two black triangles and two gray triangles from the vertices of the master triangle, so a(3) = 3+2+2 = 7.
At stage 4 we add four black triangles, so a(4) = 7+4 = 11.
At stage 5 we add two black triangles and two gray triangles from the vertices of the master triangle, so a(5) = 11+2+2 = 15.
At stage 6 we add four black triangles and four gray triangles, so a(6) = 15+4+4 = 23.
At stage 7 we add four black triangles and six gray triangles, so a(7) = 23+4+6 = 33.
At stage 8 we add eight black triangles, so a(8) = 33+8 = 41.
And so on.
Note that always we add both black triangles and gray triangles except if n is a power of 2. In this case at stage 2^k we add only 2^k black triangles, for k >= 0.
		

Crossrefs

A377346 Decimal expansion of the midradius of a truncated cuboctahedron (great rhombicuboctahedron) with unit edge length.

Original entry on oeis.org

2, 2, 6, 3, 0, 3, 3, 4, 3, 8, 4, 5, 3, 7, 1, 4, 6, 2, 3, 5, 9, 2, 0, 2, 5, 8, 0, 3, 4, 3, 2, 5, 3, 7, 1, 4, 2, 2, 2, 9, 0, 6, 7, 2, 0, 2, 6, 5, 0, 7, 5, 5, 4, 8, 3, 8, 1, 7, 6, 1, 2, 4, 0, 6, 0, 4, 0, 5, 6, 7, 4, 5, 9, 8, 9, 1, 5, 3, 0, 4, 7, 0, 7, 7, 5, 8, 7, 6, 2, 7
Offset: 1

Views

Author

Paolo Xausa, Oct 26 2024

Keywords

Examples

			2.26303343845371462359202580343253714222906720265...
		

Crossrefs

Cf. A377343 (surface area), A377344 (volume), A377345 (circumradius).
Cf. A010527 (analogous for a cuboctahedron).

Programs

  • Mathematica
    First[RealDigits[Sqrt[3 + 3/Sqrt[2]], 10, 100]] (* or *)
    First[RealDigits[PolyhedronData["TruncatedCuboctahedron", "Midradius"], 10, 100]]

Formula

Equals sqrt(12 + 6*sqrt(2))/2 = sqrt(12 + A010524)/2 = sqrt(3 + 3/sqrt(2)) = sqrt(3 + A230981).

A358614 Decimal expansion of 9*sqrt(2)/32.

Original entry on oeis.org

3, 9, 7, 7, 4, 7, 5, 6, 4, 4, 1, 7, 4, 3, 2, 9, 8, 2, 4, 7, 5, 4, 7, 4, 9, 5, 3, 6, 8, 3, 9, 7, 7, 5, 8, 4, 5, 9, 7, 7, 2, 0, 2, 1, 4, 9, 4, 9, 7, 6, 6, 6, 4, 5, 5, 8, 0, 9, 4, 1, 1, 7, 6, 3, 0, 9, 8, 9, 3, 5, 0, 9, 5, 6, 7, 4, 6, 7, 6, 0, 4, 6, 7, 6, 6, 7, 1, 4, 9, 4, 0, 2, 9, 6, 4, 9, 1, 9, 2
Offset: 0

Views

Author

Bernard Schott, Dec 05 2022

Keywords

Comments

Smallest constant M such that the inequality
|a*b*(a^2 - b^2) + b*c*(b^2 - c^2) + c*a*(c^2 - a^2)| <= M * (a^2 + b^2 + c^2)^2
holds for all real numbers a, b, c.
Equality stands for any triple (a, b, c) proportional to (1 - 3*sqrt(2)/2, 1, 1 + 3*sqrt(2)/2), up to permutation.
This constant is the answer to the 3rd problem, proposed by Ireland during the 47th International Mathematical Olympiad in 2006 at Ljubljana, Slovenia (see links).
Equivalently |(a - b)(b - c)(c - a)(a + b + c)| / (a^2 + b^2 + c^2)^2 <= M with (a,b,c) != (0,0,0).

Examples

			0.3977475644174329824...
		

Crossrefs

Programs

  • Maple
    evalf(9*sqrt(2)/32), 100);
  • Mathematica
    RealDigits[9*Sqrt[2]/32, 10, 120][[1]] (* Amiram Eldar, Dec 05 2022 *)

Formula

Equals (3/16) * A230981 = (3/32) * A010474 = (9/32) * A002193 = (9/16) * A010503.
Showing 1-3 of 3 results.