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.

A374338 Start with two vertices and draw a circle around each whose radius is the distance between the vertices. The sequence gives the number of vertices constructed after n iterations of drawing circles with this same radius around every new vertex created from all circles' intersections. See the Comments.

Original entry on oeis.org

4, 8, 14, 24, 34, 46, 62, 78, 96, 118, 140, 164, 192, 220, 250, 284, 318, 354, 394, 434, 476, 522, 568, 616, 668, 720, 774, 832, 890, 950, 1014, 1078, 1144, 1214, 1284, 1356, 1432, 1508, 1586, 1668, 1750, 1834, 1922, 2010, 2100, 2194, 2288, 2384, 2484, 2584, 2686, 2792
Offset: 1

Views

Author

Scott R. Shannon, Jul 05 2024

Keywords

Comments

Start with two vertices and, using each as the center, draw a circle around each whose radius is the distance between the vertices. These circles' intersections create two additional vertices, so after the first iteration four vertices exist. Using these four vertices as centers draw four new circles whose radius is the same as the distance between the initial two vertices. These circles' intersections create eight new vertices. Repeat this process n times; the sequence gives the number of vertices after n iterations.

Links

Crossrefs

Cf. A374337 (regions), A374339 (edges), A359569, A371373, A371254.

Formula

a(n) = A374339(n) - A374337(n) + 1, by Euler's formula.
Conjectured:
If n = 3*k + 1, k >= 0, a(n) = (3*n^2 + 5*n + 4)/3.
If n = 3*k, k >= 1, a(n) = (3*n^2 + 5*n)/3.
If n = 3*k - 1, k >= 1, a(n) = (3*n^2 + 5*n + 2)/3.

A374827 Place n equally spaced points on the circumference of a circle of radius r and then connect each pair of points with straight lines whose intersections create A007569(n) - n additional points. Draw a circle of radius r around each of the A007569(n) points. The sequence gives the total number of curved edges formed from all circle intersections.

Original entry on oeis.org

1, 2, 9, 28, 150, 636, 3290, 6192, 35145, 57380, 230494, 192588, 1055535, 1177148
Offset: 1

Views

Author

Scott R. Shannon, Jul 21 2024

Keywords

Links

Crossrefs

Cf. A374825 (vertices), A374826 (regions), A374828 (k-gons), A007569 (total circles), A183207, A374339.

Formula

a(n) = A374825(n) + A374826(n) - 1, by Euler's formula.

A384703 On a 2 X n grid of vertices, draw a circle through every unordered triple of non-collinear vertices: a(n) is the number of distinct edges in the planar graph formed from the intersections of the circles.

Original entry on oeis.org

0, 4, 54, 416, 2182, 7884, 23294, 56982, 126310, 253564, 477462, 844524, 1424316
Offset: 1

Views

Author

Scott R. Shannon and N. J. A. Sloane, Jun 07 2025

Keywords

Comments

The edges being counted are of course arcs of circles.

Links

Crossrefs

Cf. A384700 (circles), A384701 (vertices), A384702 (regions), A359571, A374827, A374339, A373108.

Formula

a(n) = A384701(n) + A384702(n) - 1 by Euler's formula, for n > 1.

A374337 Start with two vertices and draw a circle around each whose radius is the distance between the vertices. The sequence gives the number of regions constructed after n iterations of drawing circles with this same radius around every new vertex created from all circles' intersections.

Original entry on oeis.org

3, 11, 27, 55, 99, 145, 203, 277, 353, 441, 545, 651, 769, 903, 1039, 1187, 1351, 1517, 1695, 1889, 2085, 2293, 2517, 2743, 2981, 3235, 3491, 3759, 4043, 4329, 4627, 4941, 5257, 5585, 5929, 6275, 6633, 7007, 7383, 7771, 8175, 8581, 8999, 9433, 9869, 10317, 10781, 11247, 11725, 12219, 12715
Offset: 1

Views

Author

Scott R. Shannon, Jul 05 2024

Keywords

Comments

See A374338 for further details.

Links

Crossrefs

Cf. A374338 (vertices), A374339 (edges), A359570, A371374, A371253.

Formula

a(n) = A374339(n) - A374338(n) + 1, by Euler's formula.
Conjectured:
If n = 3*k + 1, k >= 0, a(n) = |(15*n^2 - 17*n - 7)/3|.
If n = 3*k, k >= 1, a(n) = (15*n^2 - 17*n - 3)/3.
If n = 3*k - 1, k >= 1, a(n) = (15*n^2 - 17*n + 7)/3.
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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