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.

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

Author

Scott R. Shannon, Jul 05 2024

Keywords

• nonn

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

• Scott R. Shannon, Image for n = 1.
• Scott R. Shannon, Image for n = 2.
• Scott R. Shannon, Image for n = 3.
• Scott R. Shannon, Image for n = 4.
• Scott R. Shannon, Image for n = 16.

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.