cp's OEIS Frontend

Two circles are either disjoint or meet in two points. Tangential contacts are not allowed. A point belongs to exactly one or two circles. Three circles may not meet at a point. The circles may have different radii.

This is in the affine plane, rather than the projective plane.

Two arrangements are considered the same if one can be continuously changed to the other while keeping all circles circular (although the radii may be continuously changed), without changing the multiplicity of intersection points, and without a circle passing through an intersection point. Turning the whole configuration over is allowed.

Several variations are possible:

- straight lines instead of circles (see A241600).

- straight lines in general position (see A090338).

- curved lines in general position (see A090339).

- allow circles to meet tangentially but without multiple intersection points (begins 1, 5, ...); more terms are needed.

- again use circles, but allow multiple intersection points (also begins 1, 5, ...); more terms are needed.

- use ellipses rather than circles.

- a question from Walter D. Wallis: what if the circles must all have the same radius?

a(1)-a(5) computed by Jon Wild.

a(n) >= A000081(n+1) - Benoit Jubin, Dec 21 2014. More precisely, there are A000081(n+1) ways to arrange n circles if no two of them meet. - N. J. A. Sloane, May 16 2017

From Daniel Forgues, Aug 08-09 2015: (Start)

A representation for the diagrams in a250001.jpg (in the same order):

a(1) = 1: {{2}};

a(2) = 3: {{2, 3}, {2, 4}, {4, 6}};

a(3) = 14: {{2, 4, 8}, {2, 3, 6}, {2, 3, 4}, {2, 3, 5}, {4, 6, 5},

{4, 6, 15}, {2, 6, 9}, {4, 6, 12}, {2, 8, 12}, {30, 42, 70},

{?, ?, ?}, {?, ?, ?}, {15, 21, 35}, {?, ?, ?}}.

In lexicographic order:

a(3) = 14: {{2, 3, 4}, {2, 3, 5}, {2, 3, 6}, {2, 4, 8}, {2, 6, 9},

{2, 8, 12}, {4, 6, 5}, {4, 6, 12}, {4, 6, 15}, {15, 21, 35},

{30, 42, 70}, {?, ?, ?}, {?, ?, ?}, {?, ?, ?}}.

The smallest integers greater than 1 are used for the representation:

(p_1)^(a_1)*...*(p_m)^(a_m), where

0 <= a_i <= n, for 1 <= i <= m;

(a_1)+...+(a_m) > 0.

Could the Venn diagram interpretation (of the k-wise, 1 <= k <= n, common divisors of k numbers from each subset) reveal a pattern?

Does this representation work for more complex diagrams? (End)

Once you get to n=5, geometric concerns mean that not all topologically-conceivable arrangements are actually circle-drawable. My program enumerated 16977 conceivable arrangements of 5 pseudo-circles, and Christopher Jones and I together have figured out how to show that 26 of these arrangements are not actually circle-drawable. So it seems that a(5) = 16951. This entry will be updated soon, and there will be a new sequence for the number of topologically-conceivable arrangements. - Jon Wild, Aug 25 2016 [The counts in this comment were amended by Jon Wild on Aug 30 2016. I apologize for taking so long to make the corrections here. - N. J. A. Sloane, Jun 11 2017]

a(n) <= 7*13^(binomial(n,3) + binomial(n,2) + 3n - 1) is a (loose) upper bound, see Reddit link. I believe XkF21WNJ's reply shaves off a factor of 13^3 from this bound for all n > 1. - Linus Hamilton, Apr 14 2019

A good upper bound for a(6) is given in sequence A288559, which counts the arrangements of pseudo-circles, i.e. the topologically conceivable arrangements mentioned above, which are not all necessarily realizable with true circles. The number of arrangements of 6 pseudo-circles was found by Andrii Shportko and Jon Wild to be 17,552,169. - Jon Wild, Jun 03 2025

In A288559, a(5) included 26 non-circularizable pseudocircle arrangements, which generated in turn 132,546 6-pseudocircle descendants. These descendants must be excluded from A250001, which means that a tighter upper bound for A250001(6) is 17,419,623. - Andrii Shportko, Jun 06 2025

Consider the rules for the arrangements of n circles described in A250001.

Right border gives A000012. - Omar E. Pol, Aug 09 2015

From Omar E. Pol, May 21 2017: (Start)

Note that T(4,1) = 155 does not include the arrangement of four circles in which there is a central circle that is surrounded by the union of three circles, because in that arrangement there are two solid regions, not one. The smallest solid region is formed by the central solid circle. The largest solid region is formed by the union of three solid circles.

T(4,2) = 15 includes the arrangement mentioned above.

1 together with the first column gives A287149.

For another version see A285996. (End) (Comment clarified on Jun 15 2017.)

Length of n-th row: 1 + (n-1)n/2 (for a configuration for T(n,(n-1)n/2), consider n circles of radius 1 and centers at (k/n,0) for 1<=k<=n).

The generating function down the column k=1 is 1+z^2 *C^2(z) *[C^2(z)+C(z^2)]/ (2*[1-z*C(z)]) = 1+ z^2 +4*z^3 +15*z^4+ 50*z^5+...where C(z) = 1+z+2*z^2+4*z^3+... is the g.f. of A000081 divided by z; eq. (78) in arXiv:1603.00077. - R. J. Mathar, Mar 05 2016

Consider the arrangements of n circles described in A250001.

Note that the sum of the 4th row must be equal to A250001(4) = 173.

In other words: not counting the regions between circles.

Consider the arrangements of n circles described in A250001.

Note that the sum of the 4th row must be equal to A250001(4) = 173.

Consider the rules for the arrangements of n circles described in A250001.

Note that T(4,1) = 156 includes the arrangement of 4 circles in which there is a little circle that is surrounded by the union of three solid circles, because the little circle is an inland island, or lake island, which does not count. So there is only one separated island. Hence T(4,2) = 14 does not include the mentioned arrangement.

Question: is 1 together with the first column of the triangle the same as A275923?

Consider the arrangements of n circles described in A250001.