A290867 Irregular triangle read by rows: the number of points that are the intersections of k semicircles in the configuration A290447(n).

0, 0, 0, 0, 1, 0, 5, 0, 15, 0, 35, 0, 70, 0, 123, 1, 0, 195, 5, 0, 285, 15, 0, 420, 25, 0, 586, 39, 2, 0, 818, 53, 4, 0, 1110, 73, 6, 0, 1451, 103, 10, 0, 1846, 142, 18, 0, 2361, 181, 26, 0, 2956, 234, 33, 2, 0, 3704, 287, 40, 4, 0, 4567, 348, 49, 8

Offset

1,7

Comments

Row lengths are A290726(n).

The first entry of each row is 0, because an intersection requires at least 2 lines.

The first row with 3 entries is for n=9, because that is the first configuration with a nontrivial intersection.

Row sums give A290447.

Links

David Applegate, Table of n, a(n) for n = 1..800

David Applegate, Triangular table T(n,k) for n = 1..100

N. J. A. Sloane, Three (No, 8) Lovely Problems from the OEIS, Experimental Mathematics Seminar, Rutgers University, Oct 05 2017, Part I, Part 2, Slides. (Mentions this sequence)

N. J. A. Sloane (in collaboration with Scott R. Shannon), Art and Sequences, Slides of guest lecture in Math 640, Rutgers Univ., Feb 8, 2020. Mentions this sequence.

Formula

Sum_{k} T(n,k) * binomial(k,2) = binomial(n,4), because there are binomial(n,4) total pairs of semicircles, and an intersection of k consists of binomial(k,2) of those pairs.

A290865(n) = binomial(n,2) + Sum_{k} T(n,k) * (k-1).

Example

Triangle begins:
  0;
  0;
  0;
  0,   1;
  0,   5;
  0,  15;
  0,  35;
  0,  70;
  0, 123,   1;
  0, 195,   5;
  0, 285,  15;
  0, 420,  25;
  0, 586,  39,   2;

Crossrefs

Cf. A290447, A290726, A290865, A290866.

Sequence in context: A214121 A024418 A167297 * A027635 A291218 A321416

Adjacent sequences: A290864 A290865 A290866 * A290868 A290869 A290870

Keyword

nonn,tabf

Author

David Applegate, Aug 12 2017

Status

approved