A342041 - OEIS

%I #10 Feb 18 2022 20:51:46

%S 1,3,1,3,1,1,4,2,1,1,5,4,1,1,1,6,7,2,1,1,1,7,7,2,1,1,1,1,8,7,3,2,1,1,

%T 1,1,9,7,5,2,1,1,1,1,1,10,7,6,2,2,1,1,1,1,1,11,7,9,3,2,1,1,1,1,1,1,12,

%U 7,13,3,2,2,1,1,1,1,1,1,13,7,13,4

%N Triangle read by rows: T(n,k) = maximum number of lines of size k on n points so that every two lines intersect in one point.

%C Rows start at n = 2, and terms range from k = 2 to k = n. (When k = 1, there can be arbitrarily many lines.)

%C If a projective plane of order k-1 exists, then for n between k^2-k+1 and k^3-2k^2+3k-2 inclusive, T(n,k) = k^2-k+1. For higher n, T(n,k) = floor((n-1)/(k-1)).

%H Reddit, <a href="https://www.reddit.com/r/math/comments/lsd8ad/what_are_these_fake_projective_planes/">What are these "fake" projective planes?</a>

%e For n = 10, k = 4, the unique arrangement with 5 lines (up to symmetry) is

%e   1111000000

%e   1000111000

%e   0100100110

%e   0010010101

%e   0001001011

%e There are no such arrangements with 6 lines. Thus T(10,4) = 5.

%e These lines are in bijection with the sets of 4 polar axes on a dodecahedron whose endpoints form a cube.

%e Table begins:

%e n\k | 2  3  4  5  6  7  8  9

%e ----+-----------------------

%e   2 | 1;

%e   3 | 3, 1;

%e   4 | 3, 1, 1;

%e   5 | 4, 2, 1, 1;

%e   6 | 5, 4, 1, 1, 1;

%e   7 | 6, 7, 2, 1, 1, 1;

%e   8 | 7, 7, 2, 1, 1, 1, 1;

%e   9 | 8, 7, 3, 2, 1, 1, 1, 1;

%K nonn,tabl

%O 2,2

%A _Drake Thomas_, Feb 26 2021