cp's OEIS Frontend

A342041 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.

%I A342041 #10 Feb 18 2022 20:51:46
%S A342041 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 A342041 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 A342041 7,13,3,2,2,1,1,1,1,1,1,13,7,13,4
%N A342041 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 A342041 Rows start at n = 2, and terms range from k = 2 to k = n. (When k = 1, there can be arbitrarily many lines.)
%C A342041 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 A342041 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 A342041 For n = 10, k = 4, the unique arrangement with 5 lines (up to symmetry) is
%e A342041   1111000000
%e A342041   1000111000
%e A342041   0100100110
%e A342041   0010010101
%e A342041   0001001011
%e A342041 There are no such arrangements with 6 lines. Thus T(10,4) = 5.
%e A342041 These lines are in bijection with the sets of 4 polar axes on a dodecahedron whose endpoints form a cube.
%e A342041 Table begins:
%e A342041 n\k | 2  3  4  5  6  7  8  9
%e A342041 ----+-----------------------
%e A342041   2 | 1;
%e A342041   3 | 3, 1;
%e A342041   4 | 3, 1, 1;
%e A342041   5 | 4, 2, 1, 1;
%e A342041   6 | 5, 4, 1, 1, 1;
%e A342041   7 | 6, 7, 2, 1, 1, 1;
%e A342041   8 | 7, 7, 2, 1, 1, 1, 1;
%e A342041   9 | 8, 7, 3, 2, 1, 1, 1, 1;
%K A342041 nonn,tabl
%O A342041 2,2
%A A342041 _Drake Thomas_, Feb 26 2021