A376832 Irregular triangle read by rows: the n-th row gives the number of points of an n X n square lattice that lie above or to the left of a line of increasing slope that passes through two lattice points one of which is the bottom-left corner of the lattice, (0, 0).

2, 1, 0, 6, 5, 3, 2, 0, 12, 11, 10, 9, 6, 5, 4, 3, 0, 20, 19, 18, 17, 16, 15, 14, 10, 9, 8, 7, 6, 5, 4, 0, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 0, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 0

Offset: 2

Author

Stefano Spezia, Dec 22 2024

Keywords

• nonn
• look
• tabf

Comments

The increasing slopes of the line are given by the Farey series of order n - 1. Specifically, they are given by the fractions A006842(n-1)/A006843(n-1) followed by their reciprocals A006843(n-1)/A006842(n-1) in reverse order, with the fraction 1/1 included only once.

Examples

			The irregular triangle begins as:
   2,  1,  0;
   6,  5,  3,  2,  0;
  12, 11, 10,  9,  6,  5,  4,  3, 0;
  20, 19, 18, 17, 16, 15, 14, 10, 9, 8, 7, 6, 5, 4, 0;
  ...
		

Links

• Stefano Spezia, Table of n, a(n) for n = 2..9081 (first 30 rows of the irregular triangle)

Crossrefs

Cf. A002378, A006842, A006843, A118403 (row lengths), A161680, A379540 (row sums).

Programs

Mathematica

A118403[n_]:=SeriesCoefficient[(1-2*x+2*x^2)*(1+x^2)/(1-x)^3,{x,0,n}]; T[n_,k_]:=If[1<=k<(A118403[n]+1)/2,n(n-1)-k+1,If[(A118403[n]+1)/2<=k<A118403[n],n(n-1)/2-k+(A118403[n]+1)/2,0]]; Table[T[n,k],{n,2,7},{k,A118403[n]}]//Flatten

Formula

T(n, k) = n*(n - 1) - k + 1 for 1 <= k < (A118403(n)+1)/2.

T(n, k) = n*(n - 1)/2 - k + (A118403(n)+1)/2 for (A118403(n)+1)/2 <= k < A118403(n).

T(n, A118403(n)) = 0.