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

2,1

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.

Links

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

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.

Example

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;
  ...

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

Crossrefs

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

Sequence in context: A358694 A047922 A276891 * A021830 A247686 A352369

Adjacent sequences: A376829 A376830 A376831 * A376833 A376834 A376835

Keyword

nonn,look,tabf

Author

Stefano Spezia, Dec 22 2024

Status

approved