A118403 -id:A118403 - cp's OEIS frontend

A118402 Row sums of triangle A118401.

1, 1, 3, 1, 5, -1, 7, -3, 9, -5, 11, -7, 13, -9, 15, -11, 17, -13, 19, -15, 21, -17, 23, -19, 25, -21, 27, -23, 29, -25, 31, -27, 33, -29, 35, -31, 37, -33, 39, -35, 41, -37, 43, -39, 45, -41, 47, -43, 49, -45, 51, -47, 53, -49, 55, -51, 57, -53, 59, -55, 61, -57, 63, -59, 65, -61, 67, -63, 69, -65, 71

Offset: 0

Paul D. Hanna, Apr 27 2006

sign

Index entries for linear recurrences with constant coefficients, signature (-1,1,1).

Cf. A118401, A118403.

a := n -> `if`(n=1, 1, (5+(-1)^n*(2*n-3))/2);
seq(a(n), n=0..70); # Peter Luschny, Aug 04 2014

Join[{1, 1}, LinearRecurrence[{-1, 1, 1}, {3, 1, 5}, 70]] (* Jean-François Alcover, Jun 13 2019 *)

{a(n)=polcoeff((1+2*x+2*x^2)*(1+x^2)/(1+x+x*O(x^n))^2/(1-x),n,x)}

G.f.: (1+2*x+2*x^2)*(1+x^2)/(1+x)^2/(1-x).
a(n^2-n+2) = A118403(n).
a(2n) = 2n+1, a(2n+1) = 3-2n, n>0. - Ralf Stephan, Aug 18 2013
a(n) = (5+(-1)^n*(2*n-3))/2 for n>1. - Peter Luschny, Aug 04 2014

A118401 Triangle, read by rows, equal to the matrix square of triangle A118400; also equals the matrix inverse of triangle A118407.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, -2, 2, 0, 1, 4, -2, 2, 0, 1, -6, 4, -2, 2, 0, 1, 8, -6, 4, -2, 2, 0, 1, -10, 8, -6, 4, -2, 2, 0, 1, 12, -10, 8, -6, 4, -2, 2, 0, 1, -14, 12, -10, 8, -6, 4, -2, 2, 0, 1, 16, -14, 12, -10, 8, -6, 4, -2, 2, 0, 1, -18, 16, -14, 12, -10, 8, -6, 4, -2, 2, 0, 1, 20, -18, 16, -14, 12, -10, 8, -6, 4, -2, 2, 0, 1

Offset: 0

Paul D. Hanna, Apr 27 2006

This triangle has an integer matrix square-root (A118400) if the main diagonal of the square-root is allowed to be signed. Even though the columns of this triangle are all the same, the columns of the matrix square-root A118400 are all different.

			Triangle begins:
1;
0, 1;
2, 0, 1;
-2, 2, 0, 1;
4,-2, 2, 0, 1;
-6, 4,-2, 2, 0, 1;
8,-6, 4,-2, 2, 0, 1;
-10, 8,-6, 4,-2, 2, 0, 1;
12,-10, 8,-6, 4,-2, 2, 0, 1;
-14, 12,-10, 8,-6, 4,-2, 2, 0, 1;
16,-14, 12,-10, 8,-6, 4,-2, 2, 0, 1; ...

Cf. A118400 (matrix square-root), A118402 (row sums), A118403 (unsigned row sums), A118407 (matrix inverse).

{T(n,k)=polcoeff(polcoeff((1+2*x+2*x^2)*(1+x^2)/(1+x)^2/(1-x*y+x*O(x^n)),n,x)+y*O(y^k),k,y)}

G.f.: A(x,y) = (1 + 2*x + 2*x^2)*(1+x^2)/(1+x)^2/(1-x*y). Column g.f.: (1 + 2*x + 2*x^2)*(1+x^2)/(1+x)^2.

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).

Original entry on oeis.org

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

Stefano Spezia, Dec 22 2024

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.

			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;
  ...

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

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

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

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.

cp's OEIS Frontend

A118402 Row sums of triangle A118401.

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A118401 Triangle, read by rows, equal to the matrix square of triangle A118400; also equals the matrix inverse of triangle A118407.

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PARI

Formula

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).

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Examples

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Mathematica

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