1,2

Thomas Garrison, Table of n, a(n) for n = 1...1000

Index entries for linear recurrences with constant coefficients, signature (4,-4,-2,5,-2).

a(n) = 3*(2^n - 1 - n*(n+1)/2) + ceiling(n^2/2).

a(n) = A000982(n) + 3*A002662(n).

a(n) ~ 3*2^n.

From Stefano Spezia, Jul 10 2022: (Start)

G.f.: x*(1 - 2*x + 4*x^2 + x^3)/((1 - x)^3*(1 - x - 2*x^2)).

a(n) = (3*2^(n+2) - 4*n^2 - 6*n - 11 - (-1)^n)/4. (End)

a(5)=61: there are 3*(2^5 - 1 - binomial(6,2)) ways to select 3 or more points on a horizontal line, 5 ways on a vertical line, 3 ways on a diagonal line with slope 1, 3 ways on a diagonal line with slope -1, 1 way on a diagonal line with slope 1/2, and 1 way on a diagonal line with slope -1/2; 48 + 5 + 6 + 2 = 61.

LinearRecurrence[{4, -4, -2, 5, -2}, {1, 2, 8, 23, 61}, 50] (* Paolo Xausa, Oct 19 2024 *)

(Python)

def a(n): return 3*((1<<n) - 1 - n*(n+1)//2)+(n**2+1)//2

Cf. A002662 (1 X n), 2*A002662 (2 X n), A355552 (4 X n), A355553 (n X n).

Sequence in context: A079460 A154144 A255942 * A180664 A294959 A290926

Adjacent sequences: A355548 A355549 A355550 * A355552 A355553 A355554

nonn,easy

Thomas Garrison, Jul 06 2022

approved