A366314 a(n) = a(n-1) + 3*a(n-2) + 9*a(n-3) with a(0)=0, a(1)=1, a(2)=4.

0, 1, 4, 7, 28, 85, 232, 739, 2200, 6505, 19756, 59071, 176884, 531901, 1594192, 4781851, 14351536, 43044817, 129136084, 387434359, 1162245964, 3486773797, 10460420920, 31380955987, 94143182920, 282429839161, 847287991804, 2541866155567, 7625598683428, 22876789076365, 68630380526752

Offset: 0

Author

Gary Detlefs, Oct 06 2023

Keywords

• nonn
• easy

Comments

This sequence could be considered a companion sequence to A103770, in that both sequences are manifested in the averaging of a Tribonacci sequence with initial seeds of x, y, z.

If f(n) is a third order recurrence with f(0)=x, f(1)=y, f(2)=z, and f(n) = (f(n-1)+f(n-2)+f(n-3))/3,n>2, then

f(n) =(A103770(n-2)*z + a(n-1)*y + A103770(n-3)x)/3^(n-2).

In the general case, these "averaging" sequences will approach a limit of (x+2*y+3*z)/6.

a(n) mod 9 repeats [1,4,7] from offset 1 = A100402(n-1)...

Examples

			Starting with initial terms of x, y, z, the sequence proceeds:
  (z + y + x)/3
  (4z + 4y +x)/9
  (16z + 7y + 4x)/27
  (37z + 28y + 16x)/81
  (121z + 85y + 37x)/243
  ....
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (1,3,9).

Crossrefs

Cf. A103770.

Programs

Maple

a:= proc(n) option remember; if n < 3 then n^2 else a(n-1)+3*a(n-2)+9*a(n-3) fi end: seq(a(n), n=0..30);

Mathematica

LinearRecurrence[{1,3,9},{0,1,4},50] (* Paolo Xausa, Nov 14 2023 *)

Formula

From Stefano Spezia, Oct 15 2023: (Start)

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

a(n) = (4*3^n + (-2 - sqrt(2)*i)*(-1 - sqrt(2)*i)^n + i*(-1 + sqrt(2)*i)^n*(2*i + sqrt(2)))/12, where i denotes the imaginary unit. (End)