A389123 Lower (1/3,1/2)-midsequence of ((2^n)) and ((3^n)); see Comments.

0, 2, 5, 16, 45, 132, 385, 1136, 3365, 10012, 29865, 89256, 267085, 799892, 2396945, 7185376, 21545205, 64613772, 193797625, 581305496, 1743741725, 5230875652, 15691927905, 47074385616, 141220360645, 423655489532, 1270955283785, 3812843481736, 11438485705965

Offset

0,2

Comments

Suppose that s=(s(n)) and t=(t(n)) are sequences of numbers and u>0 and v>0. The lower (u,v)-midsequence of s and t is floor(u*s+v*t); the upper (u,v)-midsequence of s and t is ceiling(u*s+v*t). If s and t are linearly recurrent and u and v are rational, then both midsequences are linearly recurrent.

Links

Table of n, a(n) for n=0..28.

Index entries for linear recurrences with constant coefficients, signature (5,-5,-5,6).

Formula

a(n) = floor((1/3)*2^n + (1/2)*3^n), for n>=0.

a(n) = 5*a(n-1) - 5*a(n-2) - 5*a(n-3) + 6*a(n-4), with (a(0),a(1),a(2),a(3))=(0,2,5,16).

G.f.: x*(-2 + 5*x - x^2)/(-1 + 5*x - 5*x^2 - 5*x^3 + 6*x^4).

Example

s = (2^n) = (1,2,4,8,16,...) = A000079; t = (3^n) = (1,3,9,27,81, ...) = A000244.
(a(n)) = floor((1/3)(1,2,4,8,16,...) + (1/2)(1,3,9,27,81, ...)) = (0, 2, 5, 16, 45, ...).
A389124 = ceiling((1/3)(1,2,4,8,16,...) + (1/2)(1,3,9,27,81, ...)) = (1,3,6,17,46, ...).

Mathematica

s[n_] := 2^n ; t[n_] := 3^n;
u[n_] := Floor[(s[n]/3 + t[n]/2)];
v[n_] := Ceiling[(s[n]/3 + t[n]/2)];
Table[u[n], {n, 0, 60}]  (* A389123 *)
Table[v[n], {n, 0, 60}]  (* A389124 *)
(* Also *)
LinearRecurrence[{5, -5, -5, 6}, {0, 2, 5, 16}, 30]

Crossrefs

Cf. A000079, A000244, A387778, A389124.

Sequence in context: A152428 A317890 A138573 * A148375 A075887 A306504

Adjacent sequences: A389120 A389121 A389122 * A389124 A389125 A389126

Keyword

nonn,easy

Author

Clark Kimberling, Sep 23 2025

Status

approved