A389124 Upper (1/3,1/2)-midsequence of ((2^n)) and ((3^n)); see Comments.

1, 3, 6, 17, 46, 133, 386, 1137, 3366, 10013, 29866, 89257, 267086, 799893, 2396946, 7185377, 21545206, 64613773, 193797626, 581305497, 1743741726, 5230875653, 15691927906, 47074385617, 141220360646, 423655489533, 1270955283786, 3812843481737, 11438485705966

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) = ceiling((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))=(1,3,6,17).

G.f.: (-1 + 2*x + 4*x^2 - 7*x^3)/(-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.
A389123 = 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}, {1, 3, 6, 17}, 30]

Crossrefs

Cf. A000079, A000244, A387778, A389123.

Sequence in context: A321227 A006081 A099511 * A204517 A307685 A360273

Adjacent sequences: A389121 A389122 A389123 * A389125 A389126 A389127

Keyword

nonn,easy

Author

Clark Kimberling, Sep 23 2025

Status

approved