0,2

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

Index entries for linear recurrences with constant coefficients, signature (3,-1,1,-3,1).

a(n) = ceiling((1/2)*(F(2*n) + F(2*n+1))), where F=A000045.

a(n) = 3*a(n-1) - a(n-2) + a(n-3) - 3*a(n-4) + a(n-5).

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

6*a(n) = A061347(n+1) +3*A001906(n+1) +2. - R. J. Mathar, Sep 26 2025

s = (F(2n)) = (0, 1, 3, 8, 21, ...); t = (1, 2, 5, 13, 34, ...).

u(n) = floor((1/2)(0+1, 1+2, 3+5, 8+13, 21+34, ...)) = (0, 1, 4, 10, 27, ...).

v(n) = ceiling((1/2)(0+1, 1+2, 3+5, 8+13, 21+34, ...)) = (1, 2, 4, 11, 28, ...).

f[n_] := Fibonacci[n]; s[n_] := f[2 n]; t[n_] := f[2 n + 1]; r = 1/2;

u[n_] := Floor[r*(s[n] + t[n])]

v[n_] := Ceiling[r*(s[n] + t[n])]

Table[u[n], {n, 0, 30}] (* lower, A387778 *)

Table[v[n], {n, 0, 30}] (* upper, A387779 *)

(* Also *)

LinearRecurrence[{3, -1, 1, -3, 1}, {1, 2, 4, 11, 28}, 30]

Cf. A000045, A387778, A387780, A387781.

Sequence in context: A024962 A202085 A122423 * A099016 A026122 A108629

Adjacent sequences: A387776 A387777 A387778 * A387780 A387781 A387782

nonn,easy

Clark Kimberling, Sep 10 2025

approved