A387778 Lower (1/2)-midsequence of (F(2n)) and (F(2n+1)), where F=A000045 (Fibonacci numbers); see Comments.

0, 1, 4, 10, 27, 72, 188, 493, 1292, 3382, 8855, 23184, 60696, 158905, 416020, 1089154, 2851443, 7465176, 19544084, 51167077, 133957148, 350704366, 918155951, 2403763488, 6293134512, 16475640049, 43133785636, 112925716858, 295643364939, 774004377960

Offset

0,3

Comments

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.

Links

Table of n, a(n) for n=0..29.

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

Formula

a(n) = floor((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.: x*(1 +x - x^2)/((1-x) *(1+x+x^2) *(1-3*x+x^2)).

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

Example

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, ...).

Mathematica

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}, {0, 1, 4, 10, 27}, 30]

Crossrefs

Cf. A000045, A387779, A387780, A387781.

Sequence in context: A192879 A077923 A183325 * A052982 A108672 A000495

Adjacent sequences: A387775 A387776 A387777 * A387779 A387780 A387781

Keyword

nonn,easy

Author

Clark Kimberling, Sep 08 2025

Status

approved