A387782 Lower (1/3)-midsequence of (F(2n)) and (F(2n+1)), where F=A000045 (Fibonacci numbers); see Comments.

0, 1, 2, 7, 18, 48, 125, 329, 861, 2255, 5903, 15456, 40464, 105937, 277346, 726103, 1900962, 4976784, 13029389, 34111385, 89304765, 233802911, 612103967, 1602508992, 4195423008, 10983760033, 28755857090, 75283811239, 197095576626, 516002918640

Offset

1,3

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=1..30.

Index entries for linear recurrences with constant coefficients, signature (3,-1,0,1,-3,1).

Formula

a(n) = floor((1/3)*(F(2*n) + F(2*n+1))) = floor((1/3)*F(2*n+2)), where F=A000045.

a(n) = 3*a(n-1) - a(n-2) + a(n-4) - 3*a(n-5) +a(n-6).

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

a(n) = -1/4 -(-1)^n/4 +A001906(n+1)/3 +A056594(n)/6. - R. J. Mathar, Sep 26 2025

Example

s = (F(2n)) = (0, 1, 3, 8, 21, ...); t = (1, 2, 5, 13, 34, ...).
u(n) = floor((1/3)(0+1, 1+2, 3+5, 8+13, 21+34, ...)) = (0, 1, 2, 7, 18, ...).
v(n) = ceiling((1/3)(0+1, 1+2, 3+5, 8+13, 21+34, ...)) = (1, 1, 3, 7, 19, ...).

Mathematica

f[n_] := Fibonacci[n]; s[n_] := f[2 n]; t[n_] := f[2 n + 1]; r = 1/3;
u[n_] := Floor[r*(s[n] + t[n])]
v[n_] := Ceiling[r*(s[n] + t[n])]
Table[u[n], {n, 0, 30}]  (* lower, A387782 *)
Table[v[n], {n, 0, 30}]  (* upper, A387783 *)
(* Also *)
LinearRecurrence[{3, -1, 0, 1, -3, 1}, {0, 1, 2, 7, 18, 48}, 30] (* Corrected by Georg Fischer, Sep 25 2025 *)

Crossrefs

Cf. A000045, A387778, A387780, A387783.

Sequence in context: A161870 A362126 A072338 * A182197 A022726 A192873

Adjacent sequences: A387779 A387780 A387781 * A387783 A387784 A387785

Keyword

nonn,easy

Author

Clark Kimberling, Sep 15 2025

Status

approved