A387784 Lower (2/3)-midsequence of (F(2n)) and (F(2n+1)), where F=A000045 (Fibonacci numbers); see Comments.

0, 2, 5, 14, 36, 96, 251, 658, 1722, 4510, 11807, 30912, 80928, 211874, 554693, 1452206, 3801924, 9953568, 26058779, 68222770, 178609530, 467605822, 1224207935, 3205017984, 8390846016, 21967520066, 57511714181, 150567622478, 394191153252, 1032005837280

Offset

0,2

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,-1,0,1,-3,1).

Formula

a(n) = floor((2/3)*(F(2*n) + F(2*n+1))) = floor((2/3)*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). [Corrected by Georg Fischer, Sep 25 2025]

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

a(n) = -1/4 -(-1)^n/4 +2*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((2/3)(0+1, 1+2, 3+5, 8+13, 21+34, ...)) = (0, 2, 5, 14, 36, ...).
v(n) = ceiling((2/3)(0+1, 1+2, 3+5, 8+13, 21+34, ...)) = (1, 2, 6, 14, 37, ...).

Mathematica

f[n_] := Fibonacci[n]; s[n_] := f[2 n]; t[n_] := f[2 n + 1]; r = 2/3;
u[n_] := Floor[r*(s[n] + t[n])]
v[n_] := Ceiling[r*(s[n] + t[n])]
Table[u[n], {n, 0, 30}]  (* lower, A387784 *)
Table[v[n], {n, 0, 30}]  (* upper, A387785 *)
(* Also *)
LinearRecurrence[{3, -1, 0, 1, -3, 1}, {0, 2, 5, 14, 36, 96}, 30]

Crossrefs

Cf. A000045, A387778, A387782, A387785.

Sequence in context: A299167 A182890 A102714 * A087223 A363106 A005955

Adjacent sequences: A387781 A387782 A387783 * A387785 A387786 A387787

Keyword

nonn,easy

Author

Clark Kimberling, Sep 19 2025

Status

approved