A387785 Upper (2/3)-midsequence of (F(2n)) and (F(2n+1)), where F=A000045 (Fibonacci numbers); see Comments.

1, 2, 6, 14, 37, 96, 252, 658, 1723, 4510, 11808, 30912, 80929, 211874, 554694, 1452206, 3801925, 9953568, 26058780, 68222770, 178609531, 467605822, 1224207936, 3205017984, 8390846017, 21967520066, 57511714182, 150567622478, 394191153253, 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) = ceiling((2/3)*(F(2*n) + F(2*n+1))) = ceiling((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.: (-1 + x - x^2 + 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}, {1, 2, 6, 14, 37, 96}, 30]

Crossrefs

Cf. A000045, A387778, A387782, A387783, A387784.

Sequence in context: A026598 A006864 A217420 * A071636 A263758 A100067

Adjacent sequences: A387782 A387783 A387784 * A387786 A387787 A387788

Keyword

nonn,easy

Author

Clark Kimberling, Sep 20 2025

Status

approved