A387781 - OEIS

A387781

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

2, 5, 12, 32, 83, 216, 566, 1481, 3876, 10148, 26567, 69552, 182090, 476717, 1248060, 3267464, 8554331, 22395528, 58632254, 153501233, 401871444, 1052113100, 2754467855, 7211290464, 18879403538, 49426920149, 129401356908, 338777150576, 886930094819

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OFFSET

0,1

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..28.

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

FORMULA

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

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

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

6*a(n) = 9*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((3/2)(0+1, 1+2, 3+5, 8+13, 21+34, ...)) = (1, 4, 12, 31, 82, ...).

v(n) = ceiling((3/2)(0+1, 1+2, 3+5, 8+13, 21+34, ...)) = (2, 5, 12, 32, 83, ...).