A390559 Lower (1/2)-midsequence of (2^n) and F(n), where F = A000045 (Fibonacci numbers); see Comments.

0, 1, 2, 5, 9, 18, 36, 70, 138, 273, 539, 1068, 2120, 4212, 8380, 16689, 33261, 66334, 132364, 264234, 527670, 1054049, 2106007, 4208632, 8411792, 16814728, 33615128, 67207073, 134376633, 268692570, 537286932, 1074414958, 2148572802, 4296729585, 8592786035

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..34.

Index entries for linear recurrences with constant coefficients, signature (3,-1,-1,-3,1,2).

Formula

a(n) = 3*a(n-1) - a(n-2) - a(n-3) - 3*a(n-4) + a(n-5) + 2*a(n-6) for n>=7, with (a(0),...,a(6)) = (0, 1, 2, 5, 9, 18, 36).

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

E.g.f.: (exp(-x/2)*cos(sqrt(3)*x/2) - exp(x))/3 + exp(x)*sinh(x) + exp(x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Stefano Spezia, Dec 27 2025

Example

s = A000079 = (1, 2, 4, 8, 16, 32, 64, ...).
t = A000045 = (0, 1, 1, 2, 3, 5, 8, 13, ...).
u(n) = (0, 1, 2, 5, 9, 18, 36, 70, 138, ...).
v(n) = (1, 3, 5, 10, 19, 36, 71, 139,, ...).

Mathematica

s[n_] := 2^n ; t[n_] := Fibonacci[n]; r = 1/2;
u[n_] := Floor[r*(s[n] + t[n])]
v[n_] := Ceiling[r*(s[n] + t[n])]
Table[u[n], {n, 0, 60}]  (* A390559 *)
Table[v[n], {n, 0, 60}]  (* A390560 *)

Crossrefs

Cf. A000045, A000079, A390560.

Sequence in context: A097281 A321408 A289976 * A068036 A364525 A077947

Adjacent sequences: A390556 A390557 A390558 * A390560 A390561 A390562

Keyword

nonn,easy

Author

Clark Kimberling, Nov 11 2025

Status

approved