A355324 Lower midsequence of the Fibonacci numbers (1,2,3,5,8,...) and Lucas numbers (1,3,4,7,11,...); see Comments.

1, 2, 3, 6, 9, 15, 25, 40, 65, 106, 171, 277, 449, 726, 1175, 1902, 3077, 4979, 8057, 13036, 21093, 34130, 55223, 89353, 144577, 233930, 378507, 612438, 990945, 1603383, 2594329, 4197712, 6792041, 10989754, 17781795, 28771549, 46553345, 75324894, 121878239

Offset: 0

Author

Clark Kimberling, Jul 16 2022

Keywords

• nonn
• easy

Comments

Suppose that s = (s(n)) and t = (t(n)) are integer sequences. The lower midsequence, m = m(s,t), of s and t is defined by m(n) = floor((s(n) + t(n))/2). The upper midsequence, M = M(s,t), is defined by M(n) = ceiling((s(n) + t(n))/2).

Here, s(n) = F(n+2) and t(n) = L(n+1), for n >= 0, where F = A000045 (Fibonacci numbers) and L = A000032 (Lucas numbers).

Examples

			a(0) = 1 = floor((1+1)/2);
a(1) = 2 = floor((2+3)/2);
a(2) = 3 = floor((3+4)/2).
The Fibonacci and Lucas numbers are interspersed:
1 < 2 < 3 < 4 < 5 < 7 < 8 < 11 < 13 < 18 < 21 < 29 < ...
The midsequences m and M intersperse the ordered union of the Fibonacci and Lucas sequences, A116470, as indicated by the following table:
   F    m    M    L
   1    1    1    1
   2    2    3    3
   3    3    4    4
   5    6    6    7
   8    9   10   11
  13   15   16   18
  21   25   25   29
		

Links

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

Crossrefs

Cf. A000032, A000045, A116470, A355325.

Programs

Mathematica

Table[Floor[(LucasL[n + 1] + Fibonacci[n + 2])/2], {n, 0, 50}]    (* A355324 *)
Table[Ceiling[(LucasL[n + 1] + Fibonacci[n + 2])/2], {n, 0, 50}]  (* A355325 *)

Python

from sympy import fibonacci, lucas
def A355324(n): return fibonacci(n+2)+lucas(n+1)>>1 # Chai Wah Wu, Aug 08 2022

Formula

a(n) = floor((A000045(n+2) + A000032(n+1))/2).

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

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

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

a(n) = (3*((5 - 4*sqrt(5))*(1 - sqrt(5))^n + (1 + sqrt(5))^n*(5 + 4*sqrt(5)))/2^n + 10*(cos(2*n*Pi/3) - 1))/30. - Stefano Spezia, Jul 17 2022