cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A355325 Upper midsequence of the Fibonacci numbers (1,2,3,5,8,...) and Lucas numbers (1,3,4,7,11,...); see Comments.

Original entry on oeis.org

1, 3, 4, 6, 10, 16, 25, 41, 66, 106, 172, 278, 449, 727, 1176, 1902, 3078, 4980, 8057, 13037, 21094, 34130, 55224, 89354, 144577, 233931, 378508, 612438, 990946, 1603384, 2594329, 4197713, 6792042, 10989754, 17781796, 28771550, 46553345, 75324895, 121878240
Offset: 0

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Author

Clark Kimberling, Jul 16 2022

Keywords

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 = ceiling((1+1)/2);
a(1) = 3 = ceiling((2+3)/2);
a(2) = 4 = ceiling((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
		

Crossrefs

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 *)

Formula

a(n) = ceiling((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 + 2 x - 2 x^3 - 2 x^4)/(1 - x - x^2 - x^3 + x^4 + x^5).
G.f.: ((1 + 2 x - 2 x^3 - 2 x^4)/((-1 + x) (-1 + x + x^2) (1 + x + x^2))).
a(n) = (10 + 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))/30. - Stefano Spezia, Jul 17 2022
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