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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.
2
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
0,2
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).
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
EXAMPLE
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
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jul 16 2022
STATUS
approved