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Upper midsequence of the Fibonacci numbers (1,2,3,5,8,...) and Lucas numbers (1,3,4,7,11,...); see Comments.
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%I #16 Jul 22 2022 20:52:17

%S 1,3,4,6,10,16,25,41,66,106,172,278,449,727,1176,1902,3078,4980,8057,

%T 13037,21094,34130,55224,89354,144577,233931,378508,612438,990946,

%U 1603384,2594329,4197713,6792042,10989754,17781796,28771550,46553345,75324895,121878240

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

%C 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).

%C 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).

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,1,-1,-1).

%F a(n) = ceiling((A000045(n+2) + A000032(n+1))/2).

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

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

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

%F 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

%e a(0) = 1 = ceiling((1+1)/2);

%e a(1) = 3 = ceiling((2+3)/2);

%e a(2) = 4 = ceiling((3+4)/2).

%e The Fibonacci and Lucas numbers are interspersed:

%e 1 < 2 < 3 < 4 < 5 < 7 < 8 < 11 < 13 < 18 < 21 < 29 < ...

%e The midsequences m and M intersperse the ordered union of the Fibonacci and Lucas sequences, A116470, as indicated by the following table:

%e F m M L

%e 1 1 1 1

%e 2 2 3 3

%e 3 3 4 4

%e 5 6 6 7

%e 8 9 10 11

%e 13 15 16 18

%e 21 25 25 29

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

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

%Y Cf. A000032, A000045, A116470, A355324.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Jul 16 2022