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%I #24 Aug 08 2022 16:08:57
%S 1,2,3,6,9,15,25,40,65,106,171,277,449,726,1175,1902,3077,4979,8057,
%T 13036,21093,34130,55223,89353,144577,233930,378507,612438,990945,
%U 1603383,2594329,4197712,6792041,10989754,17781795,28771549,46553345,75324894,121878239
%N Lower 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) = floor((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 + x - x^4)/(1 - x - x^2 - x^3 + x^4 + x^5).
%F G.f.: ((1 + x - x^4)/((-1 + x) (-1 + x + x^2) (1 + x + x^2))).
%F 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
%e a(0) = 1 = floor((1+1)/2);
%e a(1) = 2 = floor((2+3)/2);
%e a(2) = 3 = floor((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 *)
%o (Python)
%o from sympy import fibonacci, lucas
%o def A355324(n): return fibonacci(n+2)+lucas(n+1)>>1 # _Chai Wah Wu_, Aug 08 2022
%Y Cf. A000032, A000045, A116470, A355325.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Jul 16 2022