%I #9 Oct 20 2024 15:32:55
%S 0,1,1,1,2,3,5,8,13,20,33,53,86,140,226,366,592,958,1550,2509,4059,
%T 6568,10627,17194,27821,45015,72836,117851,190687,308537,499224,
%U 807761,1306985,2114747,3421732,5536479,8958211,14494690,23452901,37947592,61400493
%N a(n) is the integer k that minimizes |k/Fibonacci(n) - 3/5|.
%H Clark Kimberling, <a href="/A293644/b293644.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (1, 2, -1, -2, 1, 2, -1, -2, 1, 1)
%F G.f.: -(((-1 + x)^2 x (1 + x)^2 (1 + x^4))/((-1 + x + x^2) (1 - x^2 + x^4 - x^6 + x^8))).
%F a(n) = a(n-1) + 2 a(n-2) - a(n-3) - 2 a(n-4) + a(n-5) + 2 a(n-6) - a(n-7) - 2 a(n-8) + a(n-9) + a(n-10) for n >= 11.
%F a(n) = floor(1/2 + 3*Fibonacci(n)/5).
%F a(n) = A293642(n) if (fractional part of 3*Fibonacci(n)/5) < 1/2, otherwise a(n) = A293643(n).
%t z = 120; r = 3/5; f[n_] := Fibonacci[n];
%t Table[Floor[r*f[n]], {n, 0, z}]; (* A293642 *)
%t Table[Ceiling[r*f[n]], {n, 0, z}]; (* A293643 *)
%t Table[Round[r*f[n]], {n, 0, z}]; (* A293644 *)
%t LinearRecurrence[{1,2,-1,-2,1,2,-1,-2,1,1},{0,1,1,1,2,3,5,8,13,20},50] (* _Harvey P. Dale_, Oct 20 2024 *)
%Y Cf. A000045, A293642, A293643.
%K nonn,easy
%O 0,5
%A _Clark Kimberling_, Oct 14 2017