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a(n) is the least integer k such that k/Fibonacci(n) > 1/5.
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%I #8 Feb 17 2018 20:05:30

%S 0,1,1,1,1,1,2,3,5,7,11,18,29,47,76,122,198,320,517,837,1353,2190,

%T 3543,5732,9274,15005,24279,39284,63563,102846,166408,269254,435662,

%U 704916,1140578,1845493,2986071,4831564,7817634,12649198,20466831,33116029,53582860

%N a(n) is the least integer k such that k/Fibonacci(n) > 1/5.

%H Clark Kimberling, <a href="/A293637/b293637.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_15">Index entries for linear recurrences with constant coefficients</a>, signature (1, 2, -1, -2, 2, 1, -3, -1, 3, 0, -2, 1, 2, -1, -1)

%F G.f.: -((x (1 + x) (-1 + x + x^2 - x^4 + x^5 - x^7 + x^9 - x^10 + x^12))/((-1 + x) (-1 + x + x^2)(1 + x + x^2 + x^3 + x^4) (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) + 2 a(n-5) + a(n-6) - 3 a(n-7) - a(n-8) + 3 a(n-9) - 2 a(n-11) + a(n-12) + 2 a(n-13) - a(n-14) - a(n-15) for n >= 16.

%F a(n) = ceiling(Fibonacci(n)/5).

%F a(n) = A004698(n) + 1 for n > 0.

%t z = 120; r = 1/5; f[n_] := Fibonacci[n];

%t Table[Floor[r*f[n]], {n, 0, z}]; (* A004698 *)

%t Table[Ceiling[r*f[n]], {n, 0, z}]; (* A293637 *)

%t Table[Round[r*f[n]], {n, 0, z}]; (* A293638 *)

%Y Cf. A000045, A004698, A293638.

%K nonn,easy

%O 0,7

%A _Clark Kimberling_, Oct 14 2017