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a(n) = 2*a(n-1) - a(n-3) + a(floor(n/2)), where a(0) = 1, a(1) = 1, a(2) = 1.
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%I #6 Feb 10 2018 17:52:30

%S 1,1,1,2,4,8,16,30,56,100,178,308,532,902,1526,2550,4254,7038,11626,

%T 19098,31336,51224,83658,136288,221884,360642,585898,950814,1542512,

%U 2500652,4053040,6566118,10635838,17222890,27886700,45144600,73077936,118280798

%N a(n) = 2*a(n-1) - a(n-3) + a(floor(n/2)), where a(0) = 1, a(1) = 1, a(2) = 1.

%C a(n)/a(n-1) -> (1 + sqrt(5))/2, the golden ratio (A001622), so that (a(n)) has the growth rate of the Fibonacci numbers (A000045). See A298338 for a guide to related sequences.

%H Clark Kimberling, <a href="/A298402/b298402.txt">Table of n, a(n) for n = 0..1000</a>

%t a[0] = 1; a[1] = 1; a[2] = 1;

%t a[n_] := a[n] = 2*a[n - 1] - a[n - 3] + a[Floor[n/2]];

%t Table[a[n], {n, 0, 90}] (* A298402 *)

%Y Cf. A001622, A000045, A298338, A298403.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Feb 10 2018