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A298353 a(n) = a(n-1) + a(n-2) + a([(n-1)/2]), where a(0) = 1, a(1) = 2, a(2) = 3.

%I A298353 #4 Feb 10 2018 09:50:12
%S A298353 1,2,3,7,12,22,37,66,110,188,310,520,852,1409,2298,3773,6137,10020,
%T A298353 16267,26475,42930,69715,112955,183190,296665,480707,778224,1260340,
%U A298353 2039973,3302611,5344882,8651266,13999921,22657324,36663382,59330726,96004128,155351121
%N A298353 a(n) = a(n-1) + a(n-2) + a([(n-1)/2]), where a(0) = 1, a(1) = 2, a(2) = 3.
%C A298353 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 A298353 Clark Kimberling, <a href="/A298353/b298353.txt">Table of n, a(n) for n = 0..1000</a>
%t A298353 a[0] = 1; a[1] = 2; a[2] = 3;
%t A298353 a[n_] := a[n] = a[n - 1] + a[n - 2] + a[Floor[(n-1)/2]];
%t A298353 Table[a[n], {n, 0, 30}]  (* A298353 *)
%Y A298353 Cf. A001622, A000045, A298338.
%K A298353 nonn,easy
%O A298353 0,2
%A A298353 _Clark Kimberling_, Feb 10 2018