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A298339 a(n) = a(n-1) + a(n-2) + a([n/2]), where a(0) = 1, a(1) = 2, a(2) = 3.

%I A298339 #4 Feb 09 2018 21:49:51
%S A298339 1,2,3,7,13,23,43,73,129,215,367,605,1015,1663,2751,4487,7367,11983,
%T A298339 19565,31763,51695,83825,136125,220555,357695,579265,938623,1519551,
%U A298339 2460925,3983227,6448639,10436353,16892359,27336079,44240421,71588483,115848469
%N A298339 a(n) = a(n-1) + a(n-2) + a([n/2]), where a(0) = 1, a(1) = 2, a(2) = 3.
%C A298339 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 A298339 Clark Kimberling, <a href="/A298339/b298339.txt">Table of n, a(n) for n = 0..1000</a>
%t A298339 a[0] = 1; a[1] = 2; a[2] = 3;
%t A298339 a[n_] := a[n] = a[n - 1] + a[n - 2] + a[Floor[n/2]];
%t A298339 Table[a[n], {n, 0, 30}]  (* A298339 *)
%Y A298339 Cf. A001622, A000045, A298338.
%K A298339 nonn,easy
%O A298339 0,2
%A A298339 _Clark Kimberling_, Feb 09 2018