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A298342 a(n) = a(n-1) + a(n-2) + a([2n/3]), where a(0) = 1, a(1) = 1, a(2) = 1.

%I A298342 #4 Feb 09 2018 21:50:11
%S A298342 1,1,1,3,5,11,21,37,69,127,217,381,667,1117,1911,3245,5373,8999,15039,
%T A298342 24705,40861,67477,110249,180971,296593,482937,788529,1286505,2090073,
%U A298342 3401283,5532217,8974361,14574055,23658665,38342969,62182605,100822167,163301365
%N A298342 a(n) = a(n-1) + a(n-2) + a([2n/3]), where a(0) = 1, a(1) = 1, a(2) = 1.
%C A298342 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 A298342 Clark Kimberling, <a href="/A298342/b298342.txt">Table of n, a(n) for n = 0..1000</a>
%t A298342 a[0] = 1; a[1] = 1; a[2] = 1;
%t A298342 a[n_] := a[n] = a[n - 1] + a[n - 2] + a[Floor[2n/3]];
%t A298342 Table[a[n], {n, 0, 30}]  (* A298342 *)
%Y A298342 Cf. A001622, A000045, A298338.
%K A298342 nonn,easy
%O A298342 0,4
%A A298342 _Clark Kimberling_, Feb 09 2018