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A298345 a(n) = a(n-1) + a(n-2) + a([n/3]) + a([2n/3]), where a(0) = 1, a(1) = 2, a(2) = 3.

%I A298345 #4 Feb 09 2018 21:50:29
%S A298345 1,2,3,10,18,40,79,140,262,491,842,1483,2605,4368,7482,12732,21096,
%T A298345 35351,59131,97166,160744,265532,433898,712302,1167558,1901218,
%U A298345 3104389,5065229,8229240,13392126,21782952,35336664,57385990,93158035,150979406,244851226
%N A298345 a(n) = a(n-1) + a(n-2) + a([n/3]) + a([2n/3]), where a(0) = 1, a(1) = 2, a(2) = 3.
%C A298345 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 A298345 Clark Kimberling, <a href="/A298345/b298345.txt">Table of n, a(n) for n = 0..1000</a>
%t A298345 a[0] = 1; a[1] = 2; a[2] = 3;
%t A298345 a[n_] := a[n] = a[n - 1] + a[n - 2] + a[Floor[n/3]] + a[Floor[2n/3]];
%t A298345 Table[a[n], {n, 0, 30}]  (* A298345 *)
%Y A298345 Cf. A001622, A000045, A298338.
%K A298345 nonn,easy
%O A298345 0,2
%A A298345 _Clark Kimberling_, Feb 09 2018