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A298340 a(n) = a(n-1) + a(n-2) + a([n/3]), where a(0) = 1, a(1) = 1, a(2) = 1.

%I A298340 #4 Feb 09 2018 21:49:58
%S A298340 1,1,1,3,5,9,15,25,41,69,113,185,303,493,801,1303,2113,3425,5553,8993,
%T A298340 14561,23579,38165,61769,99975,161785,261801,423655,685525,1109249,
%U A298340 1794887,2904249,4699249,7603683,12303117,19906985,32210405,52117693,84328401
%N A298340 a(n) = a(n-1) + a(n-2) + a([n/3]), where a(0) = 1, a(1) = 1, a(2) = 1.
%C A298340 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 A298340 Clark Kimberling, <a href="/A298340/b298340.txt">Table of n, a(n) for n = 0..999</a>
%t A298340 a[0] = 1; a[1] = 1; a[2] = 1;
%t A298340 a[n_] := a[n] = a[n - 1] + a[n - 2] + a[Floor[n/3]];
%t A298340 Table[a[n], {n, 0, 30}]  (* A298340 *)
%Y A298340 Cf. A001622, A000045, A298338.
%K A298340 nonn,easy
%O A298340 0,4
%A A298340 _Clark Kimberling_, Feb 09 2018