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a(n) = a(n-1) + a(n-2) + a([(n+1)/2]), where a(0) = 1, a(1) = 1, a(2) = 1.
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%I #4 Feb 10 2018 09:49:43

%S 1,1,1,3,5,11,19,35,59,105,175,299,493,827,1355,2241,3655,6001,9761,

%T 15937,25873,42109,68281,110883,179657,291367,471851,764573,1237779,

%U 2004593,3244613,5252861,8501129,13759991,22267121,36036873,58313755,94366565,152696257

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

%C 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 Clark Kimberling, <a href="/A298348/b298348.txt">Table of n, a(n) for n = 0..1000</a>

%t a[0] = 1; a[1] = 1; a[2] = 1;

%t a[n_] := a[n] = a[n - 1] + a[n - 2] + a[Floor[(n+1)/2]];

%t Table[a[n], {n, 0, 30}] (* A298348 *)

%Y Cf. A001622, A000045, A298338.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Feb 10 2018