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a(n) = a(n-1) + a(n-2) + 2 a([n/2]), where a(0) = 1, a(1) = 1, a(2) = 1.
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%I #4 Feb 09 2018 21:50:35

%S 1,1,1,4,7,13,28,49,91,154,271,451,778,1285,2161,3544,5887,9613,15808,

%T 25729,42079,68350,111331,180583,293470,475609,771649,1249828,2025799,

%U 3279949,5312836,8599873,13924483,22536130,36479839,59035195,95546650,154613461

%N a(n) = a(n-1) + a(n-2) + 2 a([n/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="/A298346/b298346.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] + 2 a[Floor[n/2]];

%t Table[a[n], {n, 0, 30}] (* A298346 *)

%Y Cf. A001622, A000045, A298338.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Feb 09 2018