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A298406 a(n) = 2*a(n-1) - a(n-3) + a(floor(n/2)) + a(floor(n/3)) + ... + a(floor(n/n)), where a(0) = 1, a(1) = 1, a(2) = 1.

%I A298406 #14 Mar 31 2021 19:18:15
%S A298406 1,1,1,3,8,19,42,84,163,301,547,961,1682,2879,4902,8241,13807,22917,
%T A298406 37962,62487,102690,168096,274798,448000,729829,1186797,1928729,
%U A298406 3130905,5080360,8237339,13352743,21634097,35045477,56753250,91896553,148771833,240830555
%N A298406 a(n) = 2*a(n-1) - a(n-3) + a(floor(n/2)) + a(floor(n/3)) + ... + a(floor(n/n)), where a(0) = 1, a(1) = 1, a(2) = 1.
%C A298406 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 A298406 Clark Kimberling, <a href="/A298406/b298406.txt">Table of n, a(n) for n = 0..1000</a>
%t A298406 a[0] = 1; a[1] = 1; a[2] = 1;
%t A298406 a[n_] := a[n] = 2*a[n - 1] - a[n - 3] + Sum[a[Floor[n/k]], {k, 2, n}];
%t A298406 Table[a[n], {n, 0, 90}]  (* A298406  *)
%o A298406 (Python)
%o A298406 from functools import lru_cache
%o A298406 @lru_cache(maxsize=None)
%o A298406 def A298406(n):
%o A298406     if n <= 2:
%o A298406         return 1
%o A298406     c, j = 2*A298406(n-1)-A298406(n-3), 2
%o A298406     k1 = n//j
%o A298406     while k1 > 1:
%o A298406         j2 = n//k1 + 1
%o A298406         c += (j2-j)*A298406(k1)
%o A298406         j, k1 = j2, n//j2
%o A298406     return c+n-j+1 # _Chai Wah Wu_, Mar 31 2021
%Y A298406 Cf. A001622, A000045, A298338, A298407.
%K A298406 nonn,easy
%O A298406 0,4
%A A298406 _Clark Kimberling_, Feb 10 2018