A298406 - OEIS

%I #14 Mar 31 2021 19:18:15

%S 1,1,1,3,8,19,42,84,163,301,547,961,1682,2879,4902,8241,13807,22917,

%T 37962,62487,102690,168096,274798,448000,729829,1186797,1928729,

%U 3130905,5080360,8237339,13352743,21634097,35045477,56753250,91896553,148771833,240830555

%N 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 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="/A298406/b298406.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] = 2*a[n - 1] - a[n - 3] + Sum[a[Floor[n/k]], {k, 2, n}];

%t Table[a[n], {n, 0, 90}]  (* A298406  *)

%o (Python)

%o from functools import lru_cache

%o @lru_cache(maxsize=None)

%o def A298406(n):

%o     if n <= 2:

%o         return 1

%o     c, j = 2*A298406(n-1)-A298406(n-3), 2

%o     k1 = n//j

%o     while k1 > 1:

%o         j2 = n//k1 + 1

%o         c += (j2-j)*A298406(k1)

%o         j, k1 = j2, n//j2

%o     return c+n-j+1 # _Chai Wah Wu_, Mar 31 2021

%Y Cf. A001622, A000045, A298338, A298407.

%K nonn,easy

%O 0,4

%A _Clark Kimberling_, Feb 10 2018