%I #8 Mar 31 2021 17:25:01
%S 1,1,1,4,8,16,32,57,103,178,308,514,874,1441,2394,3926,6462,10531,
%T 17231,28001,45614,74026,120258,194903,316210,512171,830007,1343883,
%U 2176578,3523150,5704107,9231637,14942711,24181525,39135483,63328289,102482212,165828942
%N a(n) = a(n-1) + a(n-2) + a([n/2]) + a([n/3]) + ... + a([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="/A298356/b298356.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] + Sum[a[Floor[n/k]], {k, 2, n}];
%t Table[a[n], {n, 0, 30}] (* A298356 *)
%o (Python)
%o from functools import lru_cache
%o @lru_cache(maxsize=None)
%o def A298356(n):
%o if n <= 2:
%o return 1
%o c, j = A298356(n-1)+A298356(n-2), 2
%o k1 = n//j
%o while k1 > 1:
%o j2 = n//k1 + 1
%o c += (j2-j)*A298356(k1)
%o j, k1 = j2, n//j2
%o return c+n-j+1 # _Chai Wah Wu_, Mar 31 2021
%Y Cf. A001622, A000045, A298338.
%K nonn,easy
%O 0,4
%A _Clark Kimberling_, Feb 10 2018