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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.
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%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