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%I #10 Mar 31 2021 19:10:16
%S 1,2,3,15,38,83,190,356,695,1254,2267,3861,6829,11417,19340,32076,
%T 53545,87784,145048,236589,387765,631106,1028866,1670013,2716595,
%U 4404599,7148426,11582096,18776334,30404300,49256015,79735758,129111774,208972513,338277831
%N a(n) = a(n-1) + a(n-2) + 2 a(floor(n/2)) + 3 a(floor(n/3)) + ... + n a(floor(n/n)), where a(0) = 1, a(1) = 2, a(2) = 3.
%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="/A298370/b298370.txt">Table of n, a(n) for n = 0..1000</a>
%t a[0] = 1; a[1] = 2; a[2] = 3;
%t a[n_] := a[n] = a[n - 1] + a[n - 2] + Sum[k*a[Floor[n/k]], {k, 2, n}];
%t Table[a[n], {n, 0, 30}] (* A298370 *)
%o (Python)
%o from functools import lru_cache
%o @lru_cache(maxsize=None)
%o def A298370(n):
%o if n <= 2:
%o return n+1
%o c, j = A298370(n-1)+A298370(n-2), 2
%o k1 = n//j
%o while k1 > 1:
%o j2 = n//k1 + 1
%o c += (j2*(j2-1)-j*(j-1))*A298370(k1)//2
%o j, k1 = j2, n//j2
%o return c+2*(n*(n+1)-j*(j-1))//2 # _Chai Wah Wu_, Mar 31 2021
%Y Cf. A001622, A000045, A298338.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Feb 10 2018