A298370 - OEIS

%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