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A298408 a(n) = 2*a(n-1) - a(n-3) + 2*a(floor(n/2)) + 3*a(floor(n/3)) + ... + n*a(floor(n/n)), where a(0) = 1, a(1) = 1, a(2) = 1.

%I A298408 #10 Mar 31 2021 19:10:02
%S A298408 1,1,1,6,20,53,130,277,574,1115,2126,3862,7021,12341,21553,36957,
%T A298408 63111,106224,178407,296638,492231,811731,1335994,2188950,3583027,
%U A298408 5847108,9532980,15512342,25226123,40967842,66506422,107869832,174908573,283452771,459264017
%N A298408 a(n) = 2*a(n-1) - a(n-3) + 2*a(floor(n/2)) + 3*a(floor(n/3)) + ... + n*a(floor(n/n)), where a(0) = 1, a(1) = 1, a(2) = 1.
%C A298408 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 A298408 Clark Kimberling, <a href="/A298408/b298408.txt">Table of n, a(n) for n = 0..1000</a>
%t A298408 a[0] = 1; a[1] = 1; a[2] = 1;
%t A298408 a[n_] := a[n] = 2*a[n - 1] - a[n - 3] + Sum[k*a[Floor[n/k]], {k, 2, n}];
%t A298408 Table[a[n], {n, 0, 90}]  (* A298408  *)
%o A298408 (Python)
%o A298408 from functools import lru_cache
%o A298408 @lru_cache(maxsize=None)
%o A298408 def A298408(n):
%o A298408     if n <= 2:
%o A298408         return 1
%o A298408     c, j = 2*A298408(n-1)-A298408(n-3), 2
%o A298408     k1 = n//j
%o A298408     while k1 > 1:
%o A298408         j2 = n//k1 + 1
%o A298408         c += (j2*(j2-1)-j*(j-1))*A298408(k1)//2
%o A298408         j, k1 = j2, n//j2
%o A298408     return c+(n*(n+1)-j*(j-1))//2 # _Chai Wah Wu_, Mar 31 2021
%Y A298408 Cf. A001622, A000045, A298338, A298409.
%K A298408 nonn,easy
%O A298408 0,4
%A A298408 _Clark Kimberling_, Feb 10 2018