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a(1) = 1; a(n) = Sum_{k=2..n} k * a(floor(n/k)).
1

%I #16 Apr 01 2023 11:23:52

%S 1,2,5,11,16,31,38,62,80,105,116,194,207,242,287,383,400,526,545,675,

%T 738,793,816,1200,1250,1315,1423,1605,1634,1979,2010,2394,2493,2578,

%U 2683,3475,3512,3607,3724,4364,4405,4888,4931,5217,5577,5692,5739,7563,7661,8011

%N a(1) = 1; a(n) = Sum_{k=2..n} k * a(floor(n/k)).

%H Seiichi Manyama, <a href="/A359399/b359399.txt">Table of n, a(n) for n = 1..10000</a>

%F G.f. A(x) satisfies A(x) = x + (1/(1 - x)) * Sum_{k>=2} k * (1 - x^k) * A(x^k).

%o (Python)

%o from functools import lru_cache

%o @lru_cache(maxsize=None)

%o def A359399(n):

%o if n <= 1:

%o return 1

%o c, j = 0, 2

%o k1 = n//j

%o while k1 > 1:

%o j2 = n//k1 + 1

%o c += (j2*(j2-1)-j*(j-1)>>1)*A359399(k1)

%o j, k1 = j2, n//j2

%o return c+(n*(n+1)-(j-1)*j>>1) # _Chai Wah Wu_, Mar 31 2023

%Y Cf. A022825.

%K nonn

%O 1,2

%A _Seiichi Manyama_, Mar 31 2023