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A328967 a(n+1) = 1 - Sum_{k=1..n} a(floor(n/k)).

%I A328967 #12 Mar 31 2021 14:48:34
%S A328967 1,0,0,-1,0,-2,0,-3,1,-4,0,-5,3,-6,1,-8,5,-9,3,-10,8,-13,4,-14,15,-16,
%T A328967 7,-21,15,-22,14,-23,27,-28,14,-32,32,-33,20,-42,41,-43,32,-44,50,-57,
%U A328967 33,-58,71,-61,46,-75,70,-76,59,-82,98,-95,62,-96,117,-97,81,-122,131
%N A328967 a(n+1) = 1 - Sum_{k=1..n} a(floor(n/k)).
%F A328967 G.f. A(x) satisfies: A(x) = (x/(1 - x)) * (1 - Sum_{k>=1} (1 - x^k) * A(x^k)).
%t A328967 a[n_] := a[n] = 1 - Sum[a[Floor[(n - 1)/k]], {k, 1, n - 1}]; Table[a[n], {n, 1, 65}]
%t A328967 A281487[1] = 1; A281487[n_] := A281487[n] = -Sum[A281487[d], {d, Divisors[n - 1]}]; Table[A281487[n], {n, 1, 65}] // Accumulate
%o A328967 (Python)
%o A328967 from functools import lru_cache
%o A328967 @lru_cache(maxsize=None)
%o A328967 def A328967(n):
%o A328967     if n == 0:
%o A328967         return 1
%o A328967     c, j = n-1, 1
%o A328967     k1 = (n-1)//j
%o A328967     while k1 > 1:
%o A328967         j2 = (n-1)//k1 + 1
%o A328967         c += (j2-j)*A328967(k1)
%o A328967         j, k1 = j2, (n-1)//j2
%o A328967     return j-c # _Chai Wah Wu_, Mar 31 2021
%Y A328967 Cf. A003318, A281487 (first differences).
%K A328967 sign
%O A328967 1,6
%A A328967 _Ilya Gutkovskiy_, Feb 25 2020