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