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a(1) = 1; a(n+1) = -Sum_{k=1..n} (-1)^k * a(k) * floor(n/k).
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%I #6 May 21 2026 07:52:52

%S 1,1,1,3,0,1,1,3,-3,-4,0,1,-3,-5,1,4,-6,-11,-3,-5,1,5,0,1,-7,-13,-3,

%T -7,3,7,5,11,-10,-18,-6,-10,4,9,-3,-7,3,7,7,15,-8,-16,0,1,-12,-22,-3,

%U -10,7,15,5,11,7,13,3,7,-1,-1,5,11,-21,-44,-15,-29,9,20,4,9,10,21,4,4,-15,-28,7,15

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

%H Ilya Gutkovskiy, <a href="/A396228/a396228.jpg">Scatterplot of the second partial sums of A396228</a>

%F a(1) = 1, a(n) = -Sum_{k=1..n-1} Sum_{d|k} (-1)^d * a(d).

%F G.f.: x - ( x / (1 - x) ) * Sum_{n>=1} a(n) * (-x)^n / (1 - x^n).

%t a[1] = 1; a[n_] := a[n] = -Sum[(-1)^k a[k] Floor[(n - 1)/k], {k, 1, n - 1}]; Table[a[n], {n, 1, 80}]

%Y Cf. A014668, A024919, A059851, A307778, A359479, A396227.

%K sign

%O 1,4

%A _Ilya Gutkovskiy_, May 19 2026