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a(n) = Sum_{k=1..n} (-3)^(floor(n/k) - 1).
3

%I #15 Jun 07 2021 09:10:07

%S 1,-2,11,-28,81,-234,739,-2216,6545,-19594,59139,-177408,531181,

%T -1593614,4783799,-14351032,43044597,-129133854,387426799,-1162281332,

%U 3486765521,-10460293354,31381119459,-94143358440,282429356977,-847288080362,2541866366171

%N a(n) = Sum_{k=1..n} (-3)^(floor(n/k) - 1).

%H Seiichi Manyama, <a href="/A345035/b345035.txt">Table of n, a(n) for n = 1..2000</a>

%F G.f.: (1/(1 - x)) * Sum_{k>=1} x^k * (1 - x^k)/(1 + 3*x^k).

%t a[n_] := Sum[(-3)^(Floor[n/k] - 1), {k, 1, n}]; Array[a, 30] (* _Amiram Eldar_, Jun 06 2021 *)

%o (PARI) a(n) = sum(k=1, n, (-3)^(n\k-1));

%o (PARI) my(N=40, x='x+O('x^N)); Vec(sum(k=1, N, x^k*(1-x^k)/(1+3*x^k))/(1-x))

%Y Column k=3 of A345033.

%Y Cf. A014983, A345029.

%K sign

%O 1,2

%A _Seiichi Manyama_, Jun 06 2021