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

%I #16 Jun 07 2021 09:09:12

%S 1,4,11,32,87,258,745,2224,6605,19784,59151,177438,531733,1595104,

%T 4783811,14351228,43049043,129147030,387427357,1162281532,3486804959,

%U 10460413130,31381119537,94143358500,282429716209,847289143468,2541866366735,7625599086782

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

%H Seiichi Manyama, <a href="/A345029/b345029.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 A345032.

%Y Cf. A003462, A345035.

%K nonn

%O 1,2

%A _Seiichi Manyama_, Jun 06 2021