login
a(n) = Sum_{k=1..n} floor(n/(2*k-1))^k.
3

%I #13 Dec 17 2021 11:09:58

%S 1,2,4,5,7,11,13,14,21,29,31,39,41,57,87,88,90,133,135,173,253,317,

%T 319,335,398,526,756,932,934,1300,1302,1303,1991,2503,3001,3806,3808,

%U 4832,6918,7088,7090,9836,9838,13206,21860,25956,25958,25990,27097,35560,54766

%N a(n) = Sum_{k=1..n} floor(n/(2*k-1))^k.

%F G.f.: (1/(1 - x)) * Sum_{j>=1} Sum{k>=1} k^j * x^(k*(2*j-1)) * (1 - x^(2*j-1)).

%F Limit_{n->infinity} a(n)^(1/n) = exp(exp(-1)/2). - _Vaclav Kotesovec_, Dec 17 2021

%t a[n_] := Sum[Floor[n/(2*k - 1)]^k, {k, 1, n}]; Array[a, 50] (* _Amiram Eldar_, Dec 17 2021 *)

%o (PARI) a(n) = sum(k=1, n, (n\(2*k-1))^k);

%o (PARI) my(N=66, x='x+O('x^N)); Vec(sum(j=1, N, (1-x^(2*j-1))*sum(k=1, N, k^j*x^(k*(2*j-1))))/(1-x))

%Y Cf. A345176, A350145.

%K nonn

%O 1,2

%A _Seiichi Manyama_, Dec 16 2021