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G.f. A(x) satisfies: A(x) = x*exp(2*Sum_{n>=1} Sum_{k>=1} n*a(n)*x^(n*(2*k-1))/(2*k - 1)).
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%I #18 Apr 24 2019 19:44:48

%S 0,1,2,10,78,794,9870,143610,2382350,44266538,909575170,20468012850,

%T 500542618118,13218631046786,374965272837542,11372416113131346,

%U 367296622702990270,12587154399475110546,456238999451039779510,17440439387336903608866,701272672299320517560470

%N G.f. A(x) satisfies: A(x) = x*exp(2*Sum_{n>=1} Sum_{k>=1} n*a(n)*x^(n*(2*k-1))/(2*k - 1)).

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

%e G.f.: A(x) = x + 2*x^2 + 10*x^3 + 78*x^4 + 794*x^5 + 9870*x^6 + 143610*x^7 + 2382350*x^8 + 44266538*x^9 + ...

%t a[n_] := a[n] = SeriesCoefficient[x Exp[2 Sum[Sum[j a[j] x^(j (2 k - 1))/(2 k - 1), {k, 1, n - 1}], {j, 1, n - 1}]], {x, 0, n}]; a[1] = 1; Table[a[n], {n, 0, 20}]

%t a[n_] := a[n] = SeriesCoefficient[x Product[((1 + x^k)/(1 - x^k))^(k a[k]), {k, 1, n - 1}], {x, 0, n}]; a[1] = 1; Table[a[n], {n, 0, 20}]

%Y Cf. A073075, A307708, A307709.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Apr 24 2019