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G.f. A(x) satisfies: A(x) = x * exp(3 * Sum_{k>=1} (-1)^k * A(x^k) / k).
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%I #5 Jun 28 2021 15:58:23

%S 1,-3,15,-82,486,-3090,20497,-140010,979131,-6976603,50461716,

%T -369533691,2734423934,-20414010219,153571115619,-1163003999342,

%U 8859172575069,-67835214598017,521824159637718,-4030828937892966,31252886542570119,-243142210911325273,1897466281615297698

%N G.f. A(x) satisfies: A(x) = x * exp(3 * Sum_{k>=1} (-1)^k * A(x^k) / k).

%F G.f.: x / Product_{n>=1} (1 + x^n)^(3*a(n)).

%F a(n+1) = (3/n) * Sum_{k=1..n} ( Sum_{d|k} (-1)^(k/d) * d * a(d) ) * a(n-k+1).

%p a:= proc(n) option remember; `if`(n=1, 1, 3*add(a(n-k)*add(d*a(d)

%p *(-1)^(k/d), d=numtheory[divisors](k)), k=1..n-1)/(n-1))

%p end:

%p seq(a(n), n=1..23); # _Alois P. Heinz_, Jun 28 2021

%t nmax = 23; A[_] = 0; Do[A[x_] = x Exp[3 Sum[(-1)^k A[x^k]/k, {k, 1, nmax}]] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x] // Rest

%t a[1] = 1; a[n_] := a[n] = (3/(n - 1)) Sum[Sum[(-1)^(k/d) d a[d], {d, Divisors[k]}] a[n - k], {k, 1, n - 1}]; Table[a[n], {n, 1, 23}]

%Y Cf. A006964, A049075, A052757, A345884.

%K sign

%O 1,2

%A _Ilya Gutkovskiy_, Jun 28 2021