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G.f. A(x) satisfies A(x) = 1/( 1 - x*A(x)*(1 + 9*x*A(x))^(1/3) ).
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%I #15 Nov 30 2024 08:50:05

%S 1,1,5,11,95,150,2688,-111,98489,-215578,4416842,-18887063,230670421,

%T -1356589436,13381147908,-92724422022,831047516316,-6277471705749,

%U 53925750947589,-426682784513559,3602138266461603,-29250145766625450,245524688963062050

%N G.f. A(x) satisfies A(x) = 1/( 1 - x*A(x)*(1 + 9*x*A(x))^(1/3) ).

%H <a href="/index/Res#revert">Index entries for reversions of series</a>

%F a(n) = (1/(n+1)) * Sum_{k=0..n} 9^(n-k) * binomial(n+k,k) * binomial(k/3,n-k).

%F From _Seiichi Manyama_, Nov 30 2024: (Start)

%F G.f.: exp( Sum_{k>=1} A378555(k) * x^k/k ).

%F a(n) = (1/(n+1)) * [x^n] 1/(1 - x*(1 + 9*x)^(1/3))^(n+1).

%F G.f.: (1/x) * Series_Reversion( x*(1 - x*(1 + 9*x)^(1/3)) ). (End)

%o (PARI) a(n) = sum(k=0, n, 9^(n-k)*binomial(n+k, k)*binomial(k/3, n-k))/(n+1);

%Y Cf. A001002, A372125.

%Y Cf. A372013, A372136, A378555.

%K sign,changed

%O 0,3

%A _Seiichi Manyama_, Apr 20 2024