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Expansion of (1/x) * Series_Reversion( x / ( (1+x) * (1+3*x)^2 ) ).
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%I #17 Sep 09 2024 09:35:06

%S 1,7,64,667,7513,89092,1095832,13852195,178855075,2348744095,

%T 31273438804,421224534100,5728966150924,78569975545432,

%U 1085350298162608,15087689038165555,210907141968410071,2962825568825439349,41806163408065511032,592244891188614804643

%N Expansion of (1/x) * Series_Reversion( x / ( (1+x) * (1+3*x)^2 ) ).

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

%F a(n) = (1/(n+1)) * Sum_{k=0..n} 3^k * binomial(2*(n+1),k) * binomial(n+1,n-k).

%F From _Peter Bala_, Sep 08 2024: (Start)

%F a(n) = hypergeom([-n, -2*(n+1)], [2], 3).

%F a(n) = (-2)^n * Jacobi_P(n, 1, n+2, -2)/(n+1).

%F P-recursive: 2*(11*n-5)*(2*n+3)*(n+1)*a(n) = (649*n^3+354*n^2-109*n-54)*a(n-1) + 16*(n-1)*(2*n-1)*(11*n+6)*a(n-2) with a(0) = 1 and a(1) = 7. (End)

%p seq(simplify(hypergeom([-n, -2*(n+1)], [2], 3)), n = 0..20); # _Peter Bala_, Sep 08 2024

%o (PARI) my(N=30, x='x+O('x^N)); Vec(serreverse(x/((1+x)*(1+3*x)^2))/x)

%o (PARI) a(n) = sum(k=0, n, 3^k*binomial(2*(n+1), k)*binomial(n+1, n-k))/(n+1);

%Y Cf. A001764, A034015.

%K nonn,easy

%O 0,2

%A _Seiichi Manyama_, Mar 21 2024