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G.f. A(x) satisfies A(x) = ( 1 + 4*x*A(x)/(1 - x*A(x)) )^(1/2).
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%I #12 Apr 22 2024 12:45:55

%S 1,2,4,10,30,98,336,1194,4360,16258,61644,236938,921102,3615330,

%T 14307312,57024426,228701646,922283522,3737497980,15212318730,

%U 62160993642,254909413218,1048717979424,4327273358250,17903826642780,74260741616514,308724721176676

%N G.f. A(x) satisfies A(x) = ( 1 + 4*x*A(x)/(1 - x*A(x)) )^(1/2).

%F a(n) = (1/(n+1)) * Sum_{k=0..n} 4^k * binomial(n/2+1/2,k) * binomial(n-1,n-k).

%F D-finite with recurrence n*(n+1)*(n-2)*a(n) -6*(n-2)*(3*n^2-6*n+1)*a(n-2) -27*n*(n-3)*(n-4)*a(n-4)=0. - _R. J. Mathar_, Apr 22 2024

%F Conjecture: a(2n+1) = 2*A371364(). - _R. J. Mathar_, Apr 22 2024

%p A372018 := proc(n)

%p add(4^k*binomial((n+1)/2,k)*binomial(n-1,k-1),k=0..n) ;

%p %/(n+1) ;

%p end proc:

%p seq(A372018(n),n=0..60) ; # _R. J. Mathar_, Apr 22 2024

%o (PARI) a(n) = sum(k=0, n, 4^k*binomial(n/2+1/2, k)*binomial(n-1, n-k))/(n+1);

%Y Cf. A372019, A372020.

%K nonn

%O 0,2

%A _Seiichi Manyama_, Apr 15 2024