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%I #28 Mar 18 2024 12:13:20
%S 1,9,99,1175,14499,183195,2351805,30539241,400000275,5274560891,
%T 69929215641,931226954949,12446852889901,166888293332805,
%U 2243683808486451,30235162687458327,408274269493595283,5523024440001832875,74834275541765522505,1015429462194625633125
%N G.f.: sqrt( (1-x - sqrt(1-14*x+x^2)) / (6*x*(1-14*x+x^2)) ).
%C Multiply the square of each term by -3 to form a bisection of A245925.
%C Limit a(n+1)/a(n) = 7 + 4*sqrt(3).
%H G. C. Greubel, <a href="/A245927/b245927.txt">Table of n, a(n) for n = 0..850</a>
%F a(n)^2 = (-1/3)*Sum_{k=0..2*n+1} Sum_{j=0..4*n-2*k+2} (-1)^(j+k) * C(4*n-k+2,j+k)^2 * C(j+k,k)^2.
%F a(n) ~ (3-sqrt(3)) * (7+4*sqrt(3))^(n+1) / (12*sqrt(Pi*n)). - _Vaclav Kotesovec_, Aug 16 2014
%F From _Peter Bala_, Mar 17 2018: (Start)
%F a(n) = Sum_{k = 0..n} C(2*n+1,2*k)*C(2*k,k)*3^(n-k).
%F a(n) = 3^n*hypergeom([-n, -n-1/2], [1], 4/3).
%F n*(4*n-3)*(2*n+1)*a(n) = (4*n-1)*(28*n^2-14*n-5)*a(n-1) - (n-1)*(2*n-1)*(4*n+1)*a(n-2). (End)
%F a(n) = (-1)^n*hypergeom([-n, n + 3/2], [1], 4). _Peter Luschny_, Mar 17 2018
%F a(n) = (-1)^n/sqrt(-3) * P(2*n+1, sqrt(-3)), where P(n, x) denotes the n-th Legendre polynomial. - _Peter Bala_, Mar 17 2024
%e G.f.: A(x) = 1 + 9*x + 99*x^2 + 1175*x^3 + 14499*x^4 + 183195*x^5 +... where
%e A(x)^2 = (1-x - sqrt(1-14*x+x^2)) / (6*x*(1-14*x+x^2)). Explicitly,
%e A(x)^2 = 1 + 18*x + 279*x^2 + 4132*x^3 + 59949*x^4 + 860022*x^5 + 12252547*x^6 + 173756232*x^7 + 2456093529*x^8 +...+ A245924(n)*x^n +...
%p seq(add( binomial(2*n+1,2*k)*binomial(2*k,k)*3^(n-k), k = 0..n),n = 0..20); # _Peter Bala_, Mar 17 2018
%t CoefficientList[Series[Sqrt[(1-x-Sqrt[1-14x+x^2])/(6x(1-14x+x^2))],{x,0,20}],x] (* _Harvey P. Dale_, Oct 23 2015 *)
%t a[n_] := (-1)^n Hypergeometric2F1[-n, n + 3/2, 1, 4];
%t Table[a[n], {n, 0, 19}] (* _Peter Luschny_, Mar 17 2018 *)
%o (PARI) /* From definition: */
%o {a(n)=polcoeff( sqrt( (1-x - sqrt(1-14*x+x^2 +x^2*O(x^n))) / (6*x*(1-14*x+x^2 +x*O(x^n))) ), n)}
%o for(n=0, 20, print1(a(n), ", "))
%o (PARI) /* From formula for a(n)^2: */
%o {a(n)=sqrtint((-1/3)*sum(k=0, 2*n+1, sum(j=0, 4*n-2*k+2, (-1)^(j+k)*binomial(4*n-k+2,j+k)^2*binomial(j+k, k)^2)))}
%o for(n=0, 20, print1(a(n), ", "))
%Y Cf. A245925, A245926, A245924, A243947.
%K nonn,easy
%O 0,2
%A _Paul D. Hanna_, Aug 15 2014
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