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%I #43 Oct 23 2023 11:23:52
%S 35,14,14,20,35,70,154,364,910,2380,6460,18088,52003,152950,458850,
%T 1400700,4342170,13646820,43421700,139704600,454039950,1489251036,
%U 4925984196,16419947320,55124108860,186281471320,633357002488,2165672331088,7444498638115,25717358931670,89254363351090
%N a(n) = 840*(2*n)!/((n + 4)!*n!).
%F O.g.f: (140*z^3 - 70*z^2 + 14*z - 1 + (1 - 4*z)^(7/2))/(2*z^4).
%F E.g.f: 64*exp(2*z)*((-z^3 - 1/2*z^2 - 1/4*z - 3/32)*BesselI(1,2*z) + BesselI(0,2*z)*z*(z^2 + 1/4*z + 3/32))/z^3.
%F O.g.f. g(z) satisfies z^4*g(z)^2 + (-140*z^3 + 70*z^2 - 14*z + 1)*g(z) + 4096*z^3 - 2268*z^2 + 476*z - 35 = 0;
%F a(n) = Integral_{x=0..4} x^n*64*(1 - x/4)^(7/2)/(Pi*sqrt(x)). This is the integral representation as the n-th moment of a positive function on [0, 4]. The representation is unique.
%F Remark: this sequence is not monotonically growing with n, as a(0) > a(1) = a(2) < a(3) < a(4)... .
%F From _Peter Luschny_, Nov 03 2021: (Start)
%F a(n) = 14*A007272(n)/(n + 4).
%F a(n) ~ 105*4^n*(8*n - 81)/(n^(11/2)*sqrt(Pi)).
%F a(n) = 4^(n + 4)*hypergeom([9/2, 1/2 - n], [11/2], 1) / (9*Pi). (End)
%F a(n) = (-4)^(4 + n)*binomial(7/2, 4 + n)/2. - _Peter Luschny_, Nov 04 2021
%F From _Peter Bala_, Mar 10 2023: (Start)
%F a(n) = 35*binomial(2*n, n) - 56*binomial(2*n, n + 1) + 28*binomial(2*n, n + 2) - 8*binomial(2*n, n + 3) + binomial(2*n, n + 4). Thus this sequence is integral.
%F 7 divides a(n) except when n == 3 (mod 7).
%F P-recursive: (n + 4)*a(n) = 2*(2*n - 1)*a(n-1) with a(0) = 35.
%F D-finite: the o.g.f. A(x) satisfies the differential equation (1 - 4*x)*A'(x) + (4 - 2*x)*A(x) - 140 = 0, with A(0) = 35. (End)
%F From _Peter Bala_, Mar 11 2023: (Start)
%F a(n) = Sum_{k = 0..3} (-1)^k*4^(3-k)*binomial(3,k)*Catalan(n+k) = 64*Catalan(n) - 48*Catalan(n+1) + 12*Catalan(n+2) - Catalan(n+3), where Catalan(n) = A000108(n).
%F a(n) is odd if n = 2^k - 4, k >= 2, otherwise a(n) is even. (End)
%F From _Amiram Eldar_, Mar 28 2023: (Start)
%F Sum_{n>=0} 1/a(n) = 13/70 + 4*Pi/(81*sqrt(3)).
%F Sum_{n>=0} (-1)^(n+1)/a(n) = 72*log(phi)/(3125*sqrt(5)) - 103/43750, where phi is the golden ratio (A001622). (End)
%p seq(840*(2*n)!/((n + 4)!*n!),n=0..30)
%t a[n_] := 4^(n + 4) Hypergeometric2F1[9/2, 1/2 - n, 11/2, 1] / (9 Pi);
%t Table[a[n], {n, 0, 30}] (* _Peter Luschny_, Nov 03 2021 *)
%o (Sage)
%o def A348893(n): return (-4)^(4 + n)*binomial(7/2, 4 + n)/2
%o print([A348893(n) for n in range(31)]) # _Peter Luschny_, Nov 04 2021
%o (PARI) a(n)=35*binomial(2*n,n)/binomial(n+4,4) \\ _Charles R Greathouse IV_, Oct 23 2023
%Y Row 3 of array A135573.
%Y Cf. A000108, A001622, A007054, A007272, A348898, A348899.
%K nonn,easy
%O 0,1
%A _Karol A. Penson_, Nov 02 2021
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