A348893 a(n) = 840*(2*n)!/((n + 4)!*n!).
35, 14, 14, 20, 35, 70, 154, 364, 910, 2380, 6460, 18088, 52003, 152950, 458850, 1400700, 4342170, 13646820, 43421700, 139704600, 454039950, 1489251036, 4925984196, 16419947320, 55124108860, 186281471320, 633357002488, 2165672331088, 7444498638115, 25717358931670, 89254363351090
Offset: 0
Programs
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Maple
seq(840*(2*n)!/((n + 4)!*n!),n=0..30)
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Mathematica
a[n_] := 4^(n + 4) Hypergeometric2F1[9/2, 1/2 - n, 11/2, 1] / (9 Pi); Table[a[n], {n, 0, 30}] (* Peter Luschny, Nov 03 2021 *) -
PARI
a(n)=35*binomial(2*n,n)/binomial(n+4,4) \\ Charles R Greathouse IV, Oct 23 2023
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Sage
def A348893(n): return (-4)^(4 + n)*binomial(7/2, 4 + n)/2 print([A348893(n) for n in range(31)]) # Peter Luschny, Nov 04 2021
Formula
O.g.f: (140*z^3 - 70*z^2 + 14*z - 1 + (1 - 4*z)^(7/2))/(2*z^4).
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.
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;
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.
Remark: this sequence is not monotonically growing with n, as a(0) > a(1) = a(2) < a(3) < a(4)... .
From Peter Luschny, Nov 03 2021: (Start)
a(n) = 14*A007272(n)/(n + 4).
a(n) ~ 105*4^n*(8*n - 81)/(n^(11/2)*sqrt(Pi)).
a(n) = 4^(n + 4)*hypergeom([9/2, 1/2 - n], [11/2], 1) / (9*Pi). (End)
a(n) = (-4)^(4 + n)*binomial(7/2, 4 + n)/2. - Peter Luschny, Nov 04 2021
From Peter Bala, Mar 10 2023: (Start)
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.
7 divides a(n) except when n == 3 (mod 7).
P-recursive: (n + 4)*a(n) = 2*(2*n - 1)*a(n-1) with a(0) = 35.
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)
From Peter Bala, Mar 11 2023: (Start)
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).
a(n) is odd if n = 2^k - 4, k >= 2, otherwise a(n) is even. (End)
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = 13/70 + 4*Pi/(81*sqrt(3)).
Sum_{n>=0} (-1)^(n+1)/a(n) = 72*log(phi)/(3125*sqrt(5)) - 103/43750, where phi is the golden ratio (A001622). (End)