A348898 a(n) = 15120*(2*n)!/(n!*(n + 5)!); super ballot numbers, row 4 of A135573.
126, 42, 36, 45, 70, 126, 252, 546, 1260, 3060, 7752, 20349, 55062, 152950, 434700, 1260630, 3721860, 11165580, 33982200, 104778450, 326908764, 1031019948, 3283989464, 10555680420, 34214964120, 111768882792, 367755678864, 1218190686237, 4060635620790, 13615072375590
Offset: 0
Programs
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Maple
a := n -> 15120*(2*n)!/(n!*(n + 5)!): seq(a(n), n = 0..29);
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Mathematica
a[n_] := 4^(n + 5) Hypergeometric2F1[11/2, 1/2 - n, 13/2, 1] / (11 Pi); Table[a[n], {n, 0, 29}] -
PARI
a(n)=126*binomial(2*n,n)/binomial(n+5,5) \\ Charles R Greathouse IV, Oct 23 2023
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Sage
def A348898(n): return -(-4)^(5 + n)*binomial(9/2, 5 + n)/2 print([A348898(n) for n in range(30)]) # Peter Luschny, Nov 04 2021
Formula
a(n) = (1/(2*Pi))*Integral_{x=0..4} x^n*(4 - x)^(9/2)*x^(-1/2). This is the integral representation of the n-th moment of a positive function on [0, 4]. This representation is unique.
E.g.f.: x^(-4)*exp(2*x)*((-256*x^4 - 160*x^3 - 108*x^2 - 60*x - 24)*BesselI(1, 2*x) + (256*x^4 + 96*x^3 + 60*x^2 + 24*x)*BesselI(0, 2*x)).
O.g.f.: ((512*x^4 - 325*x^3 + 110*x^2 - 17*x + 1)*sqrt(1 - 4*x) - 748*x^4 + 515*x^3 - 142*x^2 + 19*x - 1)/(sqrt(1 - 4*x)*(1 + sqrt(1 - 4*x))*x^4).
a(n) = 4^(n + 5)*hypergeom([11/2, 1/2 - n], [13/2], 1) / (11*Pi).
a(n) = -(-4)^(5 + n)*binomial(9/2, 5 + n)/2. - Peter Luschny, Nov 04 2021
From Peter Bala, Mar 11 2023: (Start)
P-recursive: (n + 5)*a(n) = 2*(2*n - 1)*a(n-1) with a(0) = 126.
a(n) = Sum_{k = 0..4} (-1)^k*4^(4-k)*binomial(4,k)*Catalan(n+k) = 256*Catalan(n) - 256*Catalan(n+1) + 96*Catalan(n+2) - 16*Catalan(n+3) + Catalan(n+4), where Catalan(n) = A000108(n). Thus a(n) is an integer for all n.
a(n) is odd if n = 2^k - 5, k >= 3, otherwise a(n) is even. (End)
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = 47/630 + 44*Pi/(2187*sqrt(3)).
Sum_{n>=0} (-1)^(n+1)/a(n) = 13/218750 + 88*log(phi)/(15625*sqrt(5)), where phi is the golden ratio (A001622). (End)