This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A348898 #34 Dec 24 2024 22:11:41
%S A348898 126,42,36,45,70,126,252,546,1260,3060,7752,20349,55062,152950,434700,
%T A348898 1260630,3721860,11165580,33982200,104778450,326908764,1031019948,
%U A348898 3283989464,10555680420,34214964120,111768882792,367755678864,1218190686237,4060635620790,13615072375590
%N A348898 a(n) = 15120*(2*n)!/(n!*(n + 5)!); super ballot numbers, row 4 of A135573.
%F A348898 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.
%F A348898 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)).
%F A348898 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).
%F A348898 a(n) = 4^(n + 5)*hypergeom([11/2, 1/2 - n], [13/2], 1) / (11*Pi).
%F A348898 a(n) = -(-4)^(5 + n)*binomial(9/2, 5 + n)/2. - _Peter Luschny_, Nov 04 2021
%F A348898 From _Peter Bala_, Mar 11 2023: (Start)
%F A348898 P-recursive: (n + 5)*a(n) = 2*(2*n - 1)*a(n-1) with a(0) = 126.
%F A348898 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.
%F A348898 a(n) is odd if n = 2^k - 5, k >= 3, otherwise a(n) is even. (End)
%F A348898 From _Amiram Eldar_, Mar 28 2023: (Start)
%F A348898 Sum_{n>=0} 1/a(n) = 47/630 + 44*Pi/(2187*sqrt(3)).
%F A348898 Sum_{n>=0} (-1)^(n+1)/a(n) = 13/218750 + 88*log(phi)/(15625*sqrt(5)), where phi is the golden ratio (A001622). (End)
%p A348898 a := n -> 15120*(2*n)!/(n!*(n + 5)!): seq(a(n), n = 0..29);
%t A348898 a[n_] := 4^(n + 5) Hypergeometric2F1[11/2, 1/2 - n, 13/2, 1] / (11 Pi);
%t A348898 Table[a[n], {n, 0, 29}]
%o A348898 (Sage)
%o A348898 def A348898(n): return -(-4)^(5 + n)*binomial(9/2, 5 + n)/2
%o A348898 print([A348898(n) for n in range(30)]) # _Peter Luschny_, Nov 04 2021
%o A348898 (PARI) a(n)=126*binomial(2*n,n)/binomial(n+5,5) \\ _Charles R Greathouse IV_, Oct 23 2023
%Y A348898 Cf. A000108, A001622, A135573.
%K A348898 nonn,easy
%O A348898 0,1
%A A348898 _Peter Luschny_, Nov 03 2021