a(n) = (2*n + 1)*(32*n^4 + 64*n^3 + 88*n^2 + 56*n + 15)/15.
a(n) = Sum_{k = 0..5} binomial(10, 2*k)*binomial(n+k, 5).
a(n) = Sum_{k = 0..5} binomial(10, 2*k+1)*binomial(n+k+1/2, 5).
O.g.f.: (1 + 45*x + 210*x^2 + 210*x^3 + 45*x^4 + x^5)/(1 - x)^6 = 1/(1 - x) * T(5, (1 + x)/(1 - x)), where T(n, x) denotes the Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/(n*a(n-1)*a(n)) = 2*log(2) - 41/30.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), for n > 5. -
Vincenzo Librandi, Dec 16 2015
Sum_{k = 1..n+1} 1/(k*a(k)*a(k-1)) = 1/(51 - 3/(59 - 60/(75 - 315/(99 - ... - n^2*(4*n^2 - 1)/((2*n + 1)^2 + 2*5^2))))).
E.g.f.: exp(x)*(1 + 50*x + 400*x^2/2! + 1120*x^3/3! + 1280*x^4/4! + 512*x^5/5!).
Note that -T(10, i*sqrt(x)) = 1 + 50*x + 400*x^2 + 1120*x^3 + 1280*x^4 + 512*x^5. See
A008310. (End)