cp's OEIS Frontend

A143002 a(0) = 0, a(1) = 1, a(n+1) = 7*(2*n+1)*a(n) + n^4*a(n-1).

%I A143002 #4 Nov 30 2023 16:54:32
%S A143002 0,1,21,751,38500,2617756,225629712,23924915568,3053853073152,
%T A143002 461404969871616,81403191005875200,16580318776579814400,
%U A143002 3861255442546368921600,1019529192596773592678400
%N A143002 a(0) = 0, a(1) = 1, a(n+1) = 7*(2*n+1)*a(n) + n^4*a(n-1).
%C A143002 This is the case m = 3 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*m+1)*(2*n+1)*a(n) + n^4*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series sum {k = 1..inf} (-1)^(k+1)/k^2 for the constant 1/2*zeta(2). For other cases see A142999 (m=0), A143000 (m=1) and A143001 (m=2).
%D A143002 Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
%F A143002 a(n) = n!^2*p(n)*sum {k = 1..n} (-1)^(k+1)/(k^2*p(k-1)*p(k)), where p(n) = (n^6 + 3*n^5 + 22*n^4 + 39*n^3 + 85*n^2 + 66*n + 36)/36. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = 7*(2*n+1)*a(n) + n^4*a(n-1). The sequence b(n):= n!^2*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 7. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7+ 1^4/(21+ 2^4/(35+ 3^4/(49+...+ (n-1)^4/(7*(2*n-1)))))), for n >=2. Lim n -> infinity a(n)/b(n) = sum {k = 1..inf} (-1)^(k+1)/(k^2*p(k-1)p(k)) = 1/(7+ 1^4/(21+ 2^4/(35+ 3^4/(49+...+ n^4/(7*(2*n+1)+...))))) = 1/2*(zeta(2)-(1+1/4+1/9)). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 30].
%p A143002 p := n -> (n^6+3*n^5+22*n^4+39*n^3+85*n^2+66*n+36)/36: a := n -> n!^2*p(n)*sum ((-1)^(k+1)/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20)
%t A143002 RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==7(2n+1)a[n]+n^4 a[n-1]},a,{n,15}] (* _Harvey P. Dale_, Nov 30 2023 *)
%Y A143002 Cf. A142999, A143000, A143001.
%K A143002 easy,nonn
%O A143002 0,3
%A A143002 _Peter Bala_, Jul 18 2008