A142998 - OEIS

%I #5 Sep 14 2013 17:33:43

%S 0,1,25,809,34380,1890076,131608656,11369370384,1196133878016,

%T 150793148779776,22461588531072000,3905311348190592000,

%U 784153616550893568000,180142618195367442432000

%N a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+21)*a(n) - n^4*a(n-1).

%C This is the case m = 4 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+m^2+m+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^2 for the constant zeta(2). See A142995 for remarks on the general case.

%D Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

%F a(n) = n!^2*p(n)*sum {k = 1..n} 1/(k^2*p(k-1)*p(k)), where p(n) = (35*n^4+70*n^3+85*n^2+50*n+12)/12 = A008384(n) is the polynomial that gives the crystal ball sequence for the A_4 lattice. Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n^2+2*n+21)*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) = 0, b(1) = 21. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(21-1^4/(25-2^4/(33-3^4/(45-...-(n-1)^4/(2*n^2-2*n+21))))), for n >=2. Thus the behavior of a(n) for large n is given by lim n -> infinity a(n)/b(n) = 1/(21-1^4/(25-2^4/(33-3^4/(45-...-n^4/((2*n^2+2*n+21)-...))))) = sum {k = 1..inf} 1/(k^2*p(k-1)*p(k)) = zeta(2) - 2*(1-1/4+1/9-1/16). The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Corollary to Entry 31] (replace x by 2x+1 in the corollary and apply Entry 14).

%p p := n -> (35*n^4+70*n^3+85*n^2+50*n+12)/12: a := n -> n!^2*p(n)*sum (1/(k^2*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..20);

%Y Cf. A008384, A001819, A142995, A142996, A142997.

%K easy,nonn

%O 0,3

%A _Peter Bala_, Jul 18 2008