login
Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 9/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.
1

%I #5 May 01 2013 12:24:20

%S 1,8,584,3490568,138073441864904,236788599971507074896206759048,

%T 756988343475413525492604622110601759725560263205883476698184

%N Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 9/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

%C Numerators of the sequence s(n) of the sum resp. product of fractions f(n) is A165427(n+2), hence sum(A165427(i+1)/A225161(i),i=1..n) = product(A165427(i+1)/A225161(i),i=1..n) = A165427(n+2)/a(n) = A165421(n+3)/a(n) = A011764(n)/a(n).

%F a(n) = 9^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/9.

%e f(n) = 9, 9/8, 81/73, 6561/5977, ...

%e 9 + 9/8 = 9 * 9/8 = 81/8; 9 + 9/8 + 81/73 = 9 * 9/8 * 81/73 = 6561/584; ...

%e s(n) = 1/b(n) = 9, 81/8, 6561/584, ...

%p b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:

%p b(1):=1/9;

%p a:=n->9^(2^(n-1))*b(n);

%p seq(a(i),i=1..8);

%Y Cf. A011764, A076628, A165421, A165427, A225161.

%K nonn

%O 1,2

%A _Martin Renner_, Apr 30 2013