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%I #5 May 01 2013 12:22:43
%S 1,5,155,176855,265770796655,679134511201261085170655,
%T 4943777738415359153962876938905400001585992709055
%N Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 6/1; 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 A165424(n+2), hence sum(A165424(i+1)/A225158(i),i=1..n) = product(A165424(i+1)/A225158(i),i=1..n) = A165424(n+2)/a(n) = A173501(n+2)/a(n).
%F a(n) = 6^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/6.
%e f(n) = 6, 6/5, 36/31, 1296/1141, ...
%e 6 + 6/5 = 6 * 6/5 = 36/5; 6 + 6/5 + 36/31 = 6 * 6/5 * 36/31 = 1296/155; ...
%e s(n) = 1/b(n) = 6, 36/5, 1296/155, ...
%p b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
%p b(1):=1/6;
%p a:=n->6^(2^(n-1))*b(n);
%p seq(a(i),i=1..8);
%Y Cf. A076628, A165424, A173501, A225158.
%K nonn
%O 1,2
%A _Martin Renner_, Apr 30 2013
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