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%I #6 May 01 2013 12:19:34
%S 1,6,43,2143,5211907,30351298460743,1016966398053911225889737707,
%T 1130815308619683511655208290917557601522304473342184143
%N Denominators of the sequence of fractions f(n) defined recursively by f(1) = 7/1; 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 of fractions f(n) is A165425(n+1), hence sum(A165425(i+1)/a(i),i=1..n) = product(A165425(i+1)/a(i),i=1..n) = A165425(n+2)/A225166(n).
%F a(n) = 7^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
%F a(n) = 7^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.
%e f(n) = 7, 7/6, 49/43, 2401/2143, ...
%e 7 + 7/6 = 7 * 7/6 = 49/6; 7 + 7/6 + 49/43 = 7 * 7/6 * 49/43 = 2401/258; ...
%p b:=n->7^(2^(n-2)); # n > 1
%p b(1):=7;
%p p:=proc(n) option remember; p(n-1)*a(n-1); end;
%p p(1):=1;
%p a:=proc(n) option remember; b(n)-p(n); end;
%p a(1):=1;
%p seq(a(i),i=1..9);
%Y Cf. A100441, A165425, A225166.
%K nonn
%O 1,2
%A _Martin Renner_, Apr 30 2013
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