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Denominators of the sequence 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.
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%I #6 May 01 2013 12:08:45

%S 1,5,31,1141,1502761,2555339110801,7279526598745139799221281,

%T 58396508924557918552199410007906486608310469119041,

%U 3723292553725227196293782783863296586090351965218332181732394788182320381276998127547535467381368961

%N Denominators of the sequence 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 of fractions f(n) is A165424(n+1), hence sum(A165424(i+1)/a(i),i=1..n) = product(A165424(i+1)/a(i),i=1..n) = A165424(n+2)/A225165(n) = A173501(n+2)/A225165(n).

%F a(n) = 6^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

%F a(n) = 6^(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) = 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; ...

%p b:=n->6^(2^(n-2)); # n > 1

%p b(1):=6;

%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, A165424, A173501, A225165.

%K nonn,frac

%O 1,2

%A _Martin Renner_, Apr 30 2013