cp's OEIS Frontend

A225158 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.

%I A225158 #6 May 01 2013 12:08:45
%S A225158 1,5,31,1141,1502761,2555339110801,7279526598745139799221281,
%T A225158 58396508924557918552199410007906486608310469119041,
%U A225158 3723292553725227196293782783863296586090351965218332181732394788182320381276998127547535467381368961
%N A225158 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 A225158 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 A225158 a(n) = 6^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
%F A225158 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 A225158 f(n) = 6, 6/5, 36/31, 1296/1141, ...
%e A225158 6 + 6/5 = 6 * 6/5 = 36/5; 6 + 6/5 + 36/31 = 6 * 6/5 * 36/31 = 1296/155; ...
%p A225158 b:=n->6^(2^(n-2)); # n > 1
%p A225158 b(1):=6;
%p A225158 p:=proc(n) option remember; p(n-1)*a(n-1); end;
%p A225158 p(1):=1;
%p A225158 a:=proc(n) option remember; b(n)-p(n); end;
%p A225158 a(1):=1;
%p A225158 seq(a(i),i=1..9);
%Y A225158 Cf. A100441, A165424, A173501, A225165.
%K A225158 nonn,frac
%O A225158 1,2
%A A225158 _Martin Renner_, Apr 30 2013