cp's OEIS Frontend

A225156 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 3/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

%I A225156 #6 May 01 2013 12:07:16
%S A225156 1,2,7,67,5623,37772347,1653794703916063,
%T A225156 3104205768420613437667191487267,
%U A225156 10767416908549848056705041797805600349527548164015760674541223
%N A225156 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 3/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.
%C A225156 Numerators of the sequence of fractions f(n) is A165421(n+1), hence sum(A165421(i+1)/a(i),i=1..n) = product(A165421(i+1)/a(i),i=1..n) = A165421(n+2)/A225163(n) = A011764(n-1)/A225163(n).
%H A225156 Paul Yiu, <a href="http://math.fau.edu/yiu/RecreationalMathematics2003.pdf">Recreational Mathematics</a>, Department of Mathematics, Florida Atlantic University, 2003, Chapter 5.4,  p. 207 (Project).
%F A225156 a(n) = 3^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
%F A225156 a(n) = 3^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.
%e A225156 f(n) = 3, 3/2, 9/7, 81/67, ...
%e A225156 3 + 3/2 = 3 * 3/2 = 9/2; 3 + 3/2 + 9/7 = 3 * 3/2 * 9/7 = 81/14; ...
%p A225156 b:=n->3^(2^(n-2)); # n > 1
%p A225156 b(1):=3;
%p A225156 p:=proc(n) option remember; p(n-1)*a(n-1); end;
%p A225156 p(1):=1;
%p A225156 a:=proc(n) option remember; b(n)-p(n); end;
%p A225156 a(1):=1;
%p A225156 seq(a(i),i=1..9);
%Y A225156 Cf. A011764, A100441, A165421, A225163.
%K A225156 nonn,frac
%O A225156 1,2
%A A225156 _Martin Renner_, Apr 30 2013