A225156 - OEIS

%I #6 May 01 2013 12:07:16

%S 1,2,7,67,5623,37772347,1653794703916063,

%T 3104205768420613437667191487267,

%U 10767416908549848056705041797805600349527548164015760674541223

%N 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 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 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 a(n) = 3^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.

%F 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 f(n) = 3, 3/2, 9/7, 81/67, ...

%e 3 + 3/2 = 3 * 3/2 = 9/2; 3 + 3/2 + 9/7 = 3 * 3/2 * 9/7 = 81/14; ...

%p b:=n->3^(2^(n-2)); # n > 1

%p b(1):=3;

%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. A011764, A100441, A165421, A225163.

%K nonn,frac

%O 1,2

%A _Martin Renner_, Apr 30 2013