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%I #9 Apr 09 2014 10:16:41
%S 1,2,6,1,28,1,1,58,1,708,1,1,2,1,2836,1,1,22696,1,1,1,590122,1,12,1,1,
%T 2,1,1180246,1,9441976,1,1,1,169955586,1,2,1,2,1,2719289392,1,1,1,1,
%U 5438578786,1,32631472722,1,2,1,391577672676,1,1,2,1,1566310690708,1,1
%N a(n)=smallest positive integer such that product{k=1 to n}(1+1/a(k)) has a prime numerator.
%H Owen Whitby, <a href="/A132181/b132181.txt">Table of n, a(n) for n = 1..200</a>
%F Comments from _Owen Whitby_, May 07 2008 (Start): Successive terms a(.) can be calculated using the following recurrences for the numerator n(.) and denominator d(.) of the product.
%F a(1)=1; n(1)=1, d(1)=1 ==> a(2)=1, n(2)=2, d(2)=1 ( to start things off );
%F n(i)=2, d(i)= odd ==> a(i+1)=q-1, n(i+1)=q, d(i+1)=d(i)(q-1)/2 where q is least odd prime not dividing d(i);
%F n(i)=odd prime, d(i)=1 ==> a(i+1)=c*n(i), n(i+1)=c*n(i)+1, d(i+1)=c where c is least even integer such that c*n(i)+1 is prime;
%F n(i)=odd prime, d(i)=even ==> a(i+1)=1, n(i+1)=n(i), d(i+1)=d(i)/2;
%F n(i)=odd prime, d(i)= odd>=3 ==> a(i+1)=p-1, n(i+1)=n(i), d(i+1)=d(i)(p-1)/p where p is least prime divisor of d(i). (End)
%K nonn
%O 1,2
%A _Leroy Quet_, Nov 04 2007
%E a(10) to a(59) and list of 200 terms added by _Owen Whitby_, May 07 2008
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