Search: id:A132181
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%I A132181
%S A132181 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,
2,
%T A132181 1,1180246,1,9441976,1,1,1,169955586,1,2,1,2,1,2719289392,1,1,1,1,
%U A132181 5438578786,1,32631472722,1,2,1,391577672676,1,1,2,1,1566310690708,1,1
%N A132181 a(n)=smallest positive integer such that product{k=1 to n}(1+1/a(k))
has a prime numerator.
%H A132181 Owen Whitby, Table of n, a(n) for n = 1..200
%H A132181 Leroy Quet, Home Page
(listed in lieu of email address)
%F A132181 Comments from Owen Whitby (whitbyo(AT)acm.org), 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 A132181 a(1)=1; n(1)=1, d(1)=1 ==> a(2)=1, n(2)=2, d(2)=1 ( to start things off
);
%F A132181 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 A132181 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 A132181 n(i)=odd prime, d(i)=even ==> a(i+1)=1, n(i+1)=n(i), d(i+1)=d(i)/2;
%F A132181 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)
%Y A132181 Sequence in context: A109530 A111519 A008855 this_sequence A027642 A117214
A134301
%Y A132181 Adjacent sequences: A132178 A132179 A132180 this_sequence A132182 A132183
A132184
%K A132181 nonn
%O A132181 1,2
%A A132181 Leroy Quet Nov 04 2007
%E A132181 a(10) to a(59) and list of 200 terms added by Owen Whitby (whitbyo(AT)acm.org),
May 07 2008
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