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Search: id:A132181
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| A132181 |
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a(n)=smallest positive integer such that product{k=1 to n}(1+1/a(k)) has a prime numerator. |
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+0
1
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| 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, 1, 1180246, 1, 9441976, 1, 1, 1, 169955586, 1, 2, 1, 2, 1, 2719289392, 1, 1, 1, 1, 5438578786, 1, 32631472722, 1, 2, 1, 391577672676, 1, 1, 2, 1, 1566310690708, 1, 1
(list; graph; listen)
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OFFSET
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1,2
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LINKS
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Owen Whitby, Table of n, a(n) for n = 1..200
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FORMULA
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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.
a(1)=1; n(1)=1, d(1)=1 ==> a(2)=1, n(2)=2, d(2)=1 ( to start things off );
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);
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;
n(i)=odd prime, d(i)=even ==> a(i+1)=1, n(i+1)=n(i), d(i+1)=d(i)/2;
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)
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CROSSREFS
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Sequence in context: A109530 A111519 A008855 this_sequence A027642 A117214 A134301
Adjacent sequences: A132178 A132179 A132180 this_sequence A132182 A132183 A132184
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KEYWORD
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nonn
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AUTHOR
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Leroy Quet (q1qq2qqq3qqqq(AT)yahoo.com), Nov 04 2007
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EXTENSIONS
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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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