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A132181 a(n)=smallest positive integer such that product{k=1 to n}(1+1/a(k)) has a prime numerator. 1
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; refs; listen; history; text; internal format)
OFFSET

1,2

LINKS

Owen Whitby, Table of n, a(n) for n = 1..200

FORMULA

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)

CROSSREFS

Sequence in context: A181365 A221913 A180512 * A027642 A117214 A185972

Adjacent sequences:  A132178 A132179 A132180 * A132182 A132183 A132184

KEYWORD

nonn

AUTHOR

Leroy Quet Nov 04 2007

EXTENSIONS

a(10) to a(59) and list of 200 terms added by Owen Whitby (whitbyo(AT)acm.org), May 07 2008

STATUS

approved

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Last modified May 19 15:27 EDT 2013. Contains 225433 sequences.