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Let q(b-1)=k(b)*q(b)^p(b)-1 for b=n to 1, where q(n)=p(n+1), with p(x) as the x-th prime. If k(b) is the minimum positive integer that makes q(b-1) prime, then a(n)=q(0).
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%I #3 Mar 31 2012 14:11:24

%S 17,3486013,95546861187714408803829067490017,

%T 11521371769146027198878540116068812681880419688486700618357345699743923465881941319972521176540806894433040020358370202464446915550483588097961647367908157261449080390753803350619037518472005174420394730557755077744685894370803149416209738934082646247291

%N Let q(b-1)=k(b)*q(b)^p(b)-1 for b=n to 1, where q(n)=p(n+1), with p(x) as the x-th prime. If k(b) is the minimum positive integer that makes q(b-1) prime, then a(n)=q(0).

%C a(5) has 2884 digits.

%e if n=3, then p(n+1)=p(4)=7.

%e when b=3, q(b)=q(3)=p(4)=7 and k(b)=2 and p(b)=p(3)=5, then q(b-1)=q(2)=2*7^5-1=33613

%e when b=2, q(b)=q(2)=33613 and k(b)=26 and p(b)=p(2)=3, then q(b-1)=q(1)=26*33613^3-1=987404664412321

%e when b=1, q(b)=q(1)=987404664412321 and k(b)=98 and p(b)=p(1)=2, then q(b-1)=q(0)=98*987404664412321^3-1=95546861187714408803829067490017

%e then a(3)=q(0)=95546861187714408803829067490017

%K nonn

%O 1,1

%A _Ray G. Opao_, Apr 20 2005