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OFFSET
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1,1
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COMMENTS
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Starting with the fraction 1/1, generate the sequence of fractions A002533(i)/A002532(i) according to the rule: "add top and bottom to get the new bottom, add top and 6 times bottom to get the new top."
The prime numerators of these fractions are listed here, at locations i= 2, 3, 4, 5, 11, 17, 32, 53,... showing prime(4), prime(8), prime(21), prime(53), prime(34719),..
Is there an infinity of primes in this sequence?
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REFERENCES
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Prime Obsession, John Derbyshire, Joseph Henry Press, April 2004, p 16.
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LINKS
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FORMULA
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PROG
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(PARI) primenum(n, k, typ) = \\ k=mult, typ=1 num, 2 denom. output prime num or denom.
{ local(a, b, x, tmp, v); a=1; b=1;
for(x=1, n, tmp=b; b=a+b; a=k*tmp+a; if(typ==1, v=a, v=b); if(isprime(v), print1(v", "); ) );
print(); print(a/b+.) }
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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