OFFSET
1,1
COMMENTS
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 denominators of these fractions are listed here, at locations i= 2, 5, 13, 19, 29, 37,.. 41, 53, 59, .... equalling prime(1), prime(26), prime(148838), ..
Is there an infinity of primes in this sequence?
REFERENCES
John Derbyshire, Prime Obsession, Joseph Henry Press, April 2004, p. 16.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..17
MATHEMATICA
Select[LinearRecurrence[{2, 5}, {0, 1}, 150], PrimeQ] (* Amiram Eldar, Jun 30 2024 *)
PROG
(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+.); }
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Cino Hilliard, Oct 02 2005
EXTENSIONS
Simplified the definition, listed some A002532 indices - R. J. Mathar, Sep 16 2009
STATUS
approved