More Continued Fractions -> Primes
Leroy Quet
qqquet at mindspring.com
Tue Dec 24 22:47:41 CET 2002
(1) Let {a(k)} be a sequence where the m_th term, a(m), is the lowest
positive integer such that the continued fraction
[a(1); a(2), a(3),..., a(m)]
has a prime for a numerator, for all m >= 1.
I think that this sequence begins:
2, 1, 1, 2, 1, 3,...
--
(2) Let {a(k)} be a sequence where the m_th term, a(m), is the lowest
positive integer NOT EQUAL TO ANY ELEMENT OF {a(1), a(2),
a(3),...,a(m-1)} such that the continued fraction
[a(1); a(2), a(3),..., a(m)]
has a prime for a numerator, for all m >= 1.
I think that this sequence begins:
2, 1, 3, 4, 6,...
Is this particular sequence a rearrangement of the positive integers?
--
(1 a) and (2 a): same as (1) and (2), but all integers must be >= 2.
These sequences should all be infinite, because there are an infinite
number of primes in the sequence: b(n) = n*j + k, where j and k are
fixed relatively prime positive integers. Since reversing the
direction of the terms of a continued fraction gets a fraction with
the same numerator as the CF with the terms in their original
direction, then there is always an a(m) where the numerator of our CF
is a prime. For there is always an a(m) where the numerator of:
a(m) + k/j = (a(m)*j + k)/j
is prime, for GCD(j,k) = 1.
Thanks,
Leroy Quet
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