Prime Numbers

74 Chapter 1 PRIMES!
10 10 . The method in [Bernstein 1998] might be used in such an endeavor. Note
that the probability predicted by Theorem 1.4.9 is ρ(10) ≈ 2.77 × 10 −11 .
1.74. What is the approximate probability that a random integer (but of
size x, say) has all but one of its prime factors not exceeding B, witha
single outlying prime in the interval (B,C]? This problem has importance
for factoring methods that employ a “second stage,” which, after a first stage
exhausts (in a certain algorithm-dependent sense) the first bound B, attempts
to locate the outlying prime in (B,C]. It is typical in implementations of
various factoring methods that C is substantially larger than B, for usually
the operations of the second stage are much cheaper. See Exercise 3.5 for
related concepts.
1.75. Here is a question that leads to interesting computational issues.
Consider the number
c =1/3+
1
1/5+ 1
1/7+···
where the so-called elements of this continued fraction are the reciprocals of
all the odd primes in natural order. It is not hard to show that c is welldefined.
(In fact, a simple continued fraction—a construct having all 1’s in
the numerators—converges if the sum of the elements, in the present case
1/3 +1/5 +···, diverges.) First, give an approximate numerical value for
the constant c. Second, provide numerical (but rigorous) proof that c is not
equal to 1. Third, investigate this peculiar idea: that using all primes, that is,
starting the fraction as 1/2+ 1
1/3+··· , results in nearly the same fraction value!
Prove that if the two fractions in question were, in fact, equal, then we would
have c = 1+ √ 17 /4. By invoking more refined numerical experiments, try
to settle the issue of whether c is actually this exact algebraic value.
1.76. It is a corollary of an attractive theorem in [Bredihin 1963] that if n
is a power of two, the number of solutions
N(n) =#{(x, y, p) : n = p + xy; p ∈P; x, y ∈ Z + }
enjoys the following asymptotic relation:
N(n)
n
105
∼ ζ(3) ≈ 0.648 ....
2π4 From a computational perspective, consider the following tasks. First, attempt
to verify this asymptotic relation by direct counting of solutions. Second, drop
the restriction that n be a power of two, and try to verify experimentally,
theoretically, or both that the constant 105 should in general be replaced by
315
p∈P, p|n
(p − 1) 2
p 2 − p +1 .
,