Prime Numbers

56 Chapter 1 PRIMES!
While the prime number theorem is equivalent to the assertion that the
product of the primes in [1,x] is e (1+o(1))x , it is still of interest to have
completely explicit inequalities as in this exercise and Exercise 1.25.
For a positive integer N, let
C(N) =
(1) Show that C(N) is an integer.
(6N)!N!
(3N)!(2N)!(2N)! .
(2) Show that if p is a prime with p>(6N) 1/k ,thenp k does not divide C(N).
(3) Using Exercise 1.25 and the idea in Exercise 1.27, show that
p>C(N)/4 (6N)1/2 +(6N) 1/3 lg(1.5N)
.
p≤6N
(4) Use Stirling’s formula (or mathematical induction) to show that C(N) >
108 N /(4 √ N) for all N.
(5) Show that
p≤x p>2x for x ≥ 2 12 .
(6) Close the gap from 2 12 to 31 with a direct calculation.
1.29. Use Exercise 1.28 to show that π(x) > x/lg x, for all x ≥ 5.
Since we have the binary logarithm here rather than the natural logarithm,
this inequality for π(x) might be humorously referred to as the “computer
scientist’s prime number theorem.” Use Exercise 1.25 to show that π(x) <
2x/ ln x for all x>0. In this regard, it may be helpful to first establish the
identity
where θ(x) :=
Theorem 1.1.3.
p≤x
π(x) = θ(x)
ln x +
x
θ(t)
2 t ln 2 t dt,
ln p. Note that the two parts of this exercise prove
1.30. Here is an exercise involving a healthy mix of computation and theory.
With σ(n) denoting the sum of the divisors of n, and recalling from the
discussion prior to Theorem 1.3.3 that n is deemed perfect if and only if
σ(n) =2n, do the following, wherein we adopt a unique prime factorization
n = p t1
1 ···ptk k :
(1) Write a condition on the pi,ti alone that is equivalent to the condition
σ(n) =2n of perfection.
(2) Use the relation from (1) to establish (by hand, with perhaps some minor
machine computations) some lower bound on odd perfect numbers; e.g.,
show that any odd perfect number must exceed 10 6 (or an even larger
bound).
(3) An “abundant number” is one with σ(n) > 2n, as in the instance
σ(12) = 28. Find (by hand or by small computer search) an odd abundant
number. Does an odd abundant number have to be divisible by 3?