Prime Numbers

computationally e cient formula is

m

Z(t) = 2 n−1/2 cos(t ln n − ϑ(t))

n=1

M

+ (−1)m+1τ −1/4 (−1)j τ −j/2Φj (z) + RM (t).

j=0

Here, M is a cuto integer of choice, the Φj are entire functions defined for j ≥ 0 in terms of a function Φ0 and its derivatives, and RM (t) is the error. A practical instance is the choice M = 2, for which we need

In spite of the complexity here, it is to be stressed that the formula is immediately applicable in actual computation. In fact, the error R2 can be rigorously bounded:

|R2(t)| < 0.011t−7/4 for all t > 200.

Higher-order (M > 2) bounds, primarily found in [Gabcke 1979], are known, but just R2 has served computationalists well for two decades.

Implement the Riemann–Siegel formula for M = 2, and test against some known values such as

ζ(1/2 + 300i) ≈ 0.4774556718784825545360619

+0.6079021332795530726590749 i,

Z(1/2 + 300i) ≈ 0.7729870129923042272624525,

which are accurate to the implied precision. Using your implementation, locate the nearest zero to the point 1/2+300i, which zero should have t ≈ 299.84035. You should also be able to find, still at the M = 2 approximation level and with very little machine time, the value

ζ(1/2 + 106i) ≈ 0.0760890697382 + 2.805102101019 i,

again correct to the implied precision.

When one is armed with a working Riemann–Siegel implementation, a beautiful world of computation in support of analytic number theory opens. For details on how actually to apply ζ evaluations away from the real axis, see [Brent 1979], [van de Lune et al. 1986], [Odlyzko 1994], [Borwein et al. 2000]. We should point out that in spite of the power and importance of

the Riemann–Siegel formula, there are yet alternative means for e cient evaluation when imaginary parts are large. In fact it is possible to avoid the inherently asymptotic character of the Riemann–Siegel series, in favor of manifestly convergent expansions based on incomplete gamma function values, or on saddle points of certain integrals. Alternative schemes are discussed in [Galway 2000], [Borwein et al. 2000], and [Crandall 1999c].

1.62.For the Riemann–Siegel formula of Exercise 1.61, and for similar

prescriptions when s = σ + it is not on the half-line, it is evident that sums of the form

m

1

Sm(s) = n=1 ns ,

√

where m is an appropriate cuto (typically, m t), could be used in actual computations. Investigate the notion of calculating Sm(s) over an arithmetic progression of s values, using the nonuniform FFT algorithm we present as Algorithm 9.5.8. That is, for values

s = σ + ikτ,

for say k = 0, . . . , K − 1, we have

m

Sm(σ + ikτ ) = n1σ e−ikτ ln n,

n=1

and sure enough, this suggests a strategy of (m/K) nonuniform FFTs each of length K. Happily, the sum Sm can thus be calculated, for all k [0, K − 1], in a total of

O(m ln K)

operations, where desired accuracy enters (only logarithmically) into the implied big-O constant. This is a remarkable gain over the naive approach of doing a length-m sum K times, which would require O(mK).

Such speedups can be used not only for RH verification, but analytic methods for prime-counting. Incidentally, this nonuniform FFT approach is essentially equivalent in complexity to the parallel method in [Odlyzko and Sch¨onhage 1988]; however, for computationalists familiar with FFT, or possessed of e cient FFT software (which the nonuniform FFT could call internally), the method of the present exercise should be attractive.

1.63.Show that ψ(x), defined in (1.22), is the logarithm of the least

hypothesis.

1.64. Perform computations that connect the distribution of primes with the Riemann critical zeros by way of the ψ function defined in (1.22). Starting with the classical exact relation (1.23), obtain a numerical table of the first 2K critical zeros (K of them having positive imaginary part), and evaluate the resulting numerical approximation to ψ(x) for, say, noninteger x (2, 1000). As a check on your computations, you should find, for K = 200 zeros and denoting by ψ(K) the approximation obtained via said 2K zeros, the amazing fact that

throughout the possible x values. This means—heuristically speaking— that the first 200 critical zeros and their conjugates determine the prime occurrences in (2, 1000) “up to a handful,” if you will. Furthermore, a plot of the error vs. x is nicely noisy around zero, so the approximation is quite good in some sense of average. Try to answer this question: For a given range on x, about how many critical zeros are required to e ect an approximation as good as |ψ − ψ(K)| < 1 across the entire range? And here is another computational question: How numerically good is the approximation (based on the Riemann hypothesis)

ψ(x) = x + 2√x sin(t ln x) + O √x , t

t

with t running over the imaginary parts of the critical zeros [Ellison and Ellison 1985]? For an analogous analytic approach to actual prime-counting, see Section 3.7 and especially Exercise 3.50.

1.65. This, like Exercise 1.64, also requires a database of critical zeros of the Riemann zeta function. There exist some useful tests of any computational scheme attendant on the critical line, and here is one such test. It is a consequence of the Riemann hypothesis that we would have an exact relation (see [Bach and Shallit 1996, p. 214])

1

= 2 + γ − ln(4π),

|ρ|2

ρ

where ρ runs over all the zeros on the critical line. Verify this relation numerically, to as much accuracy as possible, by:

(1) Performing the sum for all zeros ρ = 1/2 + it for |t| ≤ T , some T of choice.

(2) Performing such a sum for |t| ≤ T but appending an estimate of the remaining, infinite, tail of the sum, using known formulae for the approximate distribution of zeros [Edwards 1974], [Titchmarsh 1986], [Ivi´c 1985].

Note in this connection Exercises 1.61 (for actual calculation of ζ values) and 8.34 (for more computations relating to the Riemann hypothesis).

1.66. There are attractive analyses possible for some of the simpler exponential sums. Often enough, estimates—particularly upper bounds—on

(2)The infinitude of prime triples in arithmetic progression (see Exercises 1.41, 1.42) is equivalent to the divergence as N → ∞ of

1

EN2 (t)EN (−2t) dt.

0

(3)The (binary) Goldbach conjecture is equivalent to

1

e−2πitN EN2 (t) dt = 0

0

for even N > 2, and the ternary Goldbach conjecture is equivalent to

1

e−2πitN EN3 (t) dt = 0

0

for odd N > 5.

(4)The infinitude of Sophie Germain primes (i.e., primes p such that 2p + 1 is likewise prime) is equivalent to the divergence as N → ∞ of

1

e2πitEN (2t)EN (−t) dt.

0

1.68. We mentioned in Section 1.4.4 that there is a connection between exponential sums and the singular series Θ arising in the Vinogradov resolution (1.12) for the ternary Goldbach problem. Prove that the Euler product form for Θ(n) converges (what about the case n even?), and is equal to an absolutely convergent sum, namely,

∞

µ(q)

Θ(n) = q=1 ϕ3(q) cq (n),

where the Ramanujan sum cq is defined in (1.37). It is helpful to observe that µ, ϕ, c are all multiplicative, the latter function in the sense that if gcd(a, b) = 1, then ca(n)cb(n) = cab(n). Show also that for su ciently large B in the assignment Q = lnB n, the sum (1.40) being only to Q (and not to ∞) causes negligible error in the overall ternary Goldbach estimate.

Next, derive the representation count, call it Rs(n), for n the sum of s primes, in the following way. It is known that for s > 2, n ≡ s (mod 2),

where now the general singular series is given from exponential-sum theory as

∞ µs(q)

Θs(n) = q=1 ϕs(q) cq (n).

(1)Take q = 2 and explain why the above estimate on EN is obvious for a = 0, 1.

(2)Let q = 3, and for a = 1, 2 explain using a vector diagram in the complex plane how the above estimate works.

(3)Let q = 4, and note that for some a values the right-hand side of the above estimate is actually zero. In such cases, use an error estimate (such as the conditional result (1.32)) to give sharp, nonzero estimates on EN (a/4) for a = 1, 3.

These theoretical examples reveal the basic behavior of the exponential sum for small q.

For a computational foray, test numerically the behavior of EN by way of the following steps:

(1) Choose N = 105, q = 31, and by direct summation over primes p ≤ N , create a table of E values for a [0, q − 1]. (Thus there will be q complex elements in the table.)

(2) Create a second table of values of π(N ) µ(q/g) , also for each a [0, q − 1].

ϕ(q/g)

(3)Compare, say graphically, the two tables. Though the former table is “noisy” compared to the latter, there should be fairly good average agreement. Is the discrepancy between the two tables consistent with theory?

(4)Explain why the latter table is so smooth (except for a glitch at the (a = 0)-th element). Finally, explain how the former table can be constructed via fast Fourier transform (FFT) on a binary signal (i.e., a certain signal consisting of only 0’s and 1’s).

Another interesting task is to perform direct numerical integration to verify (for small cases of N , say) some of the conjectural equivalences of Exercise 1.67.

1.71. Verify the following: There exist precisely 35084 numbers less than 10100 that are 4-smooth. Prove that for a certain constant c, the number of 4-smooth numbers not exceeding x is

ψ(x, 4) c ln2 x,

giving the explicit c and also as sharp an error bound on this estimate as you can. Generalize by showing that for each y ≥ 2 there is a positive number cy such that

ψ(x, y) cy lnπ(y) x, where y is fixed and x → ∞.

1.72.Carry out some numerical experiments to verify the claim after equation (1.45) that the implicit lower bound is a “good” one.

1.73.Compute by empirical means the approximate probability that a random integer having 100 decimal digits has all of its prime factors less than

1.6Research problems

1.77.In regard to the Mills theorem (the first part of Theorem 1.2.2), try

to find an explicit number θ and a large number n such that θ3j is prime

for j = 1, 2, . . . , n. For example if one takes the specific rational θ = 165/92, show that each of

θ31 , θ32 , θ33 , θ34

is prime, yet the number θ35 is, alas, composite. Can you find a simple

rational θ that has all cases up through n = 5 prime, or even further? Say a (finite or infinite) sequence of primes q1 < q2 < . . . is a “Mills sequence” if

there is some number θ such that qj = θ3j for j = 1, 2, . . . . Is it true that

any finite Mills sequence can be extended to an infinite Mills sequence (not necessarily with the same θ, but keeping the same initial sequence of primes)? If so, it would follow that for each prime p there is an infinite Mills sequence starting with p. It may be possible to settle the more general question for q1 su ciently large using the original method in [Mills 1947] (also see [Ellison and Ellison 1985, p. 31]). Of course, if the more general question is false, it may be possible to prove it so with a numerical example. In [Weisstein 2005] it is reported that a number θ slightly larger than 1.3 works in the Mills theorem. This has not yet been rigorously proved, so a research problem is to prove this conjecture.

1.78. Is there a real number θ > 1 such that the sequence ( θn ) consists entirely of primes? The existence of such a θ seems unlikely, yet the authors are unaware of results along these lines. For θ = 1287/545, the integer parts of the first 8 powers are 2, 5, 13, 31, 73, 173, 409, 967, each of which is prime. Find a longer chain. If an infinite chain were to exist, there would be infinitely many triples of primes p, q, r for which there is some α with

are unaware of such a result. It is known that there are infinitely many pairs of primes p, q of the form p = α , q = α2!; this result is in [Balog 1989].

1.79. For a sequence A = (an), let D(A) be the sequence (|an+1 − an|). For P the sequence of primes, consider D(P), D(D(P)), etc. Is it true that each of these sequences begins with the number 1? This has been verified by Odlyzko for the first 3 · 1011 sequences [Ribenboim 1996], but has never been proved in general.

1.80. Find large primes of the form (2n +1)/3, invoking possible theorems on allowed small factors, and so on. Three recent examples, due to R. McIntosh, are

p = (242737 + 1)/3, q = (283339 + 1)/3, r = (295369 + 1)/3.

These numbers are “probable primes” (see Chapter 3). True primality proofs have not been achieved (and these examples may well be out of reach, for the foreseeable future!).

1.81. Candidates Mp = 2p − 1 for Mersenne primes are often ruled out in practice by finding an actual nontrivial prime factor. Work out software for finding factors for Mersenne numbers, with a view to the very largest ones accessible today. You would use the known form of any factor of Mp and sequentially search over candidates. You should be able to ascertain, for example, that

460401322803353 | 220295923 − 1.

On the issue of such large Mersenne numbers; see Exercise 1.82.

1.82. In the numerically accessible region of 220000000 there has been at least one attempt at a compositeness proof, using not a search for factors but the Lucas–Lehmer primality test. The result (unverified as yet) by G. Spence is that 220295631 − 1 is composite. As of this writing, that would be a “genuine” composite, in that no explicit proper factor is known. One may notice that this giant Mersenne number is even larger than F24, the latter recently having been shown composite. However, the F24 result was carefully verified with independent runs and so might be said still to be the largest “genuine” composite.

These ruminations bring us to a research problem. Note first a curious dilemma, that this “game of genuine composites” can lead one to trivial claims, as pointed out by L. Washington to [Lenstra 1991]. Indeed, if C be proven composite, then 2C −1, 22C −1 −1 and so on are automatically composite. So in absence of new knowledge about factors of numbers in this chain, the idea of “largest genuine composite” is a dubious one. Second, observe that if C ≡ 3 (mod 4) and 2C + 1 happens to be prime, then this prime is a factor of 2C − 1. Such a C could conceivably be a genuine composite (i.e., no factors known) yet the next member of the chain, namely 2C − 1, would have an explicit factor. Now for the research problem at hand: Find and prove composite some number C ≡ 3 (mod 4) such that nobody knows any factors of C (nor is it easy to find them), you also have proof that 2C + 1 is prime, so you also know thus an explicit factor of 2C − 1. The di cult part of this is to be able to prove primality of 2C + 1 without recourse to the factorization of C. This might be accomplished via the methods of Chapter 4 using a factorization of C + 1.

1.83. Though it is unknown whether there are infinitely many Mersenne or Fermat primes, some results are known for other special number classes. Denote the n-th Cullen number by Cn = n2n + 1. The Cullen and related numbers provide fertile ground for various research initiatives.

One research direction is computational: to attempt the discovery of prime Cullen numbers, perhaps by developing first a rigorous primality test for the Cullen numbers. Similar tasks pertain to the Sierpi´nski numbers described below.