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A New Conjecture Related to the Riemann Hypothesis

Jack Good and Bob Churchhouse


The Riemann hypothesis

The Riemann zeta function is defined for R(s) > 1 by the series

ζ s = n = 1 n - s ζ ( s ) = n = 1 n - s

and the definition is completed for all complex values of s by analytic continuation. For example we can continue the function into the larger half-plane R(s) > 0 by means of the equation

ζ s = 1 - 2 1 - s - 1 n = 1 - 1 n - 1 n s ζ ( s ) = 1 - 2 1 - s - 1 n = 1 ( - 1 ) n - 1 n s

we can then prove the functional equation

ζ s = 2 s π s - 1 sin 1 2 s π Γ 1 - s ζ 1 - s ζ ( s ) = 2 s π s - 1 sin 1 2 s π Γ ( 1 - s ) ζ ( 1 - s ) (1)

for 0 < R(s) < 1 and finally we can use this functional equation to complete the definition of ξ(s) over the whole plane.

It follows from (1) that ξ(s) has a simple pole at s = 1 and simple zeros at s = -2, -4, -6,... It also follows that the only other zeros of ξ(s) must all lie in the strip 0 < R(s) < 1, that these must occur in conjugate pairs and that if ρ is a zero, so is 1-ρ. Thus the complex zeros of ξ(s) are symmetric about the line R(s) = ½. In 1859 Riemann [1] conjectured that all these complex zeros actually lie on the line R(s) = ½. This is the celebrated Riemann Hypothesis, it has still not been proved or disproved.

It can be shown that

f t = π - 1 2 it Γ 1 4 + 1 2 it ζ 1 2 + it f ( t ) = π - 1 2 it Γ 1 4 + 1 2 it ζ 1 2 + it

is real when t is real, so if f(t1) and f(t2) have opposite signs then ξ(s) has a zero on the line R(s) = ½ for some t between t1 and t2. In addition we can count the number of zeros of f(t) within various contours by means of integrals of the form

1 2 π i f ' z f z d z 1 2 π i f ' ( z ) f ( z ) d z

It is thus possible to find zeros of ξ(s) on the line R(s) = ½ and sometimes at least to show that such a zero has no others in its vicinity.

Evidence from the zeros

Titchmarsh and Comrie [2] computed the first 1041 zeros by hand in 1936 and found that they all lay on the line R(s) = ½. In recent years a number of people have used computers to find many more zeros and in 1966 Rosser and Schoenfeld [3] announced that the first 2,000,000 complex zeros of ξ(s) all lie on the line R(s) = ½. It might appear that this result makes it almost certain that the Riemann Hypothesis is true, but this is not necessarily so. The reason for the persistent doubt is that in the theory of the zeta function and in the closely allied theory of the distribution of prime numbers the iterated logarithm, log log x, frequently occurs in asymptotic formulae and this function increases extremely slowly. The first zero off the critical line, if there is one, might have an imaginary part whose iterated logarithm is as large as, say, 10 and, if so, it might never be practicable to find this zero by calculation.

Now the Riemann Hypothesis can be proved to be equivalent to assertions about properties of a number of functions other than the zeta function; indeed some of these functions appear, at first sight, to be totally unconnected with the zeta function. For example, it can be proved that a necessary and sufficient condition for the Riemann Hypothesis to be true is that the sum of the distances between certain rational points in the interval (0, 1) should not be too large (as a function of the number of points). Some of these equivalent assertions possess a superficial attractiveness; on closer examination they seem to be even more formidable than the original Riemann formulation of the problem.

Mertens' conjecture

One of the more approachable equivalents of the Riemann Hypothesis relates to a property of the Mobius Function. The Mobius Function, μ(n), is defined for all integers n ≥ 1 as follows

    μ(1) = 1 
    μ(n) = (-l)k if n is divisible by exactly k distinct primes 
    μ(n) = 0 if n is divisible by the square of a prime.

Thus μ(1) = 1, μ(2) = μ(3) = -1, μ(4) = 0, μ(5) = -1, μ(6) = + 1,...


M N = | n = 1 N μ n | M ( N ) = n = 1 N μ ( n )

then the following theorem is true:

Theorem 1: A necessary and sufficient condition for the Riemann Hypothesis to be true is that

M(x) = O(x½+Έ).

For a proof of this see Titchmarsh [4].

A much stronger conjecture than the one in this theorem is the Mertens Conjecture

M(x) < x½ (x > 1).                 (2)

If Mertens' Conjecture is true then the Riemann Hypothesis is also true, but not necessarily vice-versa. An even stronger conjecture than (2), known as von Sterneck's conjecture

M(x) < ½x½ (x > 1)                (3)

was disproved by Neubauer [5], using a computer in 1963. Neubauer showed that (3) was certainly false somewhere near x = 7.76 X 109.

During 1966 we used the Atlas Computer of the SRC to study the sum M(N) for all N up to about 1O8. The program was run as a background job whenever the Operations Group found it convenient. The work did not entirely overlap Neubauer's because:

  1. we kept a set of statistical counts of the behaviour of M(N) in blocks of values of N, and
  2. Neubauer used a formula of von Sterneck which allowed him to find the value of M(N) in certain intervals without having computed it in all preceding intervals.

We did not, of course, find evidence to refute the Mertens Conjecture but the counts we obtained proved to have such interesting properties that we were led to make an entirely new conjecture, much weaker than the Mertens Conjecture but which if true would imply the truth of the Riemann Hypothesis. Our new conjecture, at the same time, implies that the Mertens Conjecture is false.

Evidence from Atlas and the principal conjecture

From the definition of μ(n) we see that μ(n) = 0 when n is a multiple of 4, so there are zeros in the sequence {μ(n)} at regular intervals of 4, and similarly at regular intervals of 9 and so on. So, if we write vi for the number of values of n (in a block of length N) for which μ(n) = i we would expect v0 tv be very close to its expected value

N - N(1-2-2)(1-3-2)(1-5-2)... = N (1-6π-2)    (4)

We found, in fact, that this estimate seems to provide an astonishingly good fit. For example, the Atlas program told us that up to N = 33,000,000 there are 12,938,407 zero values of {μ(n)} and the formula above tells us that we expect 12,938,405.6. We tested (4) again at N = 25 × 106, 50 × 106, 75 × 106 and 100 × 106 and found that the biggest error was only 27 (at 75 x 106). These, and other, results have led us to make a number of conjectures of which the principal one is given below.

If we now ignore the places where μ(n) = 0 we obtain a sequence {μ'(n)} all of whose terms are +1 or -1. It is clear that, on average, there will be about as many terms of each sign. Now if the Mobius sequence {μ(n)} were a random sequence taking the values +1, 0, -1 with specified probabilities those of +1 and -1 being equal then the condition M(x) = O(x½+Ε) would be true. The same remark is still true if we delete all the zero terms from the sequence {μ(n)} provided we know that, in this sequence {μ(n)} the value + 1 and -1 are randomly distributed each with probability ½.

Now {μ(n)}, {μ'(n)} are not random sequences* but the evidence from the computer is that the behaviour of the sequence {μ'(n)} considered over very large intervals is not very different from what one would expect in a random sequence of this kind. It is therefore tempting to apply the law of the iterated logarithm (for example Feller [6]). This leads to our principal conjecture:

* As an example of the non-randomness of {μ(n)} note that we can predict starting points for long runs of zero in the sequence for {μ(n)}; for if m ≡ - 1(4), ≡ - 2(9), ≡ - 3(25), ... ≡- n(mod pn 2) then the terms μ(m + 1), μ(m + 2),. .., μ(m + n) are all zero. Thus, for example. we can predict that the sequence {μ(n)} must have a run of at least 4 consecutive zeros beginning at all points n ofthe form n = 29,348 + 44,100 K.


Lim sup M x x log log x - 1 2 = 12 π Lim sup M ( x ) x log log x - 1 2 = ( 12 ) π

This conjecture contradicts any Mertens-type conjecture of the form M(x)< Cx½ for any positive constant C and so, taking C = 1, it contradicts Mertens conjecture itself. If our conjecture is true, and it is difficult to obtain numerical evidence either way, then the Riemann Hypothesis would also be true, by the Theorem quoted above.

A more detailed account of this work, with the numerical evidence and several other conjectures is being published elsewhere [7].


1. Riemann, G. F. B. 1953. 'Collected Works, VII.' Dover Publications, New York.

2. Titchmarch, E. C. 1936. The zeros of the Riemann zeta-function, Proc. Roy. Soc. (A) 157, 261-3.

3. Rosser, J. B. and Schoenfeld, L. 1966. The first two million zeros of the Riemann zeta-function are on the critical line, in Abstracts for the Conference of Mathematicians, 8. Moscow.

4. Titchmarsh, E. C. 1951. The Theory of the Riemann Zeta Function,' Oxford, p. 315.

5. Neubauer, G. 1963. Eine empirische Untersuchung zur Mertenschen Funktion, Numerische Math. 5, 1-13.

6. Feller, W. 1968. An Introduction to Probability Theory and its Applications, p. 205, Wiley.

7. Good, I. J. and Churchhouse, R. F. 1968. The Riemann hypothesis and pseudorandom features of the Mobius sequence, Mathematics of Computation, 22, 857--861.