Consequences of progress toward the Riemann Hypothesis

Posted on 7 October 2026 by John

The Riemann Hypothesis (RH) is the conjecture that all the zeros of the Riemann zeta function ζ(s) in the critical strip, i.e. the region of the complex plane with real part between 0 and 1, have real part equal to ½.

The Quasi Riemann Hypothesis (QRH) says that there exists a constant θ < 1 such that no zeros of ζ(s) have real part greater than θ. OpenAI has published a paper claiming QRH with θ = 7/8.

The RH is so important to number theory that even partial results can have big consequences. This post will focus on one consequence: the error term in the Prime Number Theorem.

The Prime Number Theorem says that π(x), the number of primes less than x, is asymptotically equal to Li(x). We’d like to know more specifically at what rate π(x) approaches Li(x).

The best known result before the QRH announcement was

$\pi(x) = \operatorname{Li}(x) + {\cal O} \left(x \exp\!\left(-c\,\frac{(\log x)^{3/5}}{(\log\log x)^{1/5}}\right)\right)$

If the QRH holds for some θ, such as OpenAI’s assertion that θ = 7/8,

$\pi(x) = \operatorname{Li}(x) + {\cal O} \left(x^\theta\,\log x\right).$
If RH holds, θ = ½.

Incidentally, you may have seen the Prime Number Theorem stated with x/log(x) rather than Li(x). These two functions are asymptotically equal, so they give the same theorem, if you’re not interested in quantifying the rate of convergence. The function Li(x) gives better error bounds.

Irrationality exponent of π

Posted on 7 October 2026 by John

For a real number x, the irrationality index μ(x) is a way of measuring how well x can be approximated by rational numbers. If x is rational, μ(x) = 1. If x is irrational, μ(x) ≥ 2.

OpenAI recently published a proof that μ(π) = 2. Almost all real numbers have irrationality exponent 2, so the new result says π is typical in this regard. There are numbers proven to have irrationality index greater than 2 (more on that below), but π isn’t one of them.

The irrationality exponent μ(x) is defined as the supremum of the set of values ν such that

$0 < \left| x - \frac{p}{q} \right| < \frac{1}{q^\nu}$

for infinitely many coprime integers p and q with q > 0.

This means that the approximation error for approximating π with a rational number p/q is typically on the order of 1/q², just like most irrational numbers.

There are numbers with higher irrationality exponents. For example, Cahen’s constant C has irrationality exponent 3. This means C is an irrational number that has infinitely many rational approximations p/q with error less than 1/q³.

Fibonacci product

Posted on 16 September 2026 by John

The product of four consecutive Fibonacci numbers equals the product of two consecutive integers.

For example,

3 × 5 × 8 × 13 = 39 × 40.

I ran across this theorem in a note [1] that says “The product of any four consecutive Fibonacci numbers is twice a triangular number.” Since triangular numbers have the form n(n + 1)/2, twice a triangular number is the product of two consecutive integers.

The note also gives a way to find the numbers on the right hand side. We have

F_n F_n+1 F_n+2 F_n+3 = m(m + 1)

where m equals

F_n+1 F_n+2

if n is odd and

F_n F_n+3

if n is even.

In the example at the top, 3 is the 4th Fibonacci number, so n = 4. Since 4 is even, m is the product of the 4th and 7th Fibonacci numbers, i.e. m = 3 × 13 = 39.

Mathematical alchemy

Posted on 4 August 2026 by John

After writing the previous post about metallic ratios, I thought about the analogy to alchemy and the attempt to make precious metals out of base metals.

When can you make one metallic ratio out of another? Can you make the golden ratio out of the lead ratio?

Before we can make gold out of lead, we have to say what lead is.

Defining metallic ratios

The metallic ratios M(n) can be defined several ways. The most interesting definition is the number whose continued fraction representation contains all ns. A more prosaic but more convenient definition is the larger number that equals its reciprocal plus n, which can be found using the quadratic formula.

$M(n) = n + \cfrac{1}{n+\cfrac{1}{n+\cfrac{1}{n+\cdots}}} = \frac{n + \sqrt{n^2 + 4}}{2}$

The golden ratio is M(1), the silver ratio is M(2), and the bronze ratio is M(3).

Gold from silver and bronze?

Can you make the golden ratio out of the silver and bronze ratios? Not by integer arithmetic. The golden ratio involves √5, the silver ratio √2 and the bronze ratio √13. No integer operations on the latter two radicals will produce the former, though you can come arbitrarily close.

Gold from lead

The metallic ratios for n > 3 don’t have standard names, but let’s call M(4) the lead ratio. Can you make the golden ratio out of the lead ratio? Yes you can:

M(1) = (M(4) − 1)/2.

General solution

In general, when can you make M(n) out of M(m)? In abstract terms the question is when the fields

ℚ(√(n² + 4))

and

ℚ(√(m² + 4))

are the same, i.e. when adjoining √(n² + 4) to the rational numbers gives the same field as adjoining √(m² + 4) to the rational numbers. This occurs if and only if

(n² + 4)/(m² + 4)

is the square of a rational number.

Bronze from copper and tin

Can you make bronze out of copper and tin? Yes, if you define M(36) to be the copper ratio and M(393) to be the tin ratio, because

(3² + 4)/(36² + 4) = (1/10)²

and

(3² + 4)/(292² + 4) = (1/109)².

When sine of x degrees equals sine of x radians

Posted on 22 July 2026 by John

Ordinarily the sine of x radians and the sine of x degrees are very different numbers. Having your calculator in radian mode when it should be in degree mode, or vice versa, results in a major error.

But sometimes it doesn’t matter. A trivial example is when x = 0. A more interesting example is

x = 180π/(180 + π) = 3.08770208….

For that value of x,

sin(x) = sin(x°).

In this article I’ll use the common convention of using radians by default and denoting degrees with ° as above.

Note that

x = πx°/180

and so we are interested in solutions to the equation

sin(x) = sin(πx/180)

Now two angles A and B have the same sine if they differ by a multiple of 2π, or if they’re supplementary (i.e. A = π − B), or both. To put it another way, if A and B have the same sine, they are either equal mod 2π or supplementary mod 2π. This means that

sin(x) = sin(πx/180)

if and only if

x = πx/180 + 2πk

x = π − πx/180 + 2πk

for some integer k.

Therefore all solutions have the form

x = 360πk/(180 − π)

x = 180π(2k + 1)/(180 + π).

Alternative solution

The derivation above is correct, but it occurred to me later that a simpler argument would be to use the identity

sin(A) − sin(B) = 2 cos((A + B)/2) sin((A − B)/2).

Thus A and B have the same sine if

cos((A + B)/2) = 0

or if

sin((A − B)/2) = 0.

These two possibilities correspond to the two families of solutions above.

Density

When reduced modulo 2π, both families are dense in [0, 2π]. This means that for every y in [−1, 1], there is a number x such that

sin(x) = sin(x°) ≈ y

and we can make the approximation as good as we’d like.

Example 1

For example, today is July 22, so let’s set y = 0.722. We’d like to find a value of x such that the sine of x radians and the sine of x degrees both approximately equal 0.722. And let’s say our approximation tolerance is ε = 0.0001.

We can search for a value of x in the first family of solutions by looking for a value of k with

| sin(360πk/(180 − π)) − 0.722 | < 0.0001

and the smallest such k is 96343 and so

x = 360×96343 π/(180 − π) = 616093.78713621…

will do, and sin(x) = 0.72191…

Example 2

Now let’s set y = 0.2026 and look for a solution in the other family of solutions, and this time let’s set ε = 10⁻⁶. The smallest value of k such that

| sin(180π(2k + 1)/(180 + π)) − 0.2026 | < 10⁻⁶

is k = 741141. Then

sin( 4576848.310950611 ) = sin( 4576848.310950611° ) = 0.202600139…

Sum of low squares

Posted on 19 July 2026 by John

Squares, high and low

Let p be an odd prime number. Then half the numbers from 1 through p − 1 are squares and half are not. That is, for half of numbers 1 ≤ k < p, the equation

x² = k mod p

has a solution. The traditional name for these numbers is “quadratic residues” but we can just say “squares” if the context is clear. So, for example, the numbers 1, 2, and 4 are squares mod 7, and the numbers 3, 5, and 6 are not.

If k is a square mod p we will call is a low square if 0 ≤ k < p/2 and a high square if p/2 < k < p.

Signatures

Now let p > 3 be a prime congruent to 3 mod 4. Add up all the low squares mod p and take the remainder mod p. Call this the signature of p. Here’s Python code to make this explicit.

from sympy import isprime, factorint, is_quad_residue

def signature(p):
    assert(p > 3)
    assert(isprime(p))
    assert(p % 4 == 3)
    s = 0
    for k in range(1, 1 + p//2):
        if is_quad_residue(k, p):
            s += k
    return s % p

Inverse signatures

Surprisingly, the signature of each p is unique. Given the signature of p, you can uniquely determine p, and in fact you can do so easily. I ran across this in a paper [1] that presented the results in the form of a parlor trick: have someone pick a prime p such that p = 3 mod 4 and ask them to compute its signature, the sum of the low squares mod p. Then you can quickly tell them what their choice of p was.

Given a signature s, the corresponding prime p is the largest prime factor of 16s + 1.

Not only that,

p = (16s + 1)/m

where m is the smallest of the numbers {3, 7, 11, 15} such that the fraction above is a prime number. In term of Python code, both the following functions should invert the signature of p.

def inverse_signature1(s):
    n = 16*s + 1
    return max(factorint(n).keys())

def inverse_signature2(s):
    n = 16*s + 1
    for m in [3, 7, 11, 15]:
        if n % m == 0 and isprime(n // m):
            return n // m

The following code demonstrates that this is the case for numbers less than 1,000.

for n in range(7, 1000, 4):
    if isprime(n):
        s = signature(n)
        assert(n == inverse_signature1(s))
        assert(n == inverse_signature2(s))

[1] David M. Bloom. A Quadratic Residues Parlor Trick. Mathematics Magazine, Vol. 71, No. 3 (Jun., 1998), pp. 201–203.

Progress on Gilbreath’s conjecture

Posted on 11 July 2026 by John

Years ago I wrote about Gilbreath’s conjecture. It’s a simple conjecture; you could explain it to anyone who understands what prime numbers are. See the linked post for a description of the problem.

Gilbreath’s conjecture is simple, but it’s also kinda weird. As I wrote before,

Paul Erdős speculated that Gilbreath’s conjecture is true but it would be 200 years before anyone could prove it. I find Erdős’s conjecture more interesting than Gilbreath’s conjecture.

The conjecture is hard in a way that, say, solving a nasty-looking differential equation is not. Over the last three centuries, mathematics has developed quite a toolbox for solving differential equations. But Gilbreath’s conjecture is just odd enough that it’s not at all clear what kind of tool might be useful in approaching it.

Terence Tao has a new blog post announcing a paper that he and two coauthors wrote on a random model intended to mimic Gilbreath’s calculation on primes. This random model is more sophisticated than the little game Gilbreath was playing, but it’s also much more amenable to analysis by established techniques. Tao’s post gives a heuristic explanation for why Gilbreath’s conjecture is plausible, but then adds

However, it seems well beyond current technology to try to make these heuristics rigorous; even the first step … is far out of reach.

When will the decimals in a/b repeat?

Posted on 27 June 2026 by John

The previous post looked at how many digits are in the reduced fraction for the nth harmonic number. I was curious about how long the cycle of digits in a harmonic number might be.

I wrote about the period length for the digits of fractions almost a decade ago. This post includes code so I can apply it to harmonic denominators.

from sympy import lcm, factorint, n_order

def period(n):
    factors = factorint(n)
    exp2 = factors.get(2, 0)
    exp5 = factors.get(5, 0)
    r = max(exp2, exp5)

    d = n // (2**exp2 * 5**exp5)
    s = 1 if d == 1 else n_order(10, d)
    return (r, s)

This function returns two numbers: r is the number of non-repeating digits at the beginning and s is the length of the repeating part.

The following code

from functools import reduce

def lcm_range(n):
    return reduce(lcm, range(1, n + 1))

print( period( lcm_range(50) ) )

prints (5, 1275120) meaning that 1/lcm(1, 2, 3, …, 49, 50) has five non-repeating digits following by 1,275,120 digits that repeat ad infinitum. And so the decimals in the expansion of H₅₀ have a cycle length of 1,275,120.

Height of harmonic numbers

Posted on 27 June 2026 by John

The previous post looked at writing the harmonic numbers as reduced fractions and estimating the number of digits in the numerator and denominator based on asymptotics. This is a follow up post with plots.

We’ll choose our base b to be 2. And we’ll look at the total number of bits in both the numerator and denominator, which we will use as the height of the fractions.

First, let’s look at the actual and estimated heights, using the estimates from the previous post.

Next let’s look at the difference between the actual and estimated heights.

In the previous post I looked at n = 50, which was kind of a lucky choice, the error being smaller than usual. I had also looked at, but didn’t publish, n = 100, which would be an unlucky choice.

Finally, let’s look at the relative error in the estimates, and plot over a larger range of n.

The error goes to zero, as predicted by the asymptotic estimates. And it goes noisily, which you’d expect since the heights are related to the distribution of primes.

Writing down harmonic numbers

Posted on 26 June 2026 by John

The nth harmonic number is the sum of the reciprocals of the first n positive integers.

H_n = 1 + 1/2 + 1/3 + 1/4 + … + 1/n

The product of all the denominators is n!, so you could write H_n as a fraction

H_n = p/q

where p = n! H_n is an integer and q = n!.

While p/q is a way to write H_n as a fraction, it’s not the most efficient because p and n! will have common factors.

If we write H_n as a reduced fraction, the denominator will be the least common multiple of the integers 1 through n. That number is asymptotically exp(n). That estimate follows from the prime number theorem.

So for large n the denominator will be roughly exp(n), and in base b it would have around

n/log(b)

digits.

The numerator will be exp(n) H_n, and since H_n is asymptotically log(n) + γ, the numerator for large n will be roughly

exp(n) (log(n) + γ)

and will have around

(n + log log(n) ) / log(b)

digits.

Let’s see how well our asymptotic estimates work for n = 50. The 50th harmonic number is

H₅₀ = 13943237577224054960759 / 3099044504245996706400.

This fraction has 23 digits in the numerator and 22 in the denominator. We would have predicted around

(50 + log(log(50)))/log(10) = 22.3

digits in the numerator and

50/log(10) = 21.7

digits in the denominator.

Let’s try a larger example, looking at the 1000th harmonic number in binary. We’ll use the following Python code.

from fractions import Fraction

def bits(n):
    H = sum(Fraction(i, i+1) for i in range(1, n+1))
    p, q = H.numerator, H.denominator
    # subtract 2 because bin returns a string starting with 0b.
    return len(bin(p)) - 2, len(bin(q)) - 2

print(bits(1000))

This returns 1448 and 1438. We would have estimated

(1000 + log(log(1000)))/log(2) = 1445.4

bits in the numerator and

1000/log(2) = 1442.7

bits in the denominator.

Update: See the next post for plots as a function of n.

Number theory