Another Look At Elliptic Curves

It was on the 24th February 2021 that I made a post title Elliptic Curves and I've not made mention of them since despite their importance in high level mathematics. However, I was reminded of them again by a recent article in Quanta Magazine titled Big Advance on Simple-Sounding Math Problem Was a Century in the Making on the 14th October 2024. The article begins:

One morning last November, the mathematician Hector Pasten(opens a new tab) finally solved the problem that had been dogging him for more than a decade by using a time-tested productivity hack: procrastination.

He was supposed to be writing a final exam for his number theory class at the Pontifical Catholic University of Chile in Santiago. To avoid the task, he started pondering, for the umpteenth time, one of his favorite sequences: 2, 5, 10, 17, 26 and so on, the list of all numbers of the form \(n^2 + 1\) (where \(n\) is a whole number).

Mathematicians have used this sequence for over a century to probe the fraught relationship between addition and multiplication, a tension that lies at the heart of number theory. Fundamental problems about multiplication — about, say, how numbers factor into primes — suddenly become much deeper and more challenging as soon as addition enters the picture. One of math’s biggest open questions, for example, asks whether every even number larger than 2 is the sum of two primes; another asks whether there are infinitely many pairs of primes that differ by only 2, such as 11 and 13.

The \(n^2 + 1\) sequence offers a good starting point for investigating the relationship between addition and multiplication, because it combines one of the simplest types of multiplication (squaring a number) with one of the simplest types of addition (adding 1). That doesn’t mean the sequence itself is simple. Mathematicians still can’t answer elementary questions about it, such as whether it contains infinitely many primes. “It doesn’t take far to get to the boundary of our knowledge,” said Andrew Granville(opens a new tab) of the University of Montreal. When mathematicians do manage to shift this boundary even a little, the techniques they develop often illuminate much broader questions about addition and multiplication.

Pasten was trying to show that the numbers in the sequence must always have at least one prime factor that is fairly large. On the morning when he should have been writing his final exam, he finally succeeded, by figuring out how to embed information about the prime factors of \(n^2 + 1\) in the structure of an equation called an elliptic curve.

Over lunch that day, he described his proof to his wife, the mathematician Natalia Garcia-Fritz(opens a new tab). Given the surprising strength of his result, she “told me that I should probably check this many times,” Pasten said. “That afternoon I did so, and the theorems were still there.”

The elliptic curve that Pasten was focused on is \(y^2 = x^3 + 3x + 2n \) with the variable \(n\) being any integer. Curves of the form:$$ y^2=x^3+Ax+B$$have a discriminant given by:$$ \Delta=-16(4A^3+27B^2)$$Thus when \(A=3\) and \(B=2n\) we have:$$ \Delta=16 \times 108 (1+n^2)$$It was the \(n^2+1\) factor in the discriminant that caught Pasten's attention. The level of mathematics that he then applied to make his discovery is beyond my comprehension unless I take the Krell Brain Boost but here is an excerpt from the article that explains what he found:

Applying his theory of Shimura curves to this particular elliptic curve, he could show that the product of the exponents of \(n^2 + 1\) must be fairly small. This didn’t necessarily mean that all the exponents must be small, but it gave him enough control over them to be able to bring in Stewart and Yu’s older method from transcendence theory. By using the two techniques together, he was able to prove that the largest prime factor of \(n^2 + 1\) must be at least about \( (\log(\log n))^2\) — the square of the estimate Chowla and Mahler discovered in the 1930s. Pasten’s new growth rate is much higher than the previous record, though mathematicians suspect the true growth rate is higher still.

Figure 1 shows what the curve looks like when \(n=7\):

Figure 1

In my earlier post on Elliptic Curves, I mentioned that SageMath provides a way of finding the integer values of \(x\) and \(y\) that satisfy the equation of elliptic curves of the form \(y^2=x^3+ax+b\), namely:

E=EllipticCurve([a,b])

E.integral_points()

In the case of \(n=7\) there is only the point where the curve \(y^2 = x^3 + 3x + 14 \) crosses the \(x\)-axis and thus \(x=-2\) and \(y=0\). However, other values of \(n\) yield more points. For example, if \(n=11\) we have the curve \(y^2 = x^3 + 3x + 22 \) and the points:$$(2 : \pm \, 6) \text{ and } (6 : \pm \, 16)$$Figure 2 shows the graph of this curve:

Figure 2

So that's just a brief refresher on elliptic curves and a reminder of how important they are in higher mathematics.

More On Digit Equations

I've posted before about rendering numbers as digit equations, specifically:

• Forming Equations from the Digits of a Number on the 14th March 2024
• The Number Plate Game on the 29th June 2024
• Forming Digit Equations: A Game on the 5th August 2024

Yesterday the number 27586 caught my attention because it can easily be rendered as a digit equation, viz.:$$27586 \rightarrow 2+7+5=8+6$$but it has the special quality that no digits are repeated. This got me wondering what other five digit numbers have this property. In the range of numbers from 27586 to 40000, there are 371 numbers that satisfy the two criteria:

• number has no repeating digits
• number can be split into two parts and a digit equation formed from the sum of the digits on either side of the equal sign

Here are the numbers (permalink):

27586, 27603, 27630, 27801, 27810, 28019, 28037, 28046, 28064, 28073, 28091, 28109, 28136, 28145, 28147, 28154, 28156, 28163, 28165, 28174, 28190, 28307, 28316, 28349, 28361, 28367, 28370, 28376, 28394, 28406, 28415, 28451, 28459, 28460, 28495, 28514, 28541, 28569, 28596, 28604, 28613, 28631, 28640, 28679, 28697, 28703, 28730, 28901, 28910, 29038, 29047, 29056, 29065, 29074, 29083, 29137, 29146, 29148, 29157, 29164, 29173, 29175, 29184, 29308, 29317, 29368, 29371, 29380, 29386, 29407, 29416, 29461, 29470, 29478, 29487, 29506, 29560, 29605, 29614, 29641, 29650, 29704, 29713, 29731, 29740, 29803, 29830, 30126, 30148, 30159, 30214, 30216, 30241, 30249, 30416, 30418, 30425, 30429, 30452, 30461, 30517, 30519, 30526, 30562, 30571, 30618, 30627, 30645, 30654, 30672, 30681, 30719, 30728, 30746, 30764, 30782, 30791, 30829, 30847, 30856, 30865, 30874, 30892, 30948, 30957, 30975, 30984, 31026, 31048, 31059, 31206, 31260, 31408, 31426, 31462, 31480, 31509, 31527, 31572, 31590, 31628, 31682, 31729, 31756, 31765, 31792, 31857, 31875, 31958, 31967, 31976, 31985, 32014, 32016, 32041, 32049, 32104, 32106, 32140, 32160, 32401, 32409, 32410, 32418, 32481, 32490, 32519, 32546, 32564, 32591, 32647, 32674, 32748, 32784, 32849, 32867, 32876, 32894, 32968, 32986, 34016, 34018, 34025, 34029, 34052, 34061, 34106, 34108, 34126, 34160, 34162, 34180, 34205, 34209, 34218, 34250, 34281, 34290, 34502, 34520, 34601, 34610, 34658, 34685, 34759, 34768, 34786, 34795, 34869, 34896, 35017, 35019, 35026, 35062, 35071, 35107, 35109, 35127, 35170, 35172, 35190, 35206, 35219, 35246, 35260, 35264, 35291, 35602, 35620, 35701, 35710, 35769, 35796, 35879, 35897, 36018, 36027, 36045, 36054, 36072, 36081, 36108, 36128, 36180, 36182, 36207, 36247, 36270, 36274, 36405, 36450, 36458, 36485, 36504, 36540, 36702, 36720, 36801, 36810, 37019, 37028, 37046, 37064, 37082, 37091, 37109, 37129, 37145, 37154, 37156, 37165, 37190, 37192, 37208, 37248, 37280, 37284, 37406, 37415, 37451, 37459, 37460, 37468, 37486, 37495, 37514, 37541, 37569, 37596, 37604, 37640, 37802, 37820, 37901, 37910, 38029, 38047, 38056, 38065, 38074, 38092, 38146, 38157, 38164, 38175, 38209, 38245, 38249, 38254, 38267, 38276, 38290, 38294, 38407, 38416, 38425, 38452, 38461, 38469, 38470, 38496, 38506, 38524, 38542, 38560, 38579, 38597, 38605, 38614, 38641, 38650, 38704, 38740, 38902, 38920, 39048, 39057, 39075, 39084, 39147, 39156, 39158, 39165, 39167, 39174, 39176, 39185, 39246, 39264, 39268, 39286, 39408, 39417, 39426, 39462, 39471, 39480, 39507, 39516, 39561, 39570, 39615, 39624, 39642, 39651, 39705, 39714, 39741, 39750, 39804, 39840

I've looked at the first number in this sequence so let's look at the last:$$39840 \rightarrow 3+9=8+4+0$$The sequence will eventually terminate because the largest number possible will contain all ten digits but what might this number be? For a start it can't contain all the digits from 0 to 9 because the sum of these digits is 45 and can't be divided into two equal parts. So we have to drop the 1 if looking for the largest possible number. I think the largest possible number is 985647320 where we have:$$985764320 \rightarrow 9+8+5=7+6+4+3+2+0$$It's also possible to swap the 5 on the left with the 32 on the right so that we get 983276540 so that we have:$$983276540 \rightarrow 9+8+3+2=7+6+5+4+0$$However, this number is smaller than the previous and so it is not the largest possible. What about five digit numbers that satisfy the following criteria:

• number has no repeating digits
• number can be split into two parts and a digit equation formed from the product of the digits on either side of the equal sign

In this case, between 27586 and 40000, there are only 38 numbers that qualify (permalink):

29136, 29163, 29316, 29361, 29613, 29631, 31426, 31462, 31629, 31692, 31846, 31864, 32649, 32694, 34126, 34162, 34216, 34261, 34612, 34621, 34689, 34698, 36129, 36192, 36219, 36249, 36291, 36294, 36489, 36498, 36912, 36921, 38146, 38164, 38416, 38461, 38614, 38641

Let's take the first number in this list, 29136, where we have:$$29136 \rightarrow 2 \times 9 = 1 \times 3 \times 6$$The last number in the list is 38641 where we have:$$38641 \rightarrow 3 \times 8 = 6 \times 4 \times 1$$Other variations on this theme are possible such as using the sum of squares of the digits. For example, let's propose the criteria:

• number has no repeating digits
• number can be split into two parts and a digit equation formed from the sum of the squares of the digits on either side of the equal sign

These criteria yield 48 numbers in the range from 27586 to 40000. These are (permalink):

27614, 27641, 27658, 27685, 27869, 27896, 28769, 28796, 29067, 29076, 29607, 29670, 29706, 29760, 30267, 30627, 31857, 31875, 32067, 32607, 32670, 34517, 34571, 35417, 35471, 36027, 36207, 36245, 36254, 36270, 36425, 36452, 36524, 36542, 38157, 38175, 39158, 39185, 39457, 39475, 39518, 39547, 39574, 39581, 39745, 39754, 39815, 39851

Let's take the first number in the list, 27614, as an example:$$27614 \rightarrow 2^2+7^2 = 6^2+1^2+4^2$$The last number in the list, 39851, can be split as follows:$$39851 \rightarrow 3^2+9^2=8^2+5^2+1^2$$More variations are possible of course but that will do for now. Forming digit equations from the digits of a number falls most definitely into the realm of recreational mathematics and is base-10 specific but it's an interesting mental exercise and what's wrong with having fun with numbers anyway.

These types of mental exercises, especially for children, can serve as an easy entry point to number theory after which they can embark on a deeper exploration of number properties that are not base-specific and that are intrinsic to the number itself, such as primeness and the sum of a number's divisors relative to the number itself that determines whether it is deficient, perfect or abundant.

Record Runs Involving Home Primes

I just happened to notice that there is a run of six consective numbers from 27597 to 27602 inclusive that are only one step removed from their home primes. The only longer run up to one million that occurs is a run of eight numbers from 45162 to 45169.

Here is a permalink for finding these runs. Currently I'm 27586 days old and so the coming record run is not far off. Here are the factorisations for the numbers and the home primes associated with them.

• \(27597 = 3 \times 9199 \rightarrow 39199\)
  
  
• \(27598 = 2  \times13799 \rightarrow 213799 \)
  
  
• \(27599 = 11 \times 13 \times 193 \rightarrow 1113193 \)
  
  
• \(27600 = 2^4 \times 3 \times 5^2 \times  23 \rightarrow 222235523 \)
  
  
• \(27601 = 7 \times 3943 \rightarrow 73943 \)
  
  
• \(27602 = 2 \times 37 \times 373 \rightarrow 237373 \)

The run of eight numbers is listed below together with factorisations and home primes:

• \(45162 = 2 \times 3^2 \times 13 \times 193 \rightarrow 23313193 \)
  
  
• \(45163 = 19 \times 2377 \rightarrow 192377 \)
  
  
• \(45164 = 2^2 \times 7 \times 1613 \rightarrow 2271613 \)
  
  
• \(45165 = 3 \times 5 \times 3011\rightarrow 353011 \)
  
  
• \(45166 = 2 \times 11 \times 2053 \rightarrow 2112053 \)
  
  
• \(45167 = 31^2 \times 47 \rightarrow 313147\)
  
  
• \(45168 = 2^4 \times 3 \times 941 \rightarrow 22223941 \)
  
  
• \(45169 = 17 \times 2657 \rightarrow 172657 \)
  

All concatenations are applied to the prime factors in order from lowest to highest. Of course, runs formed by numbers that are concatenations of prime factors from highest to lowest are impossible because every second number is even with a smallest factor of 2. Every second concatenated number will thus be even as well.

Binary Complement

I hadn't really heard of the binary complement or 1's complement as it's alternatively called until I came across one of the sequences of which 27583 is a member. The sequence is OEIS A323067:

A323067    Primes whose binary complement (A035327) is a square.

To find the binary complement of a number, simply the invert the 0's and 1's in its binary representation. Let's use 27583 and its binary representation as an example:$$27583_{10}=110101110111111_2$$It's binary complement is:$$1010001000000_2=1584_{10}=72^2$$Thus 27583 does belong in OEIS A323067 and Table 1 shows the members of the sequence up to 40,000 together with their binary complements and square roots:

Table 1: permalink

Notice that powers of 2 plus 1 will all have binary complements equal to 0. All decimal numbers have a 1's complement or binary complement. However, it is not 1-to-1 relationship as different numbers can have the same complement. This is different to the Gray Code that also involves manipulation of the binary digits of a number to produce a new number but involves a 1-to-1 relationship.

The following source gives more information about the topic and its applications:  https://www.tutorialspoint.com/one-s-complement

The Prime Constant And Beyond

A recent Numberphile video informed me of the prime constant which firstly incodes all the primes within an infinitely long sequence of 0's and 1's as shown below:$$01101010001010001010001000001010000010001010001 \dots$$The leading zero corresponds to 1 which is not prime and the next two 1's correspond to 2 and 3 that are prime and so on. The next step is to add a decimal point in front of the leading zero to get:$$0.01101010001010001010001000001010000010001010001 \dots$$This represents a number between 0 and 1 and if we interpret this as a number in base 2 we have a constant that can be expressed as (permalink):$$ \begin{align} \text{Prime Constant } &= \frac{0}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4} + \dots \\ \\&\approx 0.414682509851112 \dots \end{align}$$Of course, we could equally well create the non-prime constant by representing every non-prime by a 1 and every prime as a 0. This gives:$$ \text{Non-Prime Constant } \approx 0.585317490148888 \dots$$The prime constant and the non-prime constant of course add to 1. The same idea can be applied to create other constants, for example a Fibonacci constant. The Fibonacci numbers are:$$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, \dots $$This generates a series of 0's and 1's as follows:$$ 11101001000010000000100000000000010000 \dots$$It can be seen that the three leading 1's are there because 1, 2 and 3 are Fibonacci numbers whereas 4 is not and so it represented by a 0 etc. This gives (permalink):$$ \text{Fibonacci Constant } \approx 0.910278797207866 \dots$$It can thus be seen that all monotonically increasing sequences like the sequence of prime numbers and the sequence of Fibonacci numbers can be represented by a constant between 0 and 1. The base 2 representation is arbitrary but simple but other bases could be used especially to represent more than one sequence. For example, base 4 could be used to represent primes, square-free semiprimes and sphenic numbers. So a 3 could represent a square-free semiprime, a 2 could represent a sphenic number, a 1 could represent a prime and 0 could represent a number that is none of these. This leads to another constant, let's call it the 1-2-3 Factor Constant, where we have a series of 0's, 1's, 2's and 3's:$$011012100210122010102210020013102220 \dots$$These again are converted to a number between 0 and 1:$$0.011012100210122010102210020013102220 \dots$$Interpreting this as a number in base 4 leads to (permalink):$$ \begin{align} \text{1-2-3 Factor Constant } &= \frac{0}{4^1}+\frac{1}{4^2}+\frac{1}{4^3} + \dots \\  \\ &\approx 0.0796530489502776 \dots \end{align}$$There is endless fun to be had in generating these sorts of constants from multiple sequences.

Magic Numbers

Many stable atoms have ‘magic numbers’ of protons and neutrons − 75 years ago, two physicists discovered their special properties

Published: October 7, 2024 7.55pm BST

This is the article that alerted me to the existence of these so-called "magic numbers".

The word magic is not often used in the context of science. But in the early 1930s, scientists discovered that some atomic nuclei – the center part of atoms, which make up all matter – were more stable than others. These nuclei had specific numbers of protons or neutrons, or magic numbers, as physicist Eugene Wigner called them.

The race to figure out what made these nuclei so stable began. Understanding these magic numbers would allow scientists to predict the properties of other nuclei, such as their mass or how long they are expected to live. With that, scientists could also predict which combinations of protons and neutrons can result in a nucleus.

The solution to the puzzle came in 1949 from two directions simultaneously. In the U.S., physicist Maria Goeppert Mayer published an explanation, at the same time as a group of scientists led by J. Hans D. Jensen in Germany found the same solution.

For their discovery, the two physicists each got a quarter of the 1963 Nobel Prize in physics. We’re two nuclear scientists whose work is built on Goeppert Mayer’s and Jensen’s discoveries 75 years ago. These magic numbers continue to play an important role in our research, only now we can study them in nuclei that live for just a fraction of a second.

Stability in the atom

The atom is a complex system of particles. It’s made up of a central nucleus consisting of protons and neutrons, called nucleons, with electrons orbiting around the nucleus.

Nobel prize-winning physicist Niels Bohr described these electrons in the atom as existing in a shell structure. The electrons circulate around the nucleus in particular energy levels, or orbits. These orbits have specific energies, and each orbit can hold only so many electrons.

Chemical reactions result from interactions between the electrons in two atoms. In Bohr’s model, if an electron orbit is not already filled, then it’s easier for the atoms to exchange or share those electrons and induce chemical reactions.

The Bohr model of the atom.

One class of elements, the noble gases, hardly ever react with other elements. In noble gases, the electrons occupy completely filled orbits, and as a result the atoms greedily hold onto their electrons instead of sharing and undergoing a chemical reaction.

In the 1930s, scientists wondered whether protons and neutrons might also occupy orbits, like electrons. But nobody could show this conclusively. For more than a decade, the scientific community was unable to describe the nucleus in terms of individual protons and neutrons. Scientists used a more simplified picture, one that treated protons and neutrons as one single system, like a drop of water.

In 1949, Goeppert Mayer and Jensen developed the so-called shell model of the nucleus. Protons and neutrons occupy particular orbits, analogous to electrons, but they also have a property called spin – similar to a spinning top. Goeppert Mayer and Jensen found that when combining the two properties in their calculations, they were able to reproduce the experimental observations.

Through some experiments, they found that nuclei with certain magic numbers of neutrons or protons are unusually stable and hold onto their nucleons more than researchers previously expected, just like how noble gases hold onto their electrons.

The magic numbers known to scientists are 2, 8, 20, 28, 50, 82 and 126. They are the same for both protons and neutrons. When a nucleus has a magic number of protons or neutrons, then the particular orbit is filled, and the nucleus is not very reactive, similar to the noble gases.

For example, the element tin has a magic number of protons. Tin always has 50 protons, and its most common isotope has 70 neutrons. Isotopes are atoms of the same element that have a different number of neutrons.

There are nine other stable isotopes of tin that can exist – it’s the element with the largest number of stable isotopes. A stable isotope will never spontaneously change into a different element, which is what happens to radioactive isotopes.

Helium, with two protons and two neutrons, is the lightest “doubly magic” nucleus. Both its neutron count and its proton count are a magic number. The forces that hold the helium-4 nucleus together are so strong that it’s impossible to attach another proton or neutron. If you tried to add another proton or neutron, the resulting atom would fall apart instantaneously.

On the other hand, the heaviest stable nucleus in existence, lead-208, is also a doubly magic nucleus. It has magic numbers of 82 protons and 126 neutrons.

Many stable isotopes have magic numbers of protons and neutrons.

Examples of magic numbers and stable nuclei exist everywhere – but scientists couldn’t explain them without the introduction of the shell model.

Stable nuclei in nature

The shell structure in nuclei tells researchers about how elements are distributed across the Earth and throughout the universe.

One of the most abundant elements on our planet and in the human body is oxygen, in particular the isotope oxygen-16.

With eight protons and eight neutrons, oxygen-16 has an extremely stable nucleus. A nearby star produced the oxygen we find on Earth through nuclear reactions in its core sometime before the solar system was formed.

Since oxygen nuclei are doubly magic, these nuclei in the star did not interact very much with other nuclei. So more oxygen was left around to eventually act as an essential ingredient for life on Earth.

In her Nobel lecture, Maria Goeppert Mayer talked about the work she did with physicist Edward Teller. The two had attempted to describe how these elements formed in stars. In the 1930s, it was impossible for them to explain why certain elements and isotopes were more abundant in stars than others. She later found that the increased abundances corresponded to nuclei with something in common: They all had magic numbers of neutrons.

With the shell model and the explanation of magic numbers, the production of elements in stars was possible and was published in 1957.

Scientists today continue to use ideas from the nuclear shell model to explain new phenomena in nuclear science. A few accelerator facilities, such as the Facility for Rare Isotope Beams, where we work, aim to create more exotic nuclei to understand how their properties change compared with their stable counterparts.

At the Facility for Rare Isotope Beams, scientists produce new isotopes by accelerating stable isotopes to about half the speed of light and smashing them at a target. Out of the pieces, we select the rarest ones and study their properties.

Possibly the most profound modern discovery is the fact that the magic numbers change in exotic nuclei like the type we create here. So, 75 years after the original discovery, the race to discover the next magic number is still on.

If the sequence of numbers  2, 8, 20, 28, 50, 82,126 is entered into the OEIS, we find A018226 :

A018226     Magic numbers of nucleons: nuclei with one of these numbers of either protons or neutrons are more stable against nuclear decay.

The OEIS comments state:

"The results of the experiment indicate that 54Ca's first excited state lies at a relatively high energy, which is characteristic of a large nuclear shell gap, thus indicating that N = 34 in 54Ca is a new magic number, as predicted theoretically by the University of Tokyo group in 2001. By conducting a more detailed comparison to nuclear theory the researchers were able to show that the N = 34 magic number is equally as significant as some other nuclear shell gaps." Link

So it seems that maybe 34 should be included as well. It's interesting that the discovery was made in my birth year, 1949, now 75 years ago. So the sequence currently may better be represented as:$$2, 8, 20, 28, 34, 50, 82,126$$

My Wordle Statistics

Figure 1 shows a screenshot of my Wordle statistics as of the 6th October 2024, a day on which I turned 75.5 years old.

Figure 1

Even though I've played 629 games, I lost 11 of them because I didn't guess the correct word in six tries or under. In order to accurately assess my average score, these losses need to be taken into account. To do that, I think the best approach is to add the number 7 to the guess distribution and include the 11 losses there. This then gives the following (number, frequency) distribution:$$(\textbf{1},0), (\textbf{2},38),(\textbf{3},196),(\textbf{4},234),(\textbf{5},113),(\textbf{6},37),(\textbf{7},11)$$The arithmetic mean can then be calculated using the formula:$$ \begin{align} \text{mean} &= \frac{\text{total of number} \times \text{frequency}}{\text{frequency total}}\\ &= \frac{2464}{629}\\ &\approx 3.92 \end{align} $$So there we have it. My average is slightly under 4 tries in successfully guessing the correct word. Google's Gemini confirms that the introduction of the 7 is the best way to calculate the average. How do I compare with the rest of the world? Well, there is an interesting site that lists statistics regarding Wordle results worldwide. Figure 2 shows a world map. 

Figure 2: data as of 22nd August 2023

Canberra, Australia, is the global city with the best Wordle average: 3.58 guesses. Sweden is the world’s best country at Wordle, with an average of 3.72. The source used to obtain this data was tweets on Twitter or X as it's now called. Clearly this introduces a huge bias because only people who are keen Wordle players will bother tweeting about their Wordle expertise and there will be a decided bias toward posting impressive success rather than bare wins (six guesses) or losses (failing to identify the word within six guesses). There's also the possibility of straight out cheating.

There's also hard mode and default mode. On every turn in the former you must use all the letters guessed on the previous turn. I haven't set Wordle to hard mode but I always play that way, although occassionally I'll slip up and forget to use a letter that I've already guessed. There's an online site that will generate your average once you input your data. Based on data input to this site, the statistics shown in Figure 3 seem more believable:

Figure 3: source

Once again however, only very dedicated players will be using this site but at least it shows that my average is higher than the national average for Australia. There is a site that lists the results for each day's Wordle and this site is based on actual attempts made that day to the Wordle site. See Figure 4.

Figure 4: source

These results most accurately depict the correct state of affairs I think. I've posted previously about Wordle: Wordle Statistics on 7th of February 2022 and More Wordle Statistics on 8th of February 2022.