The Spiral, and What It Is Not

Write 1, then 2 to its east, and keep going in an anticlockwise square spiral: 3 above 2, then 4 and 5 to the west, then 6 and 7 down, then 8 and 9 across, and outwards for as long as you like. Every whole number gets exactly one cell, and every cell gets exactly one number.

Now mark the cells holding primes.

That is the whole construction. Stanisław Ulam did it on a scrap of paper during a lecture in 1963, and what appeared was not the even haze you would expect from numbers that thin out slowly. It was a set of diagonal streaks.

The one thing this page insists on is the comparison, and it is worth being exact about what it compares. Looking at this picture, you do not assess one diagonal: you scan several hundred of them and stop at the darkest, and the darkest of several hundred lines is well above the average even when the marks were scattered by a coin. So the laboratory draws the same field twice, once with the primes and once with the same number of cells marked from a seed, measures the best line in each the same way, and prints both numbers.

Start with the drawing rather than the primes.

Walk out from the centre in a straight diagonal line, one cell at a time. Each step takes you one ring further out, and each ring is longer than the last by a fixed amount, so the numbers you pass grow at a steady second difference. That is precisely what a quadratic does. Walk along a row instead and you cross the spiral sideways, hopping between rings in a way that has no such formula. Diagonals are special in this picture before the primes are ever mentioned.

So the question becomes: are some quadratics much better at producing primes than others?

They plainly are, for a reason with no depth to it at all. Take a diagonal whose values are all even: every one of them past 2 is composite, so that whole line is blank. Take a diagonal whose values are never divisible by 3, and it has already dodged one of the commonest reasons for a number to be composite, so it should be richer than average. Every small prime that a quadratic avoids pushes its output further towards being prime, and the effect multiplies.

That gets you as far as expecting some variation between diagonals. What it does not get you is a reason for any particular line to run 157 primes long. For that, the honest answer is in layer 6.

Ring $n$ of the spiral holds the numbers from $(2n-1{)}^2+1$ up to $(2n+1{)}^2$, so it holds $8n$ of them, and its four corners are

Those four sequences are the four main diagonals, and every other diagonal is one of the same shape with a different constant. Write $u=x-y$ and $v=x+y$ for the cell at $(x,y)$: the arm heading north-east from it is

and the other three arms are $4n^2+2n+(1+u)$, $4n^2+(1-v)$ and $4n^2\pm 4n+(1+v)$. In every case the leading coefficient is 4 and $n$ is the ring number.

Now put 41 in the centre instead of 1. Every value shifts up by 40, and the north-east arm of the main diagonal becomes $4n^2-2n+41$. Substitute $n=(m+1)/2$ for odd $m$ and that is $m^2+m+41$; the opposite arm gives the same polynomial at even $m$. So the diagonal through the centre of the 41-spiral is Euler's polynomial, read outwards in both directions - which is why that one line is so much darker than anything near it, and why the laboratory offers 41 as a starting value at all.

Euler's polynomial is prime at $n=0,1,2,\dots ,39$. Forty values in a row, no exceptions. At $n=40$ it is 1681, which is $41^2$.

The diagonals. For a spiral centred on $s$ with 1 at ring 0, the number in the cell $(x,y)$ is $s-1+f(n)$ where $n=\max (|x|,|y|)$ and $f$ is one of the four quadratics above, chosen by which arm the cell lies on. Every diagonal of the spiral therefore has the form $4n^2+bn+c$.

Euler's polynomial. $n^2+n+41$ is prime for $0\le n\le 39$ and composite at $n=40$.

No polynomial is prime everywhere. If $f$ is a non-constant polynomial with integer coefficients, there is some non-negative integer $n$ with $f(n)$ not prime.

What is conjectured. For a quadratic $f(n)=a{n}^2+bn+c$ with $a>0$, $\gcd (a,b,c)=1$, $f$ not a product of two integer polynomials, and $a+b+c$ odd, the Hardy-Littlewood Conjecture F predicts that the number of $n\le N$ with $f(n)$ prime is asymptotically

with an explicit constant $C_f$ built from how $f$ behaves modulo each prime, and the counting is of inputs $n$ rather than of values $f(n)$. This is the statement that would explain the picture, and it is not a theorem. Not one instance of it is known - it is not even known that $n^2+1$ is prime infinitely often.

1. Every diagonal is a quadratic in the ring number.

Ring $n\ge 1$ begins at $s_n=(2n-1{)}^2+1$ at the cell $(n,-(n-1))$ and runs anticlockwise through four sides of $2n$ cells. On the east side, the cell $(n,y)$ carries $s_n+(y+n-1)$. Take a cell $(x,y)$ with $u=x-y$ and follow the north-east arm: far enough out the cell is $(n,n-u)$ on that east side, so its number is

If instead $y>x$ the arm runs up the top side, where the cell $(x,n)$ carries $s_n+2n+(n-1-x)$; substituting $x=n+u$ gives the same expression. The other three arms follow by the same substitution on the remaining two sides, with the south-east arm splitting into two cases because the ray $y=-x$, $x>0$ is the seam where each ring closes at $(2n+1{)}^2$ and the next begins. $■$

2. No non-constant integer polynomial is prime at every non-negative integer.

Suppose $f$ has integer coefficients, is non-constant, and $f(n)$ is prime for every $n\ge 0$. Let $p=f(0)$, which is then prime.

For any integers $m$ and $n$, $m-n$ divides $f(m)-f(n)$, since it divides $m^k-n^k$ for every $k$. Take $m=kp$ and $n=0$: then $p\mid f(kp)-f(0)$, and $p\mid f(0)$, so

But $f(kp)$ was assumed prime, and a prime divisible by $p$ is $p$ itself. So $f(kp)=p$ for every $k\ge 0$ - infinitely many solutions of $f(x)-p=0$. A non-zero polynomial has at most $\mathrm{deg}f$ roots, so $f-p$ is identically zero and $f$ is constant, contrary to assumption. $■$

So Euler's run had to stop. The proof says nothing about when, and 40 is a long way from obvious; what it forbids is the thing a reader looking at forty primes in a row is tempted to believe.

The control is the argument, and it controls for the search. Reading this picture means scanning several hundred lines and stopping at the darkest one, and the darkest of several hundred lines is far above the average even when the marks mean nothing. On the 201-cell field centred on 41 the seeded control reaches 21 marks in 113 cells, or 18.6%, against a field average of 10.5%

• that whole excess is the price of looking at every line. The primes reach 157 in 201, or 78.1%. The gap between those two numbers is what this page has to offer: it is real, it is reproducible from the seed in the URL, and it is not a theorem about anything.

Note what the control does not say. The random marks are placed uniformly, so they have no preferred direction, and the streaks you can find in them are alignments rather than structure the spiral imposed. The spiral's contribution is the one layer 5 proved: it turns lines into quadratics, which is what lets a set as structured as the primes concentrate on one at all.

Nothing here is known about the primes. The statement that would explain the picture is Conjecture F, unproved since 1923. Its weakest recognisable consequence, that some quadratic takes infinitely many prime values, is also unproved. Every dense diagonal on this page is a finite observation about a finite field, and would remain one if the field were a billion cells wide.

Why 41. The run of forty is not luck. $n^2+n+c$ is prime for every $n<c-1$ exactly when the imaginary quadratic field of discriminant $1-4c$ has class number one, a criterion of Rabinowitsch from 1913; those discriminants are known, and $c=41$ is the largest value that works. So 41 is the end of that pattern, and that much is settled - by machinery from a different part of mathematics entirely, not by anything visible in the picture.

Be careful with what it settles. It says no larger $c$ keeps $n^2+n+c$ prime all the way to $n=c-2$; it does not say that no other polynomial anywhere has a longer opening run of primes, and this page makes no claim either way about that.

What this page is for. It is a lesson in what a striking image is worth. The spiral is the most reproduced picture in recreational number theory and it is almost always shown without its control, which makes it an argument that cannot be examined. Shown with one, it becomes something better: a real effect, honestly measured, sitting on top of an artefact of the drawing, with no proof of the first and a three-line proof of the second.