Field

Number — page 2

The whole numbers, and how much structure they turn out to have.

The partition product's coefficients to q¹². A row of series coefficients computed by expanding a product, beside the same numbers obtained another way.

Every partition, hidden in a product

Multiply out one factor for each part size and the coefficient of q to the n is the number of partitions of n. Nothing is being approximated: the product is a bookkeeping device that does the counting by multiplying.

The product of (1 − qᵏ), and what survives at 12. The coefficients of the pentagonal product drawn as signed bars, with the partitions into distinct parts that Franklin's move leaves unpaired.

The terms that cancel almost everything

Multiply out the product of 1 − q, 1 − q², 1 − q³ and so on, and nearly every coefficient is zero. What survives is a single plus or minus one at 1, 2, 5, 7, 12, 15 — and the reason is a way of pairing partitions off so that each pair cancels.

p(n) to 60, against the Hardy–Ramanujan estimate. The number of partitions of each number up to sixty on a logarithmic scale, with the asymptotic estimate drawn over it and the ratio of the two tabulated.

The size of a number with no formula

There is no closed expression for the number of partitions of n. There is an expression for how large it is — with a square root in the exponent and a π in front — and it is accurate enough that rounding a few terms of its refinement gives the exact count.

Every fifth partition count divides, and the rank that says why. A row of partition counts with the ones in a congruence class marked, and a histogram of partitions sorted by rank.

Every fifth one divides

p(4) is 5, p(9) is 30, p(14) is 135, and every partition count at a number leaving four on division by five is divisible by five. Ramanujan read it off a table; the explanation is a way of splitting those partitions into five equal heaps.

The two supplements, and the residue classes that decide them. A table of odd primes with the Legendre symbols of minus one and two beside the residue of p modulo four and modulo eight.

The two supplements, and where the eight comes from

The main law relates two odd primes to each other and says nothing about −1 or about 2. Those two are settled separately, by their own counts, and the answers arrive modulo four and modulo eight — which is a clue about where the whole subject is really taking place.

Multiplication by 3 modulo 11, and the sign of the shuffle. Residues in two rows joined by strings showing where multiplication sends each one, with a strip beneath comparing the sign of the shuffle to the Legendre symbol for every multiplier.

The symbol is the sign of a shuffle

Multiplying every residue modulo p by a fixed number rearranges them. That rearrangement is a permutation, permutations have a sign, and the sign is exactly the Legendre symbol — so a question about squares becomes a question about crossings.

The Gauss sum for 13, added one root at a time. Partial sums of the p-th roots of unity signed by the Legendre symbol, drawn as a walk closing on a point at distance root p from the origin.

One sum, squared two ways

Add the p-th roots of unity, each taken with a plus or a minus according to whether its index is a square. The walk that results closes on a point at distance √p from the origin — and squaring that one number, evaluated two different ways, is the reciprocity law.

Where a congruence decides which primes a form represents, and where it does not. Rows of primes marked by whether each is represented by x squared plus n y squared, with the residue classes that decide it where such classes exist.

Which primes a form takes

A prime is the sum of two squares exactly when it is 1 modulo 4. Change the form slightly, to x² + 27y², and no congruence on p decides it at all — which is where the elementary subject ends and its successor begins.

The primes below 100,000, by remainder mod 4. A bar for each remainder on division by 4, showing how many primes below 100000 leave it. The 2 classes sharing no factor with 4 hold near-equal counts; the rest are empty or hold one prime.

Infinitely many of one kind

Euclid's argument produces a prime nobody had listed, and says nothing about what it looks like. Ask for infinitely many primes ending in 3, or leaving a remainder of 1 on division by 4, and the same construction has to be aimed — and for most targets nobody knows how to aim it.

Five families of primes, counted below 100,000. Five counting curves on logarithmic axes: every prime, the primes one more than a multiple of four, the twin pairs, the primes one more than a square, and the Mersenne primes. Two of the five families are known to be infinite and three are open questions.

Which infinitudes are proved

The primes never stop, and neither — apparently — do the twin pairs, the primes one more than a square, or the Mersenne primes. Three of those four statements are theorems and one is not, and counting the members of each family tells nobody which.

The primes 3 mod 4 against the primes 1 mod 4, out to 30,000. A plot of the difference between the counts of primes in two residue classes, against the bound, showing a persistent lead for one class and where it is lost.

Every class, and in equal shares

Euclid's argument aimed at a residue class reaches some classes and stalls at others. The theorem covering all of them is Dirichlet's, its proof abandons arithmetic entirely for analysis, and what it proves is stronger than infinitude — the classes are equal, though not at any point anybody has counted.

Two sums of reciprocals: 2.89 and climbing against 1.71 and level. Two curves against the logarithm of the bound: the sum of reciprocals of all primes, rising steadily, and the sum over the twin primes, flattening towards a limit.

The sieve that cannot finish

Sifting out the composites is the oldest method in the subject and it has a ceiling nobody has raised. The reciprocals of the twin primes add to a finite number, so no argument that measures thickness can reach them — and the inclusion–exclusion every sieve truncates goes wildly wrong before it goes right.

$The tree as words in two matrices. 6 nodes of the Stern–Brocot tree, each as the word of turns reaching it, the matrix that word multiplies out to, its two columns as fractions, and the mediant of those columns.$

Two matrices that generate the tree

A node of the Stern–Brocot tree is not really a fraction — it is the pair of fractions it lies between. Written as the columns of a matrix, the two turns of the tree become two multiplications, and the determinant that kept everything in lowest terms becomes a property of a product.

Every positive rational, in one sequence. The first 32 terms of Stern's diatomic sequence as bars, with the ratios of consecutive terms beneath. Every ratio is in lowest terms, no two agree, and each term counts the hyperbinary representations of its index.

Every rational in one sequence

The tree lists every positive fraction once and needs a tree to do it. One recursion on the whole numbers lists them in a single row — and each term of it counts something nobody was asking about, which is why the enumeration works.

$13 record approximations in 26 turns. The distance from π to each fraction the descent passes, against its denominator, on logarithmic axes. 13 of them beat every fraction with a smaller denominator.$

The fractions that beat every smaller one

Walking down the tree towards a number produces a sequence of fractions closing in on it. Most of them are steps along the way; a few are the best approximations there are — closer than every fraction with a smaller denominator — and which few is decided by where the turns change direction.

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