Field

Number — page 4

The whole numbers, and how much structure they turn out to have.
Descartes' number, perfect but for one factor. Five boxes for the prime powers of Descartes' number, each with its divisor sum factored beneath, multiplying to exactly twice the number provided 22021 is treated as a prime.

Perfect but for one factor

In 1638 Descartes wrote to Mersenne with an odd number whose divisors add up to exactly twice the number — provided one of its factors, 22021, is counted as a prime. It is not; it is 19² × 61. Nearly four centuries later that number is still the only odd one of its kind known, and a search of every odd number below ten million finds no other. It is the closest anyone has come to an odd perfect number, and what it shows is how an odd perfect number would have to be built.

How often a prime divides the class number. p 3: 39.1% against 1/p 33.3% and predicted 44.0%; p 5: 22.6% against 1/p 20.0% and predicted 24.0%; p 7: 15.3% against 1/p 14.3% and predicted 16.3%; p 11: 9.1% against 1/p 9.1% and predicted 9.9%.

Class groups drawn at random

The class number of an imaginary quadratic field measures how badly unique factorisation fails there, and it looks arbitrary: 1, 2, 4, 3, 5, 12, 1. Count how often 3 divides it and the answer is not one time in three but noticeably more — heading, by a conjecture of Cohen and Lenstra, for 44 per cent. The reason is that class groups behave like finite groups chosen at random with each group weighted by one over its number of symmetries, so groups with few symmetries, like the cyclic group of order 3, turn up more often than the naive count suggests.

How many different primes divide a number up to ten million. The share of numbers up to 10⁷ with each count of distinct prime factors (6.7%, 25.4%, 36.4%, 23.9%, 6.9%, 0.7%, 0.0%, 0.0%), against the Erdős–Kac normal curve with mean and variance 2.78.

How many primes a typical number has

A typical number near N has about log log N different prime factors, and the count is spread around that in a bell curve whose variance is log log N as well. Both facts are theorems. Neither is visible at any size anyone can count: up to ten million the average is right and the spread is less than half what the limit says.

The parities of the first 1,200 partition numbers. A 40-by-30 grid of p(n) mod 2 for n from 0 to 1199, filled where p(n) is odd; 568 are even.

Is the partition count even half the time?

The number of partitions of n is even for 50.0% of the n up to half a million, its runs of one parity are as long as a coin's, and nothing proves that the share is a half — the best theorems only show there are at least about √n of each. Modulo 5 and 7 the zeros carry Ramanujan's congruences and something more: an excess that follows whether 1 − 24n is a square.

Apéry's proof for ζ(3), as three numbers. log₁₀ of 2·lcm(1, …, n)³, of |a(n)ζ(3) − b(n)| and of their product for n up to 60: slopes 1.262, -1.546, -0.284 per step.

The race that makes ζ(3) irrational

Roger Apéry's 1978 proof that the sum of the reciprocal cubes is not a fraction comes down to a race between two numbers. A whole-number multiplier grows by a factor of ten every 0.79 steps; the gap it multiplies shrinks by a factor of ten every 0.65. The gap wins, by a margin of about seventeen per cent, and that margin is the whole proof.

The tree writes 4/17 as four unit fractions, and a search finds three. Left-neighbour chain 4/17, 3/13, 2/9, 1/5, 0/1, gaps 1/221, 1/117, 1/45, 1/5; three-term decomposition 1/5 + 1/30 + 1/510.

Three unit fractions for every four over n

Two neighbouring fractions in the tree always differ by a unit fraction, so stepping down the tree writes any fraction as a sum of them — four over n in at most four steps. Erdős and Straus asked in 1948 whether three always suffice. Three identities settle every n except those leaving remainder 1 on division by 24, a search settles every one anyone has tried, and nobody has a proof.

Seventy, and the totals its divisors can make. A strip of the totals 0 to 74 marking those reachable as sums of distinct proper divisors of 70; 70 itself and the excess 4 are not.

Abundant, and still not a sum of its parts

Seventy's proper divisors add to seventy-four, more than seventy itself — and yet no selection of them adds to exactly seventy. The reason is that the excess, four, cannot be made from the small divisors one, two and five. Numbers like seventy are called weird; there are thirty-six below twenty thousand, every one of them even, and nobody knows whether an odd one exists.

Twelve's divisors make every total up to twenty-eight. Stacked bars for each total from 1 to 28, each built from distinct divisors of 12.

Divisors that make every amount

Twelve's divisors — 1, 2, 3, 4, 6, 12 — can be chosen to add to every whole number from one to twenty-eight. Numbers like that are called practical, a test on their primes decides them, and they turn out to be counted like the primes: about 1.336 x / log x of them below x. For practical numbers Goldbach's conjecture and the twin conjecture are theorems.

Euclid's argument run twenty times. The first 20 terms of the Euclid–Mullin sequence: 2, 3, 7, 43, 13, 53, 5, 6221671, 38709183810571, 139, 2801, 11, 17, 5471, 52662739, 23003, 30693651606209, 37, 1741, 1313797957.

Euclid's proof run as a machine

Euclid proved there is no last prime by multiplying the primes on any list, adding one, and noting that the result has a prime factor not on the list. Run the proof as a machine — start from 2, and each time take the smallest prime factor of one more than the product so far — and it produces 2, 3, 7, 43, 13, 53, 5, 6221671, … a sequence that never repeats, that reaches small primes late and large ones early, and that nobody can prove reaches every prime.

Each prime of the form 4k + 1 as a point on its own circle. Points (a, b) with a > b > 0 and a² + b² prime, for the 1125 such primes below 20000, filling the sector between 0° and 45°.

Which way a prime's two squares point

A prime one more than a multiple of four is a sum of two squares in exactly one way, and the two squares make a point on a circle. Draw that point for every such prime and ask which way it faces. It faces every way equally — the angles spread evenly over the sector, which Hecke proved by giving each angle a remainder and copying Dirichlet. Measured over seventy-four thousand primes, they are even more evenly spread than random angles would be.

The same bell on numbers, on primes less one, and on squares plus one. Standardised ω histograms: integers ≤ 2·10^7, p − 1 for p ≤ 2·10^7, n² + 1 for n ≤ 2·10^6.

The same bell on thinner sets

A typical whole number has about log log N prime factors, spread in a bell curve. That was a theorem about all numbers, built on the picture of each prime dividing independently. The numbers one less than a prime, and the numbers n² + 1, are far too thin for that picture to have any obvious claim on them, and on both the same bell appears — with means shifted by constants that come straight from how often each small prime divides them. The squares plus one that are prime, the case of a single factor, number 102,205 up to n = two million against 102,302 predicted; whether there are infinitely many is open.

Primes in five thousand thin sectors. 581517 primes in (2e7, 4e7] in 4967 sectors; mean 117.08, variance ratio 0.874, 2 empty.

Sectors that shrink with the primes

Hecke proved that the angles of the primes a² + b² are evenly spread, so every fixed sector eventually gets its share. Let the sector shrink as the primes grow, to width X^(−α), and the question is open beyond small α. Measured on the 581,517 such primes between twenty and forty million, wide shrinking sectors fluctuate far less than chance and thin ones as much as chance — and the thin sectors that go empty are not scattered: they gather beside the directions of small rational slope, most of all slopes like 1/3 and 1/1 whose two terms are both odd.

The primes p for which Euclid's number p# ± 1 is itself prime. p up to 1100: p# + 1 prime for 2, 3, 5, 7, 11, 31, 379, 1019, 1021; p# − 1 prime for 3, 5, 11, 13, 41, 89, 317, 337, 991.

When Euclid's number is itself prime

Multiply the primes up to p and add one. Euclid needed only that the result has a prime factor beyond p; it is itself prime for p = 2, 3, 5, 7 and 11, and then not again until 31, and then not until 379. Up to 1,100 there are nine such primes of each sign, against thirteen that a careful estimate expects. The estimate predicts infinitely many and grows by about 1.8 every time x is multiplied by e; nobody can prove there is even one more.

The first ten Carmichael numbers and the divisions that make them. 561 = 3·11·17; 1105 = 5·13·17; 1729 = 7·13·19; 2465 = 5·17·29; 2821 = 7·13·31; 6601 = 7·23·41; 8911 = 7·19·67; 10585 = 5·29·73; 15841 = 7·31·73; 29341 = 13·37·61.

The numbers that fool every base

561 = 3 · 11 · 17 passes Fermat's test for every base it shares no factor with, and so do 1,105, 1,729 and 102 more numbers below ten million. They were proved infinite only in 1994. Counted to 10²¹ there are about twenty million, growing like x to a power that has crept from 0.21 to 0.35 — just past the bound the proofs guarantee and nowhere near the power of one that Erdős's argument predicts.

Three cycles of forms for the prime 229. Reduced forms of discriminant 229 in 3 cycles of lengths 6, 6, 2; checked: h(5) = h(13) = 1, h(229) = h(257) = 3, h(401) = 5, h(577) = 7, h(1129) = 9.

Real fields that factorise uniquely

Among the fields Q(√−d) only nine have unique factorisation, and Gauss guessed as much. Among the real fields Q(√p) he guessed the opposite — infinitely many — and the guess is still unproved. Counting cycles of reduced quadratic forms for the 16,900 primes p ≡ 1 mod 4 below 400,000 finds 79 per cent with class number one, drifting slowly down towards the 75.45 per cent that Cohen and Lenstra's heuristic predicts — and finds the reason the real fields behave so differently: their units are enormous.

A walk on Gaussian primes stopped by a moat. Step ≤ 2 from 1 + i: 720 primes reached, farthest 17 + 42i at 45.310.

A walk on Gaussian primes stopped by a moat

Start at 1 + i and step from Gaussian prime to Gaussian prime, never more than 2 at a time: the walk reaches 720 primes and stops at 17 + 42i, surrounded by a moat. Allow steps of √10 and it reaches a quarter of a million primes before stopping at distance 1,024; steps of 4 carry it to 4,313. Whether some fixed step lets a walker reach infinity is a question Basil Gordon asked in 1962, and it is open.

The largest prime factor, as a power of the number. Share of n with log P(n)/log n ≤ s: at s=0.5, n≤10⁴ 0.268, n≤10⁷ 0.272, Dickman 0.307; median e^(−1/2) = 0.6065; largest gap at 10⁷ 0.061.

The largest prime in a typical number

Count a number's prime factors and the answer is about log log n. Ask instead how big the largest one is, and the answer is a power of n with a law of its own: half of all large numbers have a prime factor above n^0.6065, and the share whose primes all stay below n^(1/u) is Dickman's ρ(u), a function defined by its own past. The same law governs the longest cycle of a random shuffle, and there it arrives at once; for the numbers themselves, at ten million, it is still visibly on its way.

A partition in two directions is a pile of cubes. Plane partition 5 4 3 1 / 4 3 2 / 3 2 1 / 1 of 29, drawn as stacked cubes; PL(29) = 3759612.

A partition piled into a corner

A partition is a row of numbers that never increases. Let it never increase down the columns too, and it becomes a pile of cubes pushed into the corner of a room. MacMahon counted the piles with one change to Euler's product — each part size k in k colours — and in a box with a product over its cells. The second direction moves the growth from e^(√n) to e^(n^(2/3)), puts ζ(3) where π was, and makes a count that grows with a box's area. One direction further, his guess fails at six.

Forty-two squares in a row. p = 366791: partial sums of (n/p) for n ≤ 200: 0, 10, 20, 30, 40, 46, 54, 60, 64, 72, 78, 78, 88, 92, 98, 102, 108, 116, 122, 126, 132; least non-residue 43.

The first number that is not a square

Modulo a prime, half the numbers are perfect squares. Count up from 1 and the first non-square is usually 2 or 3 — but modulo 366,791 the numbers 1 to 42 are all squares. How far the run of squares can last is one of the oldest open questions about primes: the truth is tiny, the proofs are huge, and the Riemann hypothesis would close most of the gap.

The numbers below a hundred as sums of three cubes. Smallest solution size per n < 100: 1:9 2:1 3:1 6:2 7:2 8:16 9:2 10:2 11:3 12:11 15:2 16:2 17:2 18:3 19:3 20:3 21:16 24:2 25:3 26:3 27:6 28:3 29:3 30:10d 33:16d 34:3 35:3 36:3 37:4 38:4 39:6d 42:17d 43:3 44:8 45:4 46:3 47:8 48:4 51:796 52:11d 53:3 54:3 55:3 56:4 57:4 60:5 61:5 62:3 63:4 64:6 65:4 66:4 69:5 70:21 71:4 72:4 73:4 74:15d 75:9d 78:55 79:35 80:4 81:3 82:14 83:4 84:8d 87:4271 88:5 89:7 90:4 91:4 92:4 93:7 96:22 97:5 98:5 99:4.

Thirty-three as three cubes

Which numbers are a sum of three cubes, allowing negatives? Nothing like the rule for two squares is known — only that numbers leaving 4 or 5 on division by 9 are impossible. Every other number below a hundred has a sum with cubes under 5,000, except eight; 33 needed sixteen-digit cubes and was found in 2019.

The largest primes of n and n + 1, side by side. Joint law of t(n), t(n+1) for 1000≤n<1e7 in 24×24 cells: corner (both near prime) 0 against 44149 expected; other ratios 0.268 to 2.582.

Two neighbours share out their small primes

A number and the next one have no prime in common, and the natural guess is that their largest primes are independent. In the limit that is believed and nearly proved. Below ten million it is visibly false: neighbours are both made of small primes only nine-tenths as often as independence predicts, a run of four only two-thirds as often, and the correlation between their largest primes is negative at distance one and positive at distance six — because a small prime that divides one of two neighbours is a small prime the other cannot have.

Ramanujan's tau, scaled by the power it grows like. τ(n)/n^5.5 for n ≤ 400; τ(1..8) = 1, -24, 252, -1472, 4830, -6048, -16744, 84480; largest |τ(n)|/n^5.5 for n ≤ 10000: 3.2796 at 4841.

The product raised to the twenty-fourth power

Euler's product (1 − q)(1 − q²)(1 − q³)… has the partition numbers for its reciprocal. Raise it to the twenty-fourth power and multiply by q, and the coefficients that come out — 1, −24, 252, −1472, 4830, −6048 — multiply like the primes they are indexed by, stay inside a bound Ramanujan guessed and Deligne proved fifty-eight years later, agree with a divisor sum modulo 691, and know, modulo 23, how a cubic equation factors.

Each prime fixes the next digits of A. A = 1.306377883863080690468614492602605712916…; digits fixed by the primes from 11 to the 254-digit one: 1, 4, 11, 30, 87, 257.

A constant that holds the primes

There is a number A = 1.3063778838… such that A cubed, A to the ninth, A to the twenty-seventh and every further power of three, rounded down, is prime: 2, 11, 1361, 2521008887 and on for ever. W. H. Mills proved in 1947 that such a number exists. It looks like a formula for primes and is the opposite of one. Its digits are read off the primes it produces, each prime fixes about as many digits as it has, and the whole argument stands on a theorem about gaps between primes — which is why the power is 3 and not 2.

Where each count of cubes stops being needed. last needing 9..6: 239, 454, 8042, 1290740; counts 114, 5721, 154467, 941641, 393997, 3922, 121, 15, 2 up to 1500000.

Nine cubes for two numbers

Every whole number is a sum of at most nine positive cubes, and only two numbers, 23 and 239, need all nine. Fifteen need eight, the last of them 454; 121 need seven, the last 8,042; six are needed for the last time at 1,290,740. The count every number needs is nine, fixed by two small accidents. The count every large number needs is somewhere between four and seven, and the computation says four — a question about how the cubes spread out, not about the cubes themselves.

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