The ledger

What the figures prove — page 4

Page 4 of 6, continuing through the fields in the same order. Every claim here was checked as the drawing was made rather than typed beside it.

Number

11 families

`approx`

10 kinds of claim · 5 placements

|qα − p| comes out below 1/N
and every convergent from the first onward breaks a record
every record-breaking fraction is a convergent
neighbouring convergents differ by a determinant of one
so p/q is within one over qN of the number itself
the constant is one this figure knows
the largest denominator is a whole number between 8 and 200
the number of boxes is a whole number between 3 and 14
there are record-breaking denominators to mark
two of the N + 1 points share a box, as they must

`continued`

63 kinds of claim · 24 placements

√34's cycle is as long as its period ×5
and its last term is twice the first term 5 ×5
the cycles of 34 use every reduced state once ×5
the period of √29 before its last term reads the same backwards ×5
the quotients rebuild 34/13 ×3
D is a whole number from 2 to 1000 and not a square ×2
D runs up to between 20 and 300 ×2
the number of terms is a whole number between 2 and 12 ×2
√2 has a convergent above √5, as Hurwitz says every irrational must
√D's own states are exactly one of the cycles, as long as its period
a row of −1 composed with itself gives a row of 1
a step from a reduced state lands on a reduced state
after the first step |k| is below √D
after the first step |k| stays below √D
and needs fewer steps than the convergents do
and that row is the method's own answer
and the first that does is the fundamental solution found by searching
and the last of them has closed on √5
between four and twelve terms are expanded
between two and five constants
D is a whole number between 2 and 1000 and not a perfect square
D is between 2 and 40
D is not a perfect square
D is not a perfect square, or the equation has only the trivial solution
D names the Pell equation and is read only by the pell, chakravala, compose, states and reduced views; the other views take `of`
dilations is between 2 and 6 and is read only by the dilate and ehrhart views
dividing by k gives the method's next row
dividing out k leaves whole numbers
Ds is a short list of equations and is read only by the kwalk, palindrome and cycles views
e has a convergent above √5, as Hurwitz says every irrational must
each convergent is closer than the one before
every constant expanded is one this family knows
every convergent stays close to the hyperbola
every period up to 1,000 is a palindrome ending in twice the first term
every reduced state is above 1 with its conjugate between −1 and 0
every reduced state lies on a cycle
every row is a near miss a² − Db² = k
neighbouring convergents differ by a determinant of one
rs is a short list of whole heights and is read only by the reeve and ehrhart views
some convergent solves the equation exactly
the composition multiplies the two misses
the constant is one this figure knows
the convergents alternate above and below the value
the cyclic method finishes
the cyclic method needs fewer steps than the convergents
the cyclic method reaches the continued fraction's fundamental solution
the denominator is a whole number between 1 and 10000000
the error view compares a list of constants and every other view expands one
the expansion closes a period and finds the solution
the expansion repeats
the golden ratio's convergents never leave the neighbourhood of √5
the method takes at least one step
the numerator is a whole number between 1 and 10000000
the period is never more than the number of reduced states
the recurrence stays in whole numbers
the solution found satisfies the equation
the tower's last convergent is a number
the view is one the family draws
there are enough convergents to draw
upTo is read only by the chakcount and periods views
π has a convergent above √5, as Hurwitz says every irrational must
π has a convergent that beats √5 by two orders of magnitude
φ has a convergent above √5, as Hurwitz says every irrational must

`descent`

11 kinds of claim · 7 placements

the side of the big square is a whole number between 2 and 200 ×2
the side of the small squares is a whole number between 1 and 200 ×2
and the discrepancy never changes size
and the new pair is strictly smaller
every step is strictly smaller than the one before
the descent carries the discrepancy to minus itself
the descent runs out of room before the discrepancy runs out
the descent takes at least three steps
the pair is one the descent can start from
the two small squares overlap and still reach the corners
the two small squares, less their overlap and plus the corners, are the big one

`factor`

77 kinds of claim · 46 placements

every factorisation of 5 has the same number of factors ×4999
the divisor sum of 12 from its factorisation matches the list ×1635
n² + n + 2 is prime for every n up to 0 exactly when 7 is on the list ×78
n² + n + 41 is prime at n = 0 ×40
9 is irreducible because it is two primes of the form 4k + 3 ×32
5 fails Euclid's lemma exactly when it is not an ordinary prime ×28
6 is called what it is ×10
the bar for 6 is its divisors laid end to end ×10
the discs cover the plane exactly for the five Euclidean fields (1) ×9
the lattice for −1 leaves points at most 0.707 from it ×9
the lattice draws every divisor of 30 exactly once ×5
the range is a whole number between 100 and 2000 ×4
9 cannot be split into two numbers of the form 4k + 1 ×3
9 is a Hilbert number ×3
3 leaves remainder 3 on division by 4 ×2
496 is perfect ×2
the number is a whole number between 4 and 100000 ×2
441 has exactly two factorisations into Hilbert irreducibles
441 is 3 × 3 × 7 × 7
6 and 220 are classified as what they are
6 and 28 sit on the line
a held term leads to a term at least as large that is again held by the perfect number
a proper multiple of a perfect or abundant number is abundant
a term held by the perfect driver never shrinks
a visible share of the range sits just above one
and at n = 40 it is 41²
and so do the norms of two and three
and the next term is again the perfect number times an odd number
below 1000, the sequences that pass 10^22 are the twelve undecided ones and 840
between one and eight whole numbers above one
between one and five starting values from 2 to ten million
both first splits divide the number
both trees end in the same multiset of primes
every divisor including the number itself adds to twice the number
every one of these monoids has an element with two factorisations
every sequence is decided or passes the bound within the step budget
for 4k + 1 it is 441
in fact at most half the divisor's norm, because the square lattice is covered by discs of radius √2/2
no divisor is drawn twice
no element of the ring has norm 2 or 3
no element of this ring has norm two or three, so neither factor can split further
rounding the exact quotient leaves a remainder of smaller norm
the cells add up to the divisor sum
the divisor count is the product of one more than each exponent
the divisor sum factorises as the two row totals multiplied
the driver actually holds somewhere on the drawn sequence
the driver is one of the first four perfect numbers
the exponent is a whole number between 2 and 7
the first odd abundant number is 945
the first step shown is a whole number between 0 and 200
the first tree's leaves multiply back to the number
the largest number is a whole number between 101 and 600
the last step shown is a whole number between 10 and 37
the lattice half-width is a whole number between 2 and 5
the Mersenne number for this exponent is prime
the nearest are at distance √(5/4), so every remainder has norm 4 × 5/4 = 5, more than N(2) = 4
the number has a factor tree to draw
the number has at least three prime factors, so its trees can differ
the number has at most three distinct primes
the number is a product of exactly two prime powers
the number is inside the range the primality test decides
the powers of two below 2^k add up to the Mersenne number
the rectangle holds every divisor
the search radius is a whole number between 3 and 8
the second tree's leaves multiply back to the number
the sequence is long enough for the rows asked for
the share at the end of the range is close to a quarter
the share falls as the threshold rises and stays positive through four
the smallest Hilbert number with two factorisations is 441
the smallest norms are 1, 4, 5
the step budget is a whole number between 10 and 800
the stretch shows the driver holding
the two conjugate factors have norms multiplying to 36
the two trees start differently
the view is one of lattice, grid, aliquot, mersenne, failure, trajectory, driver, fates, abundancy, primitive, abundantdensity, abundancydist, primitivesum, hilbert, hilbertgroup, hilbertatoms, hilbertmulti, monoids, covering, divfail, euclidgi, motzkin, hilbertlemma, euler41, rabinowitsch
there are primitive abundant numbers in the range
ω/2 is more than 1 from every integer of the field

`ferrers`

37 kinds of claim · 30 placements

the coefficient at 0 is what the theorem says ×27
the product's coefficient at 0 is p(0) ×19
the two products agree at 0 ×19
p(4) is divisible by 5 ×11
log p(10) is below π√(2n/3), as it is for every n ×10
the estimate at 10 is within a factor of two of the count ×10
the ranks modulo 5 split the 30 partitions of 9 into 5 equal classes ×4
the partitions of 8 into odd parts and into distinct parts come out equal ×3
the number of coefficients shown is a whole number between 6 and 20 ×2
the number sorted by rank leaves 4 on division by 5 ×2
the number whose partitions are sorted by rank is a whole number between 4 and 14 ×2
and changes the number of parts by exactly one
and it is the largest one that fits
and the estimate is closer at the far end than at the near one
and the partitions the move cannot touch are exactly what survives
and the ratio climbs towards that ceiling rather than away
at least one of the divisible values is in the table
doing it twice gives back what it started as
every listed partition adds to the number it partitions
the congruence drawn is the one modulo 5 or modulo 7
the conjugate is a partition of the same number
the largest number counted is a whole number between 20 and 120
the largest number in the table is a whole number between 12 and 40
the marked values are whole numbers inside the range drawn
the move gives another partition of the same number into distinct parts
the number being partitioned is a whole number between 3 and 12
the number whose distinct partitions are paired off is a whole number between 3 and 14
the numbers outside the congruence class are not all divisible either
the partition view is one of durfee, glaisher, series, euler, pentagonal, growth, congruence
the partitions listed are all of them
the parts are a descending list of whole numbers
the restricted counts are not the unrestricted one
the series drawn is the partition product or Euler's identity
the signed count of distinct-part partitions is the product's coefficient
the square, the arm and the leg account for every dot
there is a Durfee square to draw
turning the diagram over twice gives it back

`irrational`

59 kinds of claim · 30 placements

term 0 follows the pattern ×30
past the threshold the tail at level 1 lies strictly between 0 and 1 ×14
the search over divisors of 2 agrees with whether 2 is a perfect square power ×14
the derivative of order 7 at zero is a whole number ×12
and below 1/2 ×11
at q = 2 the tail is positive ×11
q! times the first 3 terms is a whole number ×11
the degree-1 integral is positive, because the integrand is ×9
the fraction 1/1 is no closer to √2 than 1/(3q²) allows ×9
the largest value of the degree-1 polynomial is (π²/4)ᵏ/k! ×9
and 4 has one such root, not several ×7
the search over divisors of 2 agrees with whether 2 is a perfect cube power ×7
Aₙ = q·e − p for the convergent 1/1 ×5
Bₙ = p − q·e for the convergent 1/0 ×5
Cₙ = p − q·e for the convergent 2/1 ×5
the 1th truncation beats the exponent 2 ×5
and it does so with room to spare: the error is below q^(−2) ×4
some truncation beats the barrier of degree 2 ×4
the search over divisors of 2 agrees with whether 2 is a perfect fourth power ×4
at x = 0.3 each deeper convergent is closer to tan x ×3
the fraction converges to tanh 0.5 ×3
√2 is 1; 2, 2, 2, …
√2: all 16 terms are certain
a fraction's denominator is not zero
and (1 + tanh ½)/(1 − tanh ½) is e
and ends below one millionth
and is below twice the peak, since sin integrates to two over the interval
and it is closer than 1/q², which is what makes the exponent exactly two
at least one of the numbers drawn has an irrational root, or the figure argues nothing
between one and three increasing barrier degrees, none above the number of truncations drawn
between two and five increasing degrees, each between 1 and 12
between two and ten whole numbers, each between 2 and 64
e follows its pattern
e or its square root
e: all 16 terms are certain
for √2, q² times the error never falls below a third
for e the lowest values drift down, slowly
golden ratio: all 16 terms are certain
its denominator is a whole number between 1 and 200
q times the error falls at every step
so it is strictly between 0 and 1, where no whole number is
the bounds fix at least as many terms as are drawn
the denominator is a whole number between 1 and 5
the fraction is at least close enough to π for the picture to be about π
the golden ratio is all ones
the largest denominator tried is a whole number between 3 and 14
the number of convergents drawn is a whole number between 4 and 9
the number of tail terms summed exactly is a whole number between 12 and 60
the number of truncations of the constructed number is a whole number between 3 and 5
the numerator is a whole number between 1 and 5
the numerator of the fraction π is supposed to be is a whole number between 2 and 400
the range drawn reaches a degree whose bound is below one, which is the whole argument
the root taken is a whole number between 2 and 4
the tail falls as the denominator grows
the view is one the family draws
there are derivatives to show
twelve levels at x = 1 give tan 1 to double precision
π: all 16 terms are certain
π's fifth term is 292

`lattice-circle`

99 kinds of claim · 57 placements

the count on the circle of squared radius 1 is 4(d₁ − d₃) ×64
3, 4, 5 is a Pythagorean triple ×10
the points on the circle of radius √25 are 4(d₁ − d₃) ×8
5 = 2² + 1² ×7
the tetrahedron of height 1 has no lattice point inside it ×7
the area is 20 interior points plus half of 7 on the edge, less one ×5
the count at dilation 1 is the Ehrhart polynomial's value ×4
the count at k = 1 matches the cubic fitted from the first value ×4
the number of columns in the grid is a whole number between 4 and 8 ×2
a group with entries only up to two fails to identify two of the classes
a polygon's boundary count and its dual's add to twelve
a prime is a sum of two squares exactly when it is one more than a multiple of four
a square root of −1 exists modulo a prime one more than a multiple of four
and a self-dual one has six boundary points, which is half of twelve
and dividing by the radius itself tames it further still
and every basis along the way spans a cell of the same area
and exactly four on it, which are its own corners
and from entries up to three onward the count stops moving
and is one more than a multiple of four
and it is the first one below it — the row above is still above
and it really is in the lattice
and Scott's inequality caps the boundary count at nine
and some attain the bound
and stays inside three times the two-thirds power throughout
and stays inside twice the cube root over the range drawn
and stopping one step earlier would not have given two squares
and the answer is one of the points the circle of that radius actually passes through
and the candidate really does square to −1
and the density is falling, which is the statement the constant qualifies
and the disc of radius two catches thirteen
and the flat ones are excluded rather than ignored
and the point it came from is on the unit circle
and the rest pair off
and they have between three and six corners
and twelve has six
at least two heights are drawn
between one and five powers are drawn
D is a whole number between 2 and 40 and not a perfect square
D is between 2 and 40
D is not a perfect square, or the equation has only the trivial solution
every edge lies at lattice distance one from the interior point
every lattice triangle in the grid satisfies the identity
every marked point really is on the circle
every one of the 2300 triples of grid points was reached
every power is again a solution
every solution up to the largest drawn is a power of the smallest
for some height the linear coefficient is negative, so it counts nothing
four of the sixteen are their own dual
n names one circle and is read only by the default view; the other views take their own parameters
no polygon with an interior point has more boundary points than Scott allows
no two slopes give the same triple
one extra for the hole repairs it, which is the Euler characteristic in disguise
one has one divisor
one of them attains Scott's bound
p names the prime the two squares are produced for and is read only by the cornacchia and reduce views
Pick's formula with one interior point makes twice the area the boundary count
Pick's theorem holds for every polygon swept
some circles miss the lattice entirely, which is the whole question
the answer satisfies the congruence the chain started from
the bound is attained only at one interior point
the bound the count runs to is a whole number between 2000 and 200000
the box the polygons are drawn from is a whole number between 3 and 4
the circle of squared radius one carries four points
the count times √(log x) over x is near Landau's constant at the top of the range
the disc of radius one catches five points
the divisors of everything up to six add to fourteen
the dual of one of the sixteen is another of the sixteen
the error divided by the cube root is the tamer ratio
the error divided by the two-thirds power is the tamer ratio
the error is not bounded by twice the square root over this range
the factorisation test and the circle agree about every number up to two hundred
the first remainder below the square root, with its partner, squares and adds to the prime
the identity as stated gives the wrong area for a region with a hole
the largest bound counted is a whole number between 200 and 4000
the largest prime to test is a whole number between 10 and 200
the largest radius counted is a whole number between 40 and 400
the largest squared radius drawn is a whole number between 12 and 64
the lattice half-width is a whole number between 1 and 16
the leading coefficient is the volume for every height drawn
the number is a whole number between 1 and 200
the number is prime
the polygon has between 3 and 16 whole-numbered corners and some area
the polygon is one of blob, triangle, comb, thin
the prime is one more than a multiple of four
the prime the lattice is built from is a whole number between 5 and 200
the prime the two squares are found for is a whole number between 5 and 4001
the second differences are twice the area
the shortest vector of the lattice has length squared exactly the prime
the slopes are proper positive fractions
the solution found satisfies the equation
the sweep found a substantial family
the sweep grid is between 3 and 6 points across
the view is one the family draws
the volumes differ while the counts do not
there are sixteen classes to pair up
there are sixteen lattice polygons with a single interior point
twice the area is a whole number, which is why halves are the only fractions in the formula
upTo names a range and is read only by the gaussian, count and density views
while a polygon with no interior point can exceed the bound, which is why it needs one

`mediant`

68 kinds of claim · 34 placements

?(1/11) is the binary fraction in the same position ×1023
0/1 and 1/13 have determinant one and nothing simpler between them ×359
3/2 is in lowest terms ×108
the circles on 0/1 and 1/7 touch ×56
s(1) and s(2) are coprime ×48
the 1th ratio's numerator agrees with the Calkin–Wilf walk ×48
the ratio 1/1 appears once ×48
0/1 and 1/7 are Farey neighbours ×24
1/1 arrives already in lowest terms ×15
and its mediant 1/1 is already in lowest terms ×13
s(2) counts the hyperbinary representations of 1 ×12
the 1th arc crossed is the descent's 1th interval ×10
run 1 of the turns is the continued fraction's quotient ×5
the sequence has one entry per coprime pair up to 7 ×2
? of the expansion of all ones is 2/3
? of the expansion of all twos is 2/5
?(1 − x) = 1 − ?(x)
?(x/(1 + x)) = ?(x)/2
√2 − 1 has every quotient two
and at most its right bound
and is closer to the target
and its denominator does too
and the path closes on the target
and their lengths add to one
circles that are not neighbours stay clear of each other
depth d cuts the interval into 2^d pieces
each path is a word of at most eight L's and R's
each record has a larger denominator than the last
how many terms are drawn is a whole number between 8 and 64
how many turns are taken is a whole number between 6 and 34
neighbouring nodes have determinant one
no fraction is produced twice
no two words produce the same matrix, so the tree never rejoins
reading the drawing left to right reads the fractions in increasing order
the depth is a whole number between 2 and 5
the depth of the tree checked is a whole number between 3 and 14
the descent passes at least three fractions within 1/q² of the target
the descent produces several record approximations
the drawing scale is a whole number between 200 and 400
the endpoint each crossed arc keeps spells the descent's word
the function never decreases
the largest denominator drawn is a whole number between 5 and 30
the length carrying half the rise falls with depth
the line crosses exactly as many arcs as the descent has intervals
the matrix for "L" has determinant ±1
the matrix for "LR" has determinant ±1
the matrix for "LRL" has determinant ±1
the matrix for "LRLR" has determinant ±1
the matrix for "LRLRL" has determinant ±1
the matrix for "R" has determinant ±1
the matrix for "RL" has determinant ±1
the matrix for "RLR" has determinant ±1
the matrix for "RLRR" has determinant ±1
the matrix for "RR" has determinant ±1
the matrix for "RRL" has determinant ±1
the matrix for "RRLR" has determinant ±1
the matrix for "the root" has determinant ±1
the mediant is at least its left bound
the mode of the mediant family is one of tree, farey, ford, matrices, diatomic, approx, tessellation, questionmark, singular
the number is one of phi, root2, pi, e
the order is a whole number between 2 and 12
the target is one of phi, root2, pi, e
the tree has 2^D − 1 nodes to depth D
the tree is full to its stated depth
the window is a sub-interval of the unit interval
two to six increasing depths, up to 18
up to which index the hyperbinary count is checked is a whole number between 4 and 24
φ − 1 has every quotient one

`necklace`

71 kinds of claim · 34 placements

Fermat's theorem holds at 3 ×549
the largest order mod 3 is λ(3) ×58
the primitive roots of 3 number φ(p − 1) ×29
the search and the rule agree at 2 ×29
1 to the power 4 is 1 modulo 15 ×28
1 to the power 13 − 1 is 1, which is Fermat's theorem ×12
the order of 1 divides 13 − 1 ×12
exactly φ(1) residues have order 1 ×8
the class of numbers sharing n/1 with n has φ(1) members ×6
the order of 2 divides λ(561) ×6
the order of 2 falls short of 561 − 1 ×6
the Wieferich primes to base 2 below 200000 ×6
the modulus is a whole number between 5 and 31 ×4
the powers of 2 run through every residue ×4
1009 has a primitive root ×3
11 to the 1008 is 1 mod 1009 ×3
Chernick's construction at k = 1 satisfies Korselt's divisibility ×2
no proper divisor of 4 works for every unit ×2
the base is a whole number between 2 and 12 ×2
the bound is a whole number between 12 and 42 ×2
the length is a whole number between 3 and 7 ×2
the number is a whole number between 6 and 24 ×2
the number of colours is a whole number between 2 and 4 ×2
a base passing every check exists exactly when the number is prime
and exactly a of them are single
and no smaller power along a prime is
and reach each of them once
and the largest order divides φ(n)
and the number of non-constant classes is (a^p − a)/p
at least one composite below 2000 survives every base
at least one product was checked against every base rather than by the criterion alone
at least one residue has order p − 1, which is what being cyclic means
at least two composites below 2000 survive base 2
at least two values of k below 12 give three primes
both verdicts occur inside the bound, so the table is deciding something
every base coprime to 1729 returns 1
every class is a single string or a full ring of p
every order divides the largest one
every other class holds exactly p strings
every string of the given length is drawn
Mertens's estimate matches the exact sum of 1/p at a million
more than one order occurs, so the picture is separating the residues
not every residue is a generator, so the list is a selection
only 1093 and 3511 keep their order
p divides a^p − a, which is what the count just showed
some of them are caught by another base, so the two columns differ
the base is coprime to the modulus
the classes account for every number from 1 to n
the classes account for every string
the classes of size one are exactly the constant strings
the criterion and the exhaustive base test give the same verdict
the drawn ring has as many beads as the base's order
the figure draws every string, so there is a ceiling on how many
the largest k searched is a whole number between 1 and 40
the length is prime
the length is prime — the whole argument needs it to be
the modulus is prime
the modulus is prime, so every non-zero residue has an order
the number has at least three divisors, so the identity is not trivial here
the number is a prime below a million
the number is a whole number below a million
the number is composite — the criterion is about composites
the only zeros are 1093 and 3511
the orbit closes
the orders account for every residue
the quotients are computed in exact whole numbers below 9,000
the units are counted, and there are φ(n) of them
the view is one the family draws
there are φ(13 − 1) generators
there is more than one base to test
λ(n) = n − 1 exactly when n is prime

`reciprocity`

45 kinds of claim · 21 placements

two primes agree modulo 3 and disagree about x² + 27y² ×158
3 is represented by x² + 1y² exactly when 3 mod 4 is one of the classes ×134
every step is a whole unit, because 1 is not a multiple of 13 ×70
the sign of multiplication by 3 is (3 | 11) ×22
(−1 | 3) is decided by 3 mod 4 ×17
(2 | 3) is decided by 3 mod 8 ×17
Gauss's lemma counts 1 folds for −1, and its parity is the symbol ×17
Gauss's lemma counts 1 folds for 2, and its parity is the symbol ×14
the modulus deciding x² + 1y² is at most 4n ×3
the modulus is a whole number between 5 and 31 ×3
3 folds gives the same answer as Euler's criterion ×2
(p | q) comes out of the count above it
(q | p) comes out of the count below the line
a prime with both symbols positive is 1 modulo 8, which is the two conditions met at once
and every one of them lands in the bottom half
and exactly half the multipliers are squares
and it has no imaginary part
and points up, which is the part Gauss took four years over
and positive, which is the part Gauss took four years over
and their product is minus one to the power of the whole rectangle
between two and five forms are drawn
both are odd primes and they are different
each form's coefficient is a whole number between 1 and 27
every moved cycle has the same length, which is the order of the multiplier
every point is counted exactly once, on one side or the other
for a prime 1 modulo 4 the sum is real
for a prime 3 modulo 4 the sum is purely imaginary
no lattice point lies on the diagonal, because the primes are coprime
the bound the congruence check runs to is a whole number between 2000 and 40000
the cycle count and the inversion count give the same sign
the drawing holds both kinds of form, which is what it is for
the first prime is a whole number between 3 and 31
the folded values are all different
the folds for 2 are the k above p/4
the largest modulus tried is a whole number between 20 and 400
the largest prime drawn is a whole number between 60 and 400
the largest prime in the table is a whole number between 11 and 97
the length of the sum, squared, is the modulus
the modulus is an odd prime
the multiplier is a whole number between 2 and 30
the multiplier is not a multiple of the modulus
the points below the line are the sum of the floors of kq/p
the second prime is a whole number between 3 and 31
the square of the sum is plus or minus the modulus
the view is one the family draws

`sieve`

73 kinds of claim · 55 placements

the class 0 shares a factor with 4, so it can hold at most the factor itself ×20
211 leaves remainder 1 on division by 2 ×10
the limit is a whole number between 10 and 400 ×5
2 classes share no factor with 4 ×4
the staircase ends at π(3000) ×4
every listed prime is already in the class 3 mod 4 ×2
so some prime factor of it is 3 mod 4 — the classes multiply, and a product of ones is one ×2
the class 3 leads at almost every bound ×2
the modulus is a whole number between 3 and 14 ×2
the number built is 3 mod 4 ×2
a Mersenne prime has a prime exponent
a square left unstruck holds a prime, and a struck one does not
an even truncation is an upper bound
and 168 below a thousand
and an odd one a lower bound
and every combination of exponents appears once
and it tracks the logarithm of a logarithm to within a twelfth throughout
and modulo three it never falls behind over this range
and over the last decade of the range the diverging sum gains more than twice as much
and some of the lower bounds are negative, which is no bound at all
and the first bound at which it does not is 26,861
and the number of them in the interval is never far below n / ln n
and the one prime it holds divides the modulus
and the ratio to x / ln x comes down as x grows
and the twin pairs track theirs
and there are almost none of them in this range
and they hold them in near-equal numbers, which is more than Dirichlet's theorem claims
between two and five odd primes are listed
both classes hold primes below the bound
each is near the share an even split would give
every class coprime to the modulus holds primes
every interval from n to 2n holds a prime
every prime below the bound lands in exactly one class
every prime factor of one more than a square of an even number is 1 mod 4
every prime factor of the constructed number is outside the list
multiplying the factors out gives exactly the sum over the numbers they build
no listed prime divides it
one gap between each consecutive pair
the bound the postulate is checked to is a whole number between 100 and 50000
the column count is a whole number between 4 and 25
the construction is the one that subtracts or the one that squares
the count runs above x / ln x over this range
the factorisation multiplies back to the number built
the factors are between 2 and 4 primes
the full inclusion-exclusion agrees with the count made by sifting
the gaps add up to the distance from 2 to the last prime
the grid of whole numbers drawn is a whole number between 24 and 144
the largest bound counted is a whole number between 5000 and 200000
the largest number sieved is a whole number between 100000 and 2000000
the largest prime sifted by is a whole number between 11 and 43
the list is between two and six primes
the logarithmic integral runs above the count at every checkpoint drawn
the marked checkpoints are whole numbers inside the range
the number built stays inside exact arithmetic
the number of intervals drawn is a whole number between 6 and 40
the numbers marked are exactly those the product builds
the partial sums swing far past the answer before settling
the pass number is between 0 and 8
the power each factor is truncated at is a whole number between 2 and 5
the primes 1 mod 4 are about half of all of them
the primes one more than a square track the conjectured count
the race is drawn modulo three or four
the range sifted is a whole number between 10000 and 1000000
the range the sum over primes is drawn to is a whole number between 1000 and 1000000
the sieve found every prime below the bound
the squares left standing below 100 are the primes
the sum of the reciprocals of the primes keeps growing
the sum over all primes has passed 2.8 and is still climbing
the view is one the family draws
there are 25 primes below 100
while the ratio to the logarithmic integral is nearer one by more than a factor of four
while the sum over the twins is still under 1.8
π(x) stays above x / ln x throughout this range

Dynamics

11 families

`attractor`

32 kinds of claim · 31 placements

above the critical parameter the trajectory visits both lobes
and each is narrower by the stretch factor
and even at the cusp, where a straight line cannot fit, the departure stays small
and it is the Lorenz system's own exponent, near nine tenths
and settles on neither fixed point
and the strips are laid out in order without overlapping
areas shrink at every step
below the critical parameter it spirals into a fixed point
between two and four starting gaps, each a whole power of ten from −2 to −12
enough returns to see a shape
every surviving strip lies inside the square
every window still holds some of the orbit
so the total width falls, and the limit has none
the closed-form fixed points really are fixed
the contraction is between 0.1 and 0.4
the geometric parameter is between 0.1 and 8
the growth rate does not depend on how close the two starts were
the Hénon parameter is between 1 and 1.45
the Jacobian determinant is the contraction
the map's maximum is in its interior, so it folds
the number of magnifications is a whole number between 1 and 4
the number of stages is a whole number between 2 and 5
the number of steps is a whole number between 500 and 40000
the orbit stays bounded, which is what makes it an attractor
the Prandtl parameter is between 0.1 and 40
the Rayleigh parameter is between 0.1 and 100
the return map is a curve, not a cloud — its typical thickness is under one per cent of its range
the step size is between 0.0005 and 0.05
the stretching factor is between 2.2 and 5
the strips double at every stage
the surviving set has a dimension between nought and one
the trajectory stays in a bounded region

`automaton`

26 kinds of claim · 25 placements

after one lap the flux at density 0.025 is min(ρ, 1 − ρ) ×40
rule 90 row 0 matches Pascal's triangle mod 2 ×40
every configuration of a 5-cell ring ends as its majority ×6
a density strictly between 0 and 1
above half density at least 2k − n cars are always stopped
an odd ring of 7 to 13 cells
an odd ring, so there is always a majority
and freezes with both values still present
and the drawn ring ends as its majority
below half density every jam is gone after at most one lap
below one half the stopped share reaches zero
every cell is on or off and nothing else
exactly five of the 256 rules conserve the number of 1s on every small ring
far from one half it is always right
near one half it is often wrong
no elementary rule classifies every ring
rule 184 never creates or destroys a car
stopped cars never fall below (2k − n)/n
the drawn table reads back as the rule number
the local vote freezes within a few steps
the number of rows is a whole number between 4 and 160
the ring's length is a whole number between 20 and 240
the rule number is a whole number between 0 and 255
the seed is single or random
the start is random or one block of cars
the steps drawn is a whole number between 10 and 160

`basins`

19 kinds of claim · 15 placements

root 1 is reached from at least one of the starting points ×5
a point beside root 1 runs to root 1 ×3
the iteration cap is a whole number between 8 and 200 ×3
the 5 roots are all distinct and all found ×2
the sample count is a whole number between 60 and 700 ×2
the span is between 0.02 and 6 ×2
and the cycle attracts, because the product of the map's derivative round it is under one
and the disc was sampled at 144 points
at least one root was found
each root cubes to one
every point called a root really is one
every point of a small disc round the cycle reaches no root
so the failing set has area rather than being a boundary
the polynomial is one the family knows
the roots found are counted
the set has as many points as the construction asks for
the two points really do map to each other
the view is one the family draws
with no imaginary part left over

`bifurcation`

29 kinds of claim · 27 placements

the 2-to-1 merging is inside its bracket ×4
the orbit at r = 2.8 has period 1 ×4
the lower parameter is between 0 and 4 ×2
the number of images is a whole number between 2 and 10 ×2
the parameter is between 3.57 and 4 ×2
the upper parameter is between 0 and 4 ×2
at least three small distances below the window
at the critical parameter f³ touches the diagonal
below the critical parameter the channel is open
each mean is taken over at least a hundred quiet phases
each merging comes at a lower parameter than the one before
the column count is a whole number between 40 and 1400
the diagram is drawn as marks rather than as a bitmap of elements
the distance below the window is small and positive
the doublings come in order
the gap ratio approaches Feigenbaum's constant from this side too
the last ratio measured is Feigenbaum's constant to within a twentieth
the number of doublings is a whole number between 3 and 5
the number of mergings is a whole number between 3 and 5
the orbit piles up beside each image of the turning point
the orbit stays between the first two images of the turning point
the orbit takes several steps to pass through the channel
the points kept per column is a whole number between 10 and 400
the quiet phases lengthen as the inverse square root of the distance
the range runs upward
the range runs upward and is wide enough to draw
the steps drawn is a whole number between 100 and 2000
the stretch drawn holds at least two quiet phases
the view is one of feigenbaum, critical, histogram, merging, channel, laminar, laminarlength

`billiard`

64 kinds of claim · 29 placements

a trajectory in this table takes exactly 4 directions and no more ×3
the path bounces correctly off side 1 ×3
a closed path is one that returns to within a tolerance
a path closing on itself was found within 1e-9 — the search at this depth found none otherwise
and arrives and leaves at the same angle to the radius
and does so after twice the sum of the two whole numbers
and its angles add to a straight angle
and its invariant is the same at every bounce
and never returns exactly to where it started
and no nearby triangle inscribed in the same three sides is shorter
and that genus is a whole number
and the direction along the wall is untouched
and the path does not close before it is supposed to
and with the obstacle the gap grows by far more than the distance does
at the value it was aimed with
between 10 and 200 bounces
between 20 and 200 bounces are followed
between 3 and 24 bounces are drawn
between 3 and 40 chords are drawn
each altitude meets its side between the corners
every chord passes the same distance from the centre
in both coordinates
on the empty table the gap grows no faster than the distance travelled
on the empty table the gap stays proportional to the distance travelled
so no part of it is below the axis
the angle of arrival equals the angle of departure at every wall
the angles of the table add to what a polygon's angles must
the ball is moving and has a wall to reach
the ball starts inside the table
the bouncing path and the folded straight line agree everywhere along their length
the drawn triangle has the angles it was asked for
the escaping trajectory's invariant is below it
the first hit is somewhere on the rim
the irrational path enters nearly every cell of the grid
the irrational path is followed for between 60 and 600 bounces
the obstacle fits inside the table with room to pass
the path is followed for a drawable number of bounces
the rational slope is a ratio of small whole numbers
the rational slope is in lowest terms
the rational slope's path returns to its exact starting state
the reflected copies drawn fit inside the panel they are drawn in
the search runs to between three and sixteen bounces
the slope is drawable
the star closes exactly
the star polygon is drawn in a single closed loop
the star wraps between 1 and 6 times
the stem is between a fifth and a whole radius deep
the stem is between a twelfth and half the cap wide
the surface the table unfolds into has a whole genus
the sweep is a grid of a workable size
the sweep produced something to compare
the table is a triangle
the table is one the family knows
the trajectory is followed for between 40 and 600 bounces
the trajectory stayed inside the table
the trapped trajectory never touches a stem wall
the trapped trajectory's invariant is above the stem's half-width and inside the cap
the triangle is acute, which is when this path exists
the two invariants sit on opposite sides of the stem's half-width
the two paths start very nearly together
the view is one the family draws
two angles of a triangle, in degrees, leaving a third
while the trajectory aimed under the half-width does go down the stem
with the obstacle the same two paths end up hundreds of times further apart

`boxcount`

85 kinds of claim · 34 placements

piece 0's mass is read off its binary digits ×1024
at depth 2 the cover's total agrees with the closed form at exponent 1.037 ×36
0 right-hand choices are shared by C(10, 0) pieces ×19
after 0 steps there are two to the power 0 pieces ×7
at the dimension itself the total is exactly one, at depth 2 ×6
the midpoint construction and the series agree at 1/2²⁰ ×6
f at q = 0 is the least value of qα + β(q) ×5
the counted slope for q = 0 is β(0) from the formula ×3
the counted slope for w = 0.6 is 2 + log₂ w ×3
α is minus the slope of β at q = 0 ×3
after 6 steps there are as many pieces as the rule makes ×2
f at q = -3 is the least value of qα + β(q) ×2
a finite number of splits undercounts
a smaller box never covers the set in fewer boxes
and above it the total shrinks to nought
and its limiting dimension is the curve's
and none is larger than the support
and the average over eight independent paths is closer
and the point of contact is the information dimension
and they make up the whole support
at q = 0 every piece counts once, which is the support's dimension
at q = 1 the curve touches the diagonal
at q = 1 the moment is the total mass, so β(1) = 0
below the dimension the total grows without bound as the cover is refined
between 20 and 400 thousand steps
between one and four sets this family draws
between two and six depths, each between 1 and 14
each count is exactly the number of pieces at that depth
how many box sizes are counted is a whole number between 3 and 6
how many times the box size is divided is a whole number between 1 and 5
how many times the mass is split is a whole number between 1 and 12
how many times the rule is applied is a whole number between 2 and 6
more splits bring the counts closer to the curve
near w = 1 detail finer than the grid is still large and the count reads low
near w = ½ the count carries a logarithm and reads high
no set of points is larger than its own scaling exponent allows
one to five exponents between −2 and 6, leaving out 1, whose sum is always the whole mass
one to four factors between ½ and 0.9
one to three increasing split counts, up to 1000
splitting in two, level times, makes two to that power pieces
the box level marked is a whole number between 2 and 7
the Brownian path's counted slope is near 1.5
the chosen cells are distinct positions of the grid
the comparison measure is uneven enough to have a curve
the contraction beats the stretch, so the formula's first case applies
the count's exponent is the curve's at the matching q
the counted box dimension closes on the closed form
the counted dimension closes on the one the rule forces
the counted dimension lies strictly between a curve's and a region's
the counted slope for q = -1 is β(-1) from the formula
the curve is drawn finer than the smallest box counted
the drawn boxes are the counted boxes
the exponent is swept between 0.05 and 1 either side of the dimension
the factor each level's raise shrinks by is between 0.3 and 0.95
the first exponent is positive, which is what makes the orbit sensitive to its start
the formula returns a value between a curve's dimension and a region's
the Hausdorff dimension never exceeds the box dimension
the horizontal contraction is the stronger one, which is what makes the carpet self-affine
the left half is a line plus the whole graph shrunk by w
the map contracts area but does not collapse it
the map's parameter is in the range that has an attractor
the marked columns are the counted columns
the masses add to one
the measure lives on the interval or on the Cantor set
the number of box sizes is a whole number between 4 and 10
the number of columns is a whole number between 2 and 6
the number of pieces and the ratio give the dimension the rule forces
the number of rows is a whole number between 2 and 5
the number of sampled points is a whole number between 50000 and 1000000
the number of times the rule is applied is a whole number between 1 and 7
the orbit stays on the attractor rather than escaping
the raises at a fine level have the variance the construction prescribes
the rows carry different numbers of cells, so the two dimensions differ
the rows carry equal numbers, so the two dimensions agree
the sample is finer than the smallest box counted
the second is negative, which is what collapses the attractor onto a set of no area
the seed of the Brownian path is a whole number between 1 and 99
the set is one built by a rule, or there is no exponent to pivot about
the set is one this family draws
the share of the mass the left piece of each split carries is between 0.02 and 0.98
the shares of the measure drawn for comparison is between 0.02 and 0.98
the two exponents add to the logarithm of the area factor, which the map fixes
the view is one the family draws
with equal shares every point scales alike
α is minus the slope of β at q = -2

`circle-map`

69 kinds of claim · 35 placements

inside the plateau at Ω = 0.4645 there are two orbits ×10
the nonlinearity is between 0 and 0.95 ×4
some parameter carries a period-2 orbit with rotation number 1/2 ×3
the rotation number there is exactly 1/2 ×3
inside the plateau at Ω = -0.1528 there are two orbits ×2
the number of parameters sampled is a whole number between 60 and 900 ×2
and at least one is outside it
and both boundary circles still stay where they are
and it rises rather than falling
and the reason is that this map does not preserve area
and the repelling one's is at least one
and the second really is outside it
and they turn in opposite directions
and would change an exponent-one total a great deal, so that one does not
at least two of the panels are inside the plateau
both boundary circles stay where they are
doubling the range barely changes this total, so it converges
each crossing is a fixed point of the map
each orbit is followed for between 60 and 600 steps
how many images are drawn either way is a whole number between 6 and 20
in the sorted coordinates the map is the rotation, to the orbit's own resolution
inside a plateau the rotation number does not depend on where the orbit started
most sampled parameters carry an orbit
no orbit leaves the annulus
no point of the annulus comes near to staying where it is
one fixed point is a centre and the other a saddle
one orbit attracts and the other repels
outside, the rotation number has moved off the rational
the angle is a number or one of the named constants
the angular shift changes sign across the annulus at every angle
the attracting orbit is most strongly attracting in the middle of the plateau
the attracting orbit's multiplier is at most one
the boundaries still turn in opposite directions
the conserved quantity stays put along every orbit drawn
the drawn images account for less than the whole circle
the exponent the lengths fall at is between 1.05 and 4
the first parameter really is inside the plateau
the gaps take at most three distinct values, as the three-distance theorem says
the largest nonlinearity is between 0.4 and 1.2
the length of the orbit is a whole number between 200 and 3000
the lengths sum to something finite
the locked share of the parameter axis grows with the nonlinearity
the loop and its image cross exactly twice
the loop and its image enclose the same area
the lower parameter is between 0 and 1
the map locks onto one step in two over an interval of parameters
the map preserves area everywhere on the annulus
the measured curve is flat over a stretch at more than one rational
the number of nonlinearity levels is a whole number between 6 and 40
the number of parameters per level is a whole number between 40 and 320
the number of steps drawn is a whole number between 10 and 60
the number of steps is a whole number between 3 and 120
the orbit leaves no large gap
the orbit repeats exactly when the angle is rational
the outward push is a sensible size
the parameter reaches the intended rotation number
the period is a whole number between 1 and 4
the perturbation is a sensible size
the perturbation is between a fiftieth and two fifths
the plateau has positive width
the ratio of consecutive lengths rises towards one, which is the smoothness condition
the rotation number never falls as the parameter rises
the rotation number p/q is between nought and one
the rotation per period, plus one is a whole number between 1 and 5
the two orbits are closer together near the edge of the plateau
the upper parameter is between 0 and 1
the view is one the family draws
the window runs upward and is wide enough to draw
with no nonlinearity the curve is a straight line and has no plateaus to speak of

`cobweb`

137 kinds of claim · 71 placements

T2: bin 1 matches the arcsine share ×90
bin 1 holds the share the density says it should ×60
at r = 4 the 0-th iterate has 2^0 folds ×21
T2 composed 1 times has 2 fixed points ×16
T2 composed 1 times is T2 ×16
step 0 is cos(3^0 θ) ×13
returns after 1 steps against the trace ×10
the orbits of exact period 1, walked and by inversion ×10
the points of exact period 1 make whole orbits ×10
at level 1 the admissible intervals number F(3) = 2 ×8
the number of steps is a whole number between 2 and 400 ×7
every point over 1 comes back after 1 doublings ×6
the trace is 1 + (1 + √2)ⁿ + (1 − √2)ⁿ at n = 1 ×6
0.00000 returns to itself after 2 steps ×5
a full two-branch fold has 2^1 points fixed by its 1th power ×5
after 0 doublings the overlap is within 1/2^2 of a sixth ×5
both maps have the same number of points of period dividing 1 ×5
0.6154 is fixed at r = 2.6 ×4
after 0 doublings the lowest frequency present is 1 ×4
T2(cos θ) = cos 2θ ×4
the random point visits quarter 1 a quarter of the time ×4
the total energy of f composed 0 times is a quarter ×4
in the window of three, at 3.8300, the estimate is log of the golden ratio ×3
the folds of f^6 at r = 3.5, sampled and from the preimages of 1/2 ×3
the longest word counted is a whole number between 3 and 10 ×3
the numerator of the starting point is a whole number between 1 and 10000000 ×3
the parameter is between 0 and 2 ×3
at r = 2.4 the error shrinks by the slope at the fixed point each step ×2
the longest iterate counted is a whole number between 6 and 22 ×2
the number of steps drawn is a whole number between 3 and 20 ×2
the number of times the map is applied is a whole number between 2 and 5 ×2
the orbit at r = 2.4 converges far enough to measure a rate ×2
√2 is fixed by the Newton step
0, 1/3 and 1 are a cycle of three
A covers A and B, B covers all three, C covers B and C
a new crossing has a period above one that divides 2
a point with no two 1s together never enters the top quarter
after the computed orbit is dead the exact one is still moving
and carries the right end to the right end
and it never reaches zero, because an odd denominator cannot be halved away
and never grows
and not before
and shrinks every distance by a factor under one
and stays there, because zero is fixed
and the constant the ratio approaches is 1/(2√2)
and the repeat found really is a repeat
and then separate visibly, at a step the figure marks
at 1,000 digits more than 99% of strings are within 0.05 of a half
at r = 4 the folds double and the estimate is log 2
below the end of the doublings the deeper estimate is the smaller
between 6 and 60 steps are taken
between one and six logistic parameters from 3 to 4
between one and three logistic parameters, each with an attracting fixed point
deleting the first digit of an admissible word leaves it admissible
each letter of the itinerary is the next binary place of the starting point
each Newton error is the square of the last over twice the current guess
each point of the cycle is carried to another
each point of the cycle lies in its letter's piece
each step of a closed word's point lands in that word's piece
every bin is visited
every corner of the staircase lies on the curve or on the diagonal
every crossing of the map is also a crossing of the composed map
every exact iterate is a whole numerator over the same denominator
every piece is stretched
every start reaches the same point
five Newton steps from 1.9 reach twelve decimal places
how many steps of the orbit are drawn is a whole number between 3 and 12
its denominator is a whole number between 3 and 10000000
no point is named by two closed words
one point drawn per step, and the start
T₂∘T₃, T₃∘T₂ and T₆ agree everywhere
the 2 new crossings of period 2 are whole orbits of 2 points
the average of one long orbit is the average over the whole interval
the change of coordinate fixes the left end
the change of coordinate is increasing, so it is reversible
the column count is a whole number between 20 and 241
the computed orbit reaches exactly zero inside the drawn window, having run out of binary places
the contraction is one the family knows
the coordinate change carries the tent map to the logistic map
the coordinate is one whose invariant density this figure knows
the cycle closes after the word's length
the cycle is one the family draws
the deeper estimate never falls as r rises
the degree is a whole number between 2 and 5
the denominator is odd, or the exact orbit reaches zero as surely as the float one does
the error after n steps is inside k to the n times the error at the start
the exact orbit closes up inside the drawn window
the exact shares add to one
the fixed point attracts, so there is a rate to measure
the folds grow by the golden ratio at the last step
the highest frequency drawn is a whole number between 16 and 96
the left piece covers the right, and the right covers both
the longest cycle counted is a whole number between 2 and 8
the longest word counted is between two and eight
the lower parameter is between 3 and 4
the map is one the family knows
the map is one this figure knows
the map itself brings the point back
the map itself has a fixed point
the map sends the interval into itself
the number of binary places the starting point has is a whole number between 3 and 8
the number of bins is a whole number between 8 and 80
the number of drawn steps is a whole number between 2 and 400
the number of levels is a whole number between 4 and 10
the number of stages is a whole number between 3 and 7
the number of times the map is composed is a whole number between 2 and 4
the orbit approaches the fixed point exactly when the slope is shallow
the orbit at r = 3.83 settles on a cycle of three
the orbit of 1/3 lives in the two middle quarters
the orbit of 1/3 never enters [0, ¼)
the orbit stays inside the unit interval
the point a repeating word names has that word as its itinerary
the point found by bisection is fixed
the points returning after 4 steps are as many as the trace of the matrix power
the quadratic curve is on or off
the random point's running average reaches a quarter
the range runs upward
the share near a half grows with the length
the slope at the fixed point is 2 − r
the solved point comes back after exactly the word's steps
the starting point has between three and eight binary places
the starting point is between 0 and 1
the starting point is inside the interval
the straight pieces of the composed map are the allowed words of 4 letters
the transported orbit is the orbit of the transported point
the transported orbit obeys the tent map's own recurrence
the two agree exactly for at least eight steps
the upper parameter is between 3 and 4
the view is one the family draws
the word is not a shorter word repeated
the word is two to eight letters of L and R
the word, read round, never puts L after L
the words of each length and the points of that period are equally many
two depths, the deeper at most 20
two logistic parameters above one
two points are at most k times as far apart after the map as before
x² and x² − 1 composed in the two orders differ

`collatz`

44 kinds of claim · 30 placements

step 1 follows the rule ×178
every one of the 2 patterns of length 1 occurs exactly once ×12
the orbit of 27 reaches one ×7
a power of two is never a power of three
a starting number is a whole number between 3 and 1000000
across a large sample of starts, half the steps are odd ones
and 27 is the famous long one
and by the largest run drawn it is above nine tenths
and it is the one at one
and no two residues give the same parity string
and reaches a few thousandths within the range drawn
and the average fall per step is the value the pairing predicts
at least one pattern gives a whole number
between two and four odd offsets, each at most eleven
each convergent beats every fraction with a smaller denominator
every number the reverse tree reaches does come back to one
every orbit trends downwards in the logarithm
every pattern of the length was solved
most parity strings shrink the number by the end of the run
no pattern with a negative denominator gives a positive whole solution
one is already at one
one length per start
so the residues and the parity strings match one for one
the continued fraction of log₂3 starts 1, 1, 1, 2
the denominator is always odd, so the solution stays where the map is defined
the depth of the tree is a whole number between 3 and 12
the largest number of steps is a whole number between 4 and 15
the largest start is a whole number between 20 and 4000
the longest pattern length checked is a whole number between 6 and 14
the longest pattern searched is a whole number between 8 and 20
the map is a bijection at every length checked
the multiplier condition predicts the drop, on a large member of the class
the number of convergents listed is a whole number between 5 and 10
the number of steps is a whole number between 3 and 7
the ordinary rule has exactly one cycle at these lengths
the ordinary rule is among those searched
the parity of the first steps depends only on the start modulo two to the k
the pattern length is a whole number between 4 and 10
the relative gap shrinks along the list
the rule with minus one has at least three, which no drift argument distinguishes
the share provably driven down grows with the number of steps allowed
the starting number is a whole number between 1 and 100000
the tree is still growing at the depth drawn
the view is one the family draws

`complex-set`

11 kinds of claim · 13 placements

and neither is four tenths
and so is minus one
one is not
the critical point is in this Julia set exactly when its parameter is in the Mandelbrot set
the escape time separates into bands
the iteration cap is a whole number between 20 and 400
the Julia parameter is a complex number
the sample count is a whole number between 60 and 700
the set is the Mandelbrot set or a Julia set
the span is between 0.0005 and 4
zero is in the Mandelbrot set

`divergence`

32 kinds of claim · 23 placements

doubling 1 times returns 0/1 to itself ×120
of total length (2/3)^0 ×11
after 0 steps there are 2 to the 0 surviving pieces ×6
slope 3: box-counting dimension equals 1 − κ/λ ×4
the number of steps is a whole number between 10 and 200 ×3
a periodic point lies within the interval
a random start survives s/(s − 2) steps on average
a rational turn has only its periodic points
an irrational turn is transitive and nothing else
and at the step z nears q, z and p are at least half the far distance apart
and below the onset of chaos it is negative
at r = 4 the exponent is log 2, which is exactly computable
doubling on the line is sensitive and nothing else
the column count is a whole number between 40 and 1200
the dense orbit enters the interval
the dense orbit later comes near the far point
the doubling map has all three
the largest period is a whole number between 2 and 8
the lower parameter is between 0 and 4
the measured escape rate is log(s/2)
the number of strips is a whole number between 4 and 64
the orbit enters every one of the 16 strips within 400 steps
the orbits separate exactly when the parameter is past the onset of chaos
the parameter is between 0 and 4
the range runs upward and is wide enough to draw
the slope is above 2, so the middle escapes
the starting point is between 0 and 1
the two orbits begin the stated distance apart
the two starts are close together
the upper parameter is between 0 and 4
the view is one of lyapunov, dense, periodic, banks, table, cantor, survival, transient, kg
z and p start inside the same small interval

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