A curated atlas of mathematical beauty, with quadratic polynomials as its central thread.
How this collection was researched
The strongest selections combine a distinctive silhouette, a meaningful mathematical construction, an explanation that survives scrutiny, and an interaction that reveals something new. “Best” is an editorial judgment, not a ranking established by the literature. Color variants and parameter presets belong inside one exploration.
Three parallel research tracks examined quadratic dynamics, unusual arithmetic geometry, and substitution sequences. Original papers, author manuscripts, official mathematical references, and exact computations informed the final selection. Seven additions replace repetitive sequence charts and overlapping patterns. Classical prime spirals remain alongside rarer discoveries.
Understand the rule
Every picture is calculated from its stated construction. Coordinates, connectivity, and color each have an explicit role.
Separate evidence
An exact identity, a published theorem, a finite experiment, and an unresolved conjecture support different kinds of claims.
Follow the connections
One quadratic family links the Mandelbrot set, Julia portraits, and logistic bifurcations; other quadratics organize primes and tangent circles.
What the research changed
The central Mandelbrot cardioid has an exact fixed-point parametrization. Julia portraits examine starting points while Mandelbrot examines parameters. The logistic plot is conjugate to a real quadratic map. Apollonian circles emerge from the Descartes relation; Conway’s river turns the sign of a quadratic form into a path through hyperbolic geometry. Rauzy’s tile uses a contracting algebraic projection, and the Fibonacci word uses a precise parity-dependent turn rule. Their individual notes below cite the supporting sources.
Finite images remain finite: dark escape pixels do not prove boundedness; dense prime diagonals do not prove infinitely many primes; a dot cloud does not certify a fractal boundary. Each exploration states its particular limitation, rather than asking the image to stand in for a proof.
Explore the notebook
01 / 40 · Quadratic worlds
Mandelbrot: the parameter universe
One critical orbit maps an entire universe of quadratic rules.
z₀=0, zₙ₊₁=zₙ²+c; each pixel is a parameter c
Why the shape emerges
Every pixel selects a different quadratic rule and follows its critical starting point, zero. The central cardioid contains rules with an attracting fixed point; neighboring bulbs support longer attracting cycles. Their boundaries gather parameters where behavior changes, creating the intricate outline. Squaring doubles angles and changes magnitudes; adding c shifts the result before the next iteration. Outside colors measure escape speed; dark pixels that have not escaped are only a finite approximation to the set. [1]
A closer mathematical look
A fixed point w satisfies w=w²+c. Its multiplier is λ=2w, so c=λ/2−λ²/4. As λ goes around the unit circle, this equation traces the cardioid boundary. The attracting side has |λ|<1. This gives an algebraic explanation for the central heart shape; it does not settle all questions about the surrounding boundary. [1]
Construction and experiment
Choose c from the pixel’s real and imaginary coordinates.
Iterate zero under z²+c and test for escape.
Color escaped points by escape time; inspect a parameter’s corresponding Julia portrait.
Try it: Start with the whole set, then compare the seahorse valley and the miniature near c=−1.75488.
What the image cannot establish. This is the parameter plane, not a picture of one orbit. Failure to escape within the cap does not prove membership. Analytic tests can certify the interiors of the main cardioid and period-two bulb; the remaining dark pixels are unresolved by this finite computation.
A single parameter transforms the fate of every starting point.
zₙ₊₁=zₙ²+c; c fixed, each pixel is a starting point z₀
Why the shape emerges
Here the quadratic rule stays fixed while each pixel supplies a different starting point. Squaring has two inverse branches, so repeated pullbacks create paired lobes, filaments, or separated dust. The critical orbit of zero determines whether the filled Julia set is connected; the Mandelbrot set collects exactly the parameters for which it is. Comparing the rabbit, basilica, dendrite, and circle reveals genuinely different orbit structures behind their silhouettes, rather than different color treatments of one shape. [1]
A closer mathematical look
The two inverse solutions of z²+c=w explain the paired structure of pullbacks. At c=0 the filled set is the unit disk and its Julia boundary is the unit circle. At c=−1 the critical orbit is exactly 0→−1→0. The rabbit preset approximates an algebraic parameter; the dust preset’s critical orbit escapes in the reference computation. Named portraits change the rule itself. [1]
Construction and experiment
Fix c using a named mathematical preset.
Iterate every pixel as a separate starting value.
Compare the surviving region and its boundary with the critical orbit.
What the image cannot establish. The non-escaping region approximates the filled Julia set; its boundary is the Julia set. A finite iteration cap does not prove boundedness. The rabbit parameter is a numerical approximation; only the exact algebraic center has exactly periodic critical orbit. Color is an escape-time convention.
A real parabola grows a branching atlas of long-term states.
xₙ₊₁=r xₙ(1−xₙ), 2.8≤r≤4
Why the shape emerges
Each vertical slice chooses one growth parameter and plots states retained after an initial settling period. A stable fixed point splits into a two-cycle, then further doublings create the familiar branching structure. Wider bands contain many visited states, while periodic windows open clear lanes inside complicated regions. This is another quadratic polynomial: the coordinate change z=r(½−x) gives zₙ₊₁=zₙ²+r(2−r)/4, connecting the diagram directly to the real axis of the Mandelbrot family. [2][3]
A closer mathematical look
The nonzero fixed point is x*=1−1/r. Substituting into the derivative r(1−2x) gives 2−r. Its absolute value reaches one at r=3, explaining the first split. The coordinate change z=r(½−x) transforms the map into z²+c with c=r(2−r)/4. Periodic windows remain after the first accumulation of doublings, so a dense vertical band is not by itself a proof of chaos. [2][3]
Construction and experiment
Sample r evenly between 2.8 and 4 with a nonexceptional initial state.
Discard the selected number of transient iterations.
Plot the next 64 states at that r without connecting different columns.
Try it: Compare the first split near r=3 with the period-three window near r=3.83; increase the transient count.
What the image cannot establish. This is a finite orbit sample, not a proof of an attractor, period, or chaos. Near bifurcations transients can be long, and floating-point orbits eventually repeat. Periodic windows occur beyond the first period-doubling accumulation; not every parameter above it is chaotic.
Add roots of unity in quadratic order to draw a complex exponential sum.
Sₘ=Σₖ₌₀ᵐ⁻¹ exp(2πi a k²/N)
Why the shape emerges
Every segment has unit length, but its direction is chosen by a quadratic residue. Consecutive directions change by a phase proportional to 2k+1, so nearby steps sometimes reinforce one another and sometimes wrap into tight curls. Accumulating these vectors draws the partial sums of a quadratic exponential sum in the complex plane. The final point is the complete Gauss sum; the intermediate curls reveal cancellation that the final total alone hides. [19]
Construction and experiment
Compute the phase 2πak²/N for each k from zero to N−1.
Add the corresponding unit vector to the running complex sum.
Connect consecutive partial sums and color by k.
Try it: Keep modulus 257 and vary a from 1 to 7 to rearrange the curlicues.
What the image cannot establish. Classical magnitude formulas require hypotheses on the modulus and coefficient; no universal √N claim is made here.
A polynomial draws its own diagonal through the integers.
f(n) = n² + n + c
Why the shape emerges
A square spiral assigns each integer a lattice address. Here only values produced by one quadratic appear, so the path isolates a sequence with growing gaps instead of showing every prime. At c = 41, the first forty values are prime; the next is 41². Changing the constant slides the polynomial through different divisibility obstacles. The bright and dark points distinguish primes from composites, while the diagonal shape comes from the placement rule. [7]
Construction and experiment
Evaluate n²+n+c for consecutive nonnegative n.
Place each value at its square-spiral address.
Color primes bright and composites dark.
Try it: Set c to 17, then 41; find where each initial prime run fails.
What the image cannot establish. No nonconstant integer polynomial produces only positive primes forever. The shape is caused by spiral coordinates, not a proof about prime density.
Each cell is a different polynomial, with the linear coefficient on the horizontal axis and the constant on the vertical axis. Its brightness records the fraction of tested inputs that produce positive primes. Dark bands arise when small primes divide many outputs; parity alone erases large regions. Islands such as n²+n+17 and n²+n+41 emerge from fewer local obstructions. Increasing the sample can change the ranking, which is why this is an experimental atlas. [8][7]
Construction and experiment
Choose coefficients b and c at every cell.
Test the first N values for ordinary primality.
Map the measured proportion to the gradient.
Try it: Expand the coefficient radius to 20 to include c = 41; compare longer samples.
What the image cannot establish. This is a finite prime count, not a conjectural asymptotic density or a guarantee of future primes.
Each row observes the same quadratic through a different modulus. Polynomial remainders repeat after m inputs, so row m has an exact horizontal period dividing m. Rows with small periods become bands; combinations of periods make larger woven motifs. Zero-valued cells identify divisibility by that row’s modulus. At c = 41, small prime rows contain unusually few obstacles, helping explain the polynomial’s celebrated opening run without suggesting that the pattern can guarantee primality. [7]
Construction and experiment
Evaluate the quadratic at each horizontal input.
Reduce that value by every row’s modulus.
Color each remainder relative to its modulus.
Try it: Compare c = 40 and c = 41; inspect the modulus-2 row.
What the image cannot establish. A nonzero remainder only rules out one divisor. Composite moduli are included and do not constitute independent primality tests.
A finite deterministic map cannot keep discovering new residues forever. Each orbit eventually repeats, giving a cycle with a tree of incoming paths. Here angular position records the residue and distance from the central circle records the number of steps before reaching a cycle. Points on the inner ring are periodic; outward branches are temporary states. Shared colors identify the same eventual cycle, making the map’s complete long-term behavior visible at once. [9]
Construction and experiment
Follow every starting residue until its first repeat.
Measure its tail length and identify its cycle.
Place tails outward and draw the transition arrows as segments.
Try it: Compare c = 0, 1, and 2 at modulus 97.
What the image cannot establish. These are finite modular dynamics; the layout is not a complex Julia set.
Four tangent circles satisfy a quadratic equation in their signed curvatures. Keeping three circles fixed gives two possible fourth circles; switching between them uses an integer linear rule. Starting from bends −1, 2, 2, 3 therefore keeps every bend integral. Applying the same rule to curvature-times-center coordinates locates each new circle exactly. Repeating the operation fills triangular gaps with smaller tangent circles, creating the nested lace. [25][26][27]
A closer mathematical look
The seed uses signed bends (−1,2,2,3). With three bends fixed, the Descartes quadratic has two roots for the fourth; their sum is twice the sum of the other three bends. Replacing one root therefore preserves integers. The same replacement in bend-times-center coordinates preserves exact rational centers. An important research correction: the old local-global conjecture for Apollonian bends is false; additional reciprocity obstructions can exclude congruence-allowed families. [25][26][27]
Construction and experiment
Begin with signed bends (−1,2,2,3) and their exact rational centers.
Replace one circle using the linear bend and bend-center rules.
Fill successive gaps; color by logarithmic bend and inspect circle labels.
Try it: Start at bend360, then increase the depth to discover circles accumulating near tangency points.
What the image cannot establish. The enclosing circle has negative bend. The display stops at a bend and depth cutoff. Congruence-allowed integers need not all occur as bends.
A quadratic sign change threads a river through rational hyperbolic triangles.
|ps−qr|=1; Q(p,q)=p²−Dq²; roots ±√D
Why the shape emerges
Rational pairs with determinant one form neighboring vertices of the Farey tessellation. In the disk model their connecting geodesics become arcs meeting the boundary at right angles. Evaluate p²−Dq² at each rational label: positive and negative regions are separated by a path in the dual tree. This Conway river approaches the two irrational roots. Its bright broken path differs from the smooth geodesic sharing those same endpoints. [28][29]
A closer mathematical look
A rational p/q becomes a boundary point through the Cayley map. A determinant of ±1 singles out neighboring fractions. The quadratic form Q(p,q)=p²−Dq² is evaluated on primitive pairs. For nonsquare D its zero directions are irrational; positive and negative regions meet along a dual-tree path. The displayed straight dual links are a drawing convention, while the rational adjacency and sign decisions are exact. [28][29]
Construction and experiment
Build unimodular rational triangles and map them into the Poincaré disk.
Mark dual edges crossing between opposite signs of p²−Dq².
Compare the resulting river with the smooth axis joining −√D and +√D.
Try it: Compare D=2,3,5,13 to see different paths through the same rational tessellation.
What the image cannot establish. Square D inputs use D+1. Thin arcs are Farey edges; the bright broken river lies on the dual tree. Neither the dual links nor their chosen triangle centers are claimed to be hyperbolic incenters.
Project the letter counts of an endless substitution word into three interlocking regions.
1→12, 2→13, 3→1; z=c₁+(α−1)c₂+c₃/α, where α³−α²−α−1=0 and Im(α)<0.
Why the shape emerges
Three substitution rules generate an endless word. Each prefix has three integer letter counts; a projection built from the contracting roots of a cubic equation turns those counts into a point. The average growth disappears, leaving a bounded tile. Applying the substitution to a whole prefix rotates and contracts its projection, explaining the repeated detail. Each color records the final letter of a prefix and separates one of the tile’s three pieces. [33][34][35][36]
A closer mathematical look
Let M be the substitution matrix with rows (1,1,1), (1,0,0), (0,1,0). Its polynomial x³−x²−x−1 has one expanding real root and a contracting conjugate pair. With w=(1,α−1,1/α), the identity wM=αw means substitution rotates and contracts the projected prefix. The average letter-frequency direction disappears. Points are plotted after adding a letter, and that final letter determines the color, matching the cited Sage convention. [33][34][35][36]
Construction and experiment
Start with 1 and repeatedly apply 1→12, 2→13 and 3→1.
For every nonempty prefix, count its letters and compute z=c₁+(α−1)c₂+c₃/α.
Plot Re(z), Im(z) as a point and color it by the prefix’s final letter.
Try it: Compare 1,000, 10,000 and 50,000 points; watch the same three regions gain finer boundary detail.
What the image cannot establish. The finite point cloud approximates the infinite tile. Dot size affects apparent gaps and cannot establish whether a region contains a hole.
An exact odd-even drawing rule builds repeating bays and corridors from a binary word.
0→01, 1→0. Draw one step; at a 0 turn left 90° for even position k and right 90° for odd k; at a 1 stay straight.
Why the shape emerges
A word grows by replacing zero with zero-one and one with zero. Each digit draws a unit step, and a zero then turns left or right according to its position’s parity. Longer words concatenate earlier words, so related blocks of instructions reappear at increasing scales. Their interactions create repeating bays and corridors. The gradient follows the order of the segments along the path, while every corner remains on the integer lattice. [37][38][39]
A closer mathematical look
The order of operations matters: draw a unit edge first, then decide the next turn. Positions start at one. Integer coordinates and a four-direction heading make every displayed edge exact. The recursion of the word organizes related drawing blocks, but a finite image cannot prove a global topological property. The original 2009 preprint treats universal non-self-intersection as an open problem; this atlas makes no such universal assertion. [37][38][39]
Construction and experiment
Generate the binary word 0100101001001… using 0→01 and 1→0.
Read positions starting at one and draw a unit segment before deciding the next turn.
After a zero turn left at even positions and right at odd positions; after a one continue straight.
Try it: Compare 1,597, 4,181 and 10,946 segments to find related bays at larger scales.
What the image cannot establish. This shows an exact finite prefix. Changing the initial direction rotates the picture; exchanging left and right reflects it. A finite drawing alone does not prove a property of the infinite limit.
Primes seem irregular on a number line. Arrange the integers on a square spiral and diagonal streaks emerge. Along each diagonal, the numbers follow a quadratic rule; some of these rules encounter fewer small prime divisors. [4]
Construction and experiment
Place 1 at the center and count outward in a square spiral.
Test each integer for primality.
Illuminate the primes; leave composite numbers faint.
Try it: Raise the integer count. Notice how the diagonal streaks persist even as individual gaps change.
What the image cannot establish. A prime-rich diagonal is not an unbroken line of primes, and the picture is not a proof of an infinite prime family.
Let the square root of each integer set both its distance and its number of turns. Perfect squares land on the same ray because their square roots are integers. Prime-rich quadratic sequences trace curved paths through this arrangement. [5]
Construction and experiment
Give integer n a radius of √n.
Turn through √n complete revolutions.
Keep only prime-numbered points.
Try it: Switch between this and the Ulam spiral. The same primes create different curves because the coordinate rule changes.
What the image cannot establish. The spiral reveals structure in the mapping; it does not predict which arbitrary integers are prime.
A full turn is 2π radians. Since 44 is close to 7 full turns, integers separated by 44 can nearly align. Other rational approximations to 2π create patterns at other scales. Filtering for primes removes many residue classes, sharpening some arms. [6]
Construction and experiment
Use n as both radius and angle in radians.
Plot the integers in polar coordinates.
Highlight only primes and inspect the near alignments.
Try it: Change the number of integers to reveal how the apparent arms depend on scale.
What the image cannot establish. These arms reflect near rotations and divisibility, not an exact periodicity of primes.
Residues sit at equal angular intervals around a circle, and each chord connects an input to its cubic image. The odd polynomial pairs residue k with −k and pairs their outputs the same way. With the circle oriented here, this gives mirror symmetry across its vertical axis. Cubing can permute all residues for some prime moduli but merge several inputs for others. Adding the linear term changes the collision structure rather than merely rotating the endpoints, allowing the same circle to move between sparse braids and concentrated bundles. [9]
Construction and experiment
Place residues 0 through m−1 on a circle.
Compute k³+ck modulo m for each input.
Join inputs to outputs and compare collisions.
Try it: Set c=0 and compare prime moduli 41 and 43: cubing permutes residues only for 41.
What the image cannot establish. The figure shows a finite polynomial map, not a real cubic curve. Prime moduli and composite moduli have different permutation rules.
A Gaussian integer occupies the lattice point with coordinates a and b. Away from the axes it is prime precisely when its squared distance from the origin is an ordinary prime. On the axes, ordinary primes congruent to 3 modulo 4 survive as Gaussian primes. Multiplying by i rotates a point a quarter turn, and conjugation reflects it, explaining the exact symmetry. The radial gradient encodes distance, while every displayed point passes the arithmetic test. [10]
Construction and experiment
Enumerate integer real and imaginary parts.
Apply the norm test off-axis and the special axis test.
Plot surviving primes with a radial gradient.
Try it: Find 3 on the real axis and compare it with the missing ordinary prime 5.
What the image cannot establish. An ordinary prime can factor in the Gaussian integers: 5=(2+i)(2−i).
Replacing the square lattice basis with two vectors separated by 120 degrees makes a triangular lattice. Its norm is still squared Euclidean distance, now written a²−ab+b². Prime norms give Eisenstein primes, while rational primes congruent to 2 modulo 3 also survive along the six unit directions. The sixfold symmetry follows multiplication by the ring’s units. This is not merely a rotated Gaussian picture: a different norm changes which ordinary primes split into factors. [11]
Construction and experiment
Embed (a,b) as (a−b/2,√3·b/2).
Keep prime norms and the inert-prime axis associates.
Clip to a circular norm bound.
Try it: Compare the sixfold symmetry with the Gaussian field’s fourfold symmetry.
What the image cannot establish. Coordinates use ω at 120 degrees, hence the minus sign in the norm. The radius cap preserves complete rotational shells.
Equal radii reveal how many ways an integer splits.
r₂(n)=#{(a,b)∈ℤ² : a²+b²=n}
Why the shape emerges
Every point on a circular shell has the same squared distance and therefore represents the same integer as a sum of two squares. Here color depends on the number of lattice points on that shell, rather than its radius. Sign changes and coordinate swaps account for much of the symmetry, but some integers support several genuinely different representations. A bright circle can therefore have many arithmetic witnesses even beside a radius with no integer points at all. [10]
Construction and experiment
Count all signed, ordered representations within the radius.
Assign each point its shell’s representation count.
Color shells by that count.
Try it: Compare the shells for 25 and 65: both allow more than one representation pattern.
What the image cannot establish. Counts include signs, order, and zero coordinates. Colors saturate at 32 representations.
Plot each integer against the count of residues coprime to it. The ratio φ(n)/n depends on which primes divide n, not on how large their exponents are. Integers with the same collection of prime divisors therefore lie on the same ray through the origin. Primes gather near the upper diagonal, while integers divisible by several small primes fall lower. The fan is an image of multiplicative structure, with slope recording the fraction of surviving residues. [12]
Construction and experiment
Factor each integer into distinct prime divisors.
Compute its totient exactly.
Plot (n,φ(n)) and color by φ(n)/n.
Try it: Follow powers of 2 along the ray y=x/2.
What the image cannot establish. Finite sampling makes some rays look denser than others; the color represents a ratio, not primality.
Unit steps build a spiral whose spoke lengths are square roots.
z₁=1; zₖ₊₁=zₖ+i zₖ/|zₖ|; |zₖ|=√k
Why the shape emerges
Each new outer edge has length one and stands at a right angle to the previous radius. Pythagoras therefore adds one to the squared radius at every step, giving spoke lengths √1, √2, √3 and onward. As the radius grows, the turn made by a unit edge becomes smaller. Those shrinking turns create the broad spiral, while the colored spokes reveal the sequence of right triangles. [13]
Construction and experiment
Start with a unit spoke on the positive x-axis.
Add a perpendicular unit edge at its tip.
Join the new tip to the origin and color by triangle number.
Try it: Compare 17 triangles with 180 to see when the construction winds over earlier turns.
What the image cannot establish. Later triangles overlap. This is the discrete construction, not a fitted logarithmic spiral.
A fixed skip on a circular dial splits into gcd-controlled stars.
j ↦ j+k (mod N); number of cycles=gcd(N,k)
Why the shape emerges
Place equally spaced integers around a circle and connect every vertex to the one a fixed number of steps ahead. Repeated addition eventually returns to its starting point. If the skip and vertex count share a divisor, the walk visits only one residue class before closing, so several separate stars coexist. The circular symmetry comes from addition modulo N; the number of components is governed by the greatest common divisor. [14]
Construction and experiment
Place N vertices evenly on a unit circle.
Draw each chord j→j+k modulo N.
Color chords by their starting index to reveal interlaced cycles.
Try it: Set 96 vertices and skip 36, then change the vertex count to 97.
What the image cannot establish. A shared divisor creates a compound star figure. A skip divisible by N leaves isolated vertices with zero-length edges.
A point is visible from the origin precisely when its coordinates are coprime.
(x,y) visible from (0,0) ⇔ gcd(|x|,|y|)=1
Why the shape emerges
An integer point disappears behind a nearer lattice point whenever both coordinates share a common factor. Removing those hidden points leaves precisely the primitive integer directions. Rays crowd together as the viewing window grows, but the nearest points along simple rational slopes remain conspicuous. Reflection creates the symmetric starburst. Color measures direction around the origin, so the gradient follows angle rather than distance or the size of the coordinates. [15]
Construction and experiment
Enumerate integer points in a square centered on the origin.
Retain exactly those with coordinate gcd equal to one.
Draw a ray from the origin and color it by polar angle.
Try it: Look for (2,3), then notice why (4,6) contributes no new ray.
What the image cannot establish. Lines may cross many pixels; visual brightness is not a statistical density estimate.
A rational slope folds a straight line into a closed square orbit.
y′/x′=p/q; fold(u)=1−|1−(u mod 2)|
Why the shape emerges
Imagine reflecting the square whenever a moving point reaches a wall. In the reflected copies, the path continues as one straight line. Folding it back creates the crisscrossed embroidery. When the velocity components have rational ratio p/q, both reflected coordinates repeat together after a common period, so the orbit closes. The starting point is chosen away from corners, and the gradient records time around one complete orbit. [16]
Construction and experiment
Reduce the slope p/q and start inside the unit square.
Trace a straight line through reflected copies, recording every wall crossing.
Fold coordinates into the original square and color by orbit order.
Try it: Compare slopes 13/21 and 13/20; tiny arithmetic changes rearrange the weaving.
What the image cannot establish. Rational slope is the condition here, not a rational angle in degrees. The generic interior start avoids corner ambiguity.
Binary digits guide a continuous traversal of a square lattice.
Index d∈[0,4ᵏ) ↦ rotated/reflected base-4 quadrants
Why the shape emerges
At each scale, four smaller traversals are rotated and reflected so their endpoints meet. Reading an integer index in base four determines which nested quadrants to enter. That arithmetic address turns a one-dimensional count into a continuous path through every point of a finite square grid. The repeated corner motif comes from orientation changes needed to keep neighboring pieces connected, while the gradient exposes the traversal’s order. [17]
Construction and experiment
Split each index into successive pairs of binary digits.
Rotate or reflect the local quadrant before adding its position.
Connect consecutive grid points and color by index.
Try it: Compare this connected labyrinth with the jumps in Morton’s digit zipper.
What the image cannot establish. The displayed finite polygonal path is not itself space-filling; that property concerns the normalized infinite limit.
A ratio of integers decides when a polar flower closes.
r=cos(pθ/q); (x,y)=(r cos θ,r sin θ)
Why the shape emerges
Two angular rhythms interact: the direction turns once per full revolution, while the radius oscillates at frequency p/q. Rational frequency makes the rhythms realign after finitely many turns, closing a layered flower. Negative radii place points on the opposite ray, which changes the closure period when both reduced integers are odd. The gradient follows the full tracing period, revealing repeated crossings and the order in which petals are drawn. [18]
Construction and experiment
Reduce the frequency fraction p/q.
Sample the polar radius and convert signed polar coordinates to Cartesian points.
Trace the exact parity-dependent closure interval, coloring by elapsed angle.
Try it: Compare 7/4, 7/3, and 6/4; reducing the fraction reveals shared shapes.
What the image cannot establish. This is a continuous geometric curve controlled by integers, rather than a sequence of integer-coordinate points.
Quadratic-residue differences decide which vertices are neighbors.
Vertices a∈Fₚ; join a,b iff b−a is a nonzero square, p≡1 mod 4
Why the shape emerges
A quadratic-residue rule chooses which pairs of labels connect. Every vertex has the same arithmetic neighborhood translated around the finite field, creating rotational symmetry in the circular drawing. For a prime congruent to one modulo four, minus one is a square, so the adjacency rule works in both directions and the graph is undirected. The resulting rosette combines rigid symmetry with the irregular-looking selection of square residues. [20]
Construction and experiment
Choose the first prime p≥the slider value with p≡1 modulo 4.
List the nonzero quadratic residues modulo p.
Connect each pair whose difference is on the list and color by starting vertex.
Try it: Try 5, 13, and 41 to watch equal-degree neighborhoods form progressively denser rosettes.
What the image cannot establish. The slider chooses a prime at or above its value. This implementation uses prime fields, not arbitrary prime-power fields.
Binary words become vertices; flipping one bit creates an edge.
v(b)=Σⱼ(bⱼ−½)(cos(πj/d),sin(πj/d)); edges differ in one bit
Why the shape emerges
Every binary word labels a vertex of a high-dimensional cube. To see it in the plane, assign one evenly spaced direction to each bit and add the directions selected by that word. Flipping a single bit moves along exactly one of those directions, creating families of parallel edges. The rich constellation is a projection of binary adjacency: point colors show the number of set bits, and edge colors identify which bit changes. [21]
Construction and experiment
Enumerate every binary word of the selected length.
Project its bits onto evenly spaced planar directions.
Connect words differing in exactly one bit; color edges by that bit position.
Try it: Increase dimensions from 3 to 7 and follow one family of parallel edges.
What the image cannot establish. Distinct high-dimensional vertices can overlap in this planar projection. A visual crossing is not an additional graph vertex.
A small circle rolls inside a larger one while a point fixed inside the small circle traces the path. The rolling center and the marker rotate at different rates, and the ratio of integer radii makes them realign after a finite number of turns. Their superposition creates nested loops and repeated lobes. The marker offset is fixed at sixty-five percent of the small radius; color reveals one complete rolling period. [22]
Construction and experiment
Choose integer radii, raising R to r+1 if needed.
Keep the marker distance d=0.65r from the small circle center.
Trace a period 2πr/gcd(R,r), coloring by elapsed parameter.
Try it: Try radii 13 and 5, then 12 and 5, to compare different closure symmetries.
What the image cannot establish. If the inner radius is not smaller, the outer radius automatically becomes r+1. Curves are sampled numerically.
A prime field builds a geometry where every pair of lines meets.
Points [x:y:z], lines [a:b:c]; incidence ax+by+cz≡0 mod p
Why the shape emerges
A point is represented by a nonzero triple over a finite field, with scalar multiples treated as the same point. Lines use the same kind of triples and meet points through one dot-product congruence. This creates equally many points and lines, with exactly p+1 incidences at each. The drawing places point labels on the outer ring and line labels on the inner ring; threads indicate incidence rather than ordinary Euclidean lines. [23]
Construction and experiment
Choose the first prime at or above the slider value.
Normalize each nonzero field triple so its first nonzero coordinate is one.
Put points outside, lines inside, and join pairs whose dot product vanishes modulo p.
Try it: Set p=2 to see the seven points and seven lines of the Fano plane as a 14-vertex incidence graph.
What the image cannot establish. The two rings are a graph layout. The inner vertices represent abstract projective lines, not points in the projective plane.
Consecutive Fibonacci remainders close into an exact finite orbit.
(a,b)↦(b,a+b) mod M; start (0,1)
Why the shape emerges
Remembering two consecutive Fibonacci remainders gives enough information to compute both the next pair and the previous pair. Because only finitely many pairs exist modulo M and the update is reversible, the orbit must return to its initial pair. Its length is the Pisano period. Plotting both remainders as coordinates exposes geometric order that a list of last digits hides; the gradient follows one complete return cycle. [24]
Construction and experiment
Start from the consecutive Fibonacci pair (0,1).
Update the pair modulo M until (0,1) returns.
Plot both residues as Cartesian coordinates and color by position in the cycle.
Try it: Use modulus 10 and verify that the orbit closes after 60 updates.
What the image cannot establish. The plot shows residue pairs and connecting index order. It does not display the exponentially growing Fibonacci magnitudes.
Read the sequence as a turtle path: turn by each term times 90°, then move one unit.
R(n)=(−1)^(number of overlapping 11 pairs in binary n), starting at n=0.
Why the shape emerges
The parity of overlapping binary pairs chooses a left or right quarter-turn at every step. Neighboring integers share many binary digits, so the path contains structured repetitions even when the sign stream looks irregular. The displayed geometry belongs to this chosen turtle interpretation, not to the sequence alone. Its crossings represent repeated lattice visits; they do not mean the underlying terms have become periodic. [30]
Construction and experiment
Generate the sequence from its stated rule.
Start at the origin facing right.
Turn by value × 90°, then draw one unit; repeated visits can overlap.
Try it: Check n=7: binary 111 has two overlapping pairs and gives +1.
What the image cannot establish. A turtle path is not a plot of the sequence’s partial sums.
Read the sequence as a turtle path: turn by each term times 90°, then move one unit.
For k≥1 remove factors of 2: odd part ≡1 mod4 gives +1; ≡3 mod4 gives −1.
Why the shape emerges
Repeated folding builds a turn pattern containing reflected and reversed copies of earlier instructions. Encoding the two crease directions as plus and minus one turns that pattern into left and right quarter-turns. Equal step lengths then reveal the dragon’s nested structure. Rotation or reflection depends on the chosen turning convention, while the recursive arrangement survives. More terms add detail rather than smoothing the corners. [31]
Construction and experiment
Generate the sequence from its stated rule.
Start at the origin facing right.
Turn by value × 90°, then draw one unit; repeated visits can overlap.
Try it: Compare 255, 511 and 1023 turns to see successive folding stages.
What the image cannot establish. Signed encoding 2b−1 of the regular paperfolding bits; orientation is conventional.
Follow successive terms on an index-versus-value graph.
Begin 1,2,3; choose least unused x with gcd(x,a(n−2))>1 and gcd(x,a(n−1))=1.
Why the shape emerges
The next value must share a factor with the term two places back while avoiding every factor of the immediate predecessor. These competing requirements often alternate even and odd choices, then force a much larger unused candidate. The line plot turns those jumps into geyser-like spikes. Such visual bursts follow the greedy factor constraints; the precise large-scale distribution requires mathematics beyond observing a finite picture. [32]
Construction and experiment
Generate the sequence from its stated rule.
Place each term at its index and numerical value.
Join neighboring terms; color follows position in the sequence.
Try it: Inspect 4,9,8,15: compare gcds with the previous two terms.
What the image cannot establish. Known to be a permutation, but a short plot does not establish conjectured growth laws.
Place residues around a circle and connect each one to its multiple. Adjacent chords change direction by a regular amount, so their dense envelope forms a smooth curve. Doubling gives a cardioid-like envelope; higher multipliers add lobes. [40]
Construction and experiment
Space N points evenly on a circle.
Multiply each point’s index by m, modulo N.
Draw a chord from the original point to the result.
Try it: Move the multiplier from 2 to 3 to 4. Watch the envelope gain another lobe each time.
What the image cannot establish. The smooth envelope is approached as the point count grows; a finite plot contains only straight chords.
Each entry in Pascal’s triangle adds the two above it. Modulo 2, two odd entries cancel to an even entry. Repeated across rows, this local rule produces the nested gaps of the Sierpiński triangle. Other moduli create their own repeating structures. [41]
Construction and experiment
Start with a single 1 and fill each row by addition.
Reduce each entry modulo the selected number.
Color the nonzero remainders and darken the zeros.
Try it: Compare moduli 2, 3, and 5, then try the composite modulus 6. The pattern changes with arithmetic.
What the image cannot establish. The exact Sierpiński correspondence described here is for modulus 2.
Reversed trajectories · odd steps turn left, even steps right
T(n) = n/2 if even; 3n + 1 if odd
Why the shape emerges
Different starting values often merge into the same Collatz trajectory. Reverse those finite paths and draw a small turn for each odd or even step: shared arithmetic becomes a branching tree. The organic silhouette comes from the drawing angle, not from a geometric theorem. [42]
Construction and experiment
Follow each starting value until it reaches 1.
Reverse the path so every branch grows from 1.
Turn left for an odd predecessor, right for an even one.
Try it: Change the turn angle. The arithmetic stays the same while the tree’s silhouette changes.
What the image cannot establish. Whether every positive integer eventually reaches 1 is still unproved. All starting values in this demo do reach 1.
Try stepping backward by an ever larger distance. When that would land at a nonpositive or previously visited value, step forward instead. Drawing each jump as a semicircle turns this simple memory rule into interlocking arches. [43]
Construction and experiment
Start at a₀ = 0.
At step n, subtract n if the result is positive and unseen; otherwise add n.
Connect consecutive terms with alternating upper and lower semicircles.
Try it: Scrub slowly through the first 30 steps. Follow the sequence 0, 1, 3, 6, 2, 7, 13…
What the image cannot establish. Forward moves may revisit earlier terms. The no-repeat test applies only to the proposed backward move.
The golden angle is an irrational fraction of a full turn. New seeds avoid repeatedly landing on the same rays, distributing points evenly. Fibonacci rational approximations to that angle explain the pairs of visible spiral families at different scales. [44]
Construction and experiment
Increase each seed’s radius in proportion to √n.
Advance its angle by the chosen fixed rotation.
Color seeds from the center outward.
Try it: Compare 137.51° with 144°. The latter is exactly 2/5 of a full turn and produces five rays.
What the image cannot establish. The slider approximates the golden angle to two decimal places. This is a geometric growth model, not a universal law of plants.
Each reduced fraction p/q supports a circle of radius 1/(2q²)
center = (p/q, 1/2q²), radius = 1/2q²
Why the shape emerges
Put a circle above every reduced fraction. Its radius shrinks with the square of the denominator. Two circles are tangent precisely when |ps − qr| = 1 for fractions p/q and r/s. The arithmetic keeps all circle interiors disjoint. [45]
Construction and experiment
List reduced fractions between 0 and 1.
Center each circle directly above its fraction.
Use radius 1/(2q²), making every circle touch the baseline.
Try it: Raise the denominator limit. Smaller fractions fill the gaps without overlapping existing circles.
What the image cannot establish. Circles for 0/1 and 1/1 extend beyond the unit interval; the viewport shows the portion near [0, 1].
Dates distinguish original publications from hosted versions where known. Undated reference pages are marked explicitly. Research accessed 13 September 2026. Source links accompany the claims they support above.
Eric W. Weisstein, MathWorld. Phyllotaxis. Undated reference.
Eric W. Weisstein, MathWorld. Ford Circle. Undated reference.
Validation and limits
All 40 explorations were exercised across control endpoints, reveal states, and two canvas sizes. Prime counts and greatest-common-divisor arithmetic were checked independently. New generators were checked against known critical orbits, the logistic change of coordinates, exact Apollonian tangency identities, unimodular Farey triangles, published substitution prefixes, and integer Fibonacci edges. These checks support the displayed finite constructions; they are not new proofs of the cited general theorems.
Curves outside number theory are included where their arithmetic or symbolic rules illuminate the collection. No claim of completeness is made. Historical large-scale prime counts are attributed research results, not calculations silently presented as having run in your browser.
arithmos