The Bellissima Curve

Thursday, 29th May, 2025 by Krilling for Company in Arithmetic, Fibonacci Sequence, Geometry, Mathematics and tagged bell curve, bellissima curve, golden gaps, recreational math, recreational mathematics, recreational maths, sampling integers | Leave a comment

The bell curve is a shape that appears when you make a graph by counting all possible sums of a range of integers like 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The smallest sum you can get is 1; the largest is 55 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10. But there’s only one sum of 1 and only one sum of 55. Other sums are more common:

• 10 = 1 + 2 + 3 + 4
• 10 = 1 + 2 + 7
• 10 = 1 + 3 + 6
• 10 = 1 + 4 + 5
• 10 = 2 + 3 + 5
• 10 = 2 + 8
• 10 = 3 + 7
• 10 = 4 + 6
• 10 = 10

So there are nine sums of 10. If you graph count-sums with a bigger set of consecutive integers from 1, 2, 3…, you get this shape:

Bell curve from sum-counts with 1, 2, 3, 4, 5, 6, 7, 8, 9, 10…
(open in separate window for full-sized image)

It’s a bell curve. Et c’est une belle curve, a “beautiful curve” in French. But I’ve found what I call bellissime curve — Italian for “most beautiful curves” — by sampling different sets of integers. With the set (1, 3, 5, 7, 9, 11, 13, 15, 17, 19…), you get what you could call a slightly wrinkled bell curve:

Wrinkled bell-curve from sum-counts with 1, 3, 5, 7, 9, 11, 13, 15, 17, 19…
(open in separate window for full-sized image)

After that, as you leave bigger gaps in the sampled sets, the curves start to overlap and add extra beauty:

Overlapping bell curves from sum-counts with 1, 4, 7, 10, 13, 16, 19, 22, 25, 28…

Bellissima curves from sum-counts with 1, 5, 9, 13, 17, 21, 25, 29, 33, 37…

Bellissima curves from sum-counts with 1, 6, 11, 16, 21, 26, 31, 36, 41, 46…

Bellissima curves from sum-counts with 1, 7, 13, 19, 25, 31, 37, 43, 49, 55…

With the set (3, 6, 9, 15, 18, 21…), the bell is back:

Bell curve from sum-counts with 3, 6, 9, 15, 18, 21…

But with (4, 7, 10, 13, 6, 19…), separated by the same distance, you get this:

Bell curve from sum-counts with 4, 7, 10, 13, 6, 19…

When you sample the Fibonacci numbers, (1, 2, 3, 5, 8…), you get this graph:

Caterpillar curve from sum-counts of Fibonacci numbers 1, 2, 3, 5, 8, 13, 21, 34, 55, 89…

When you sample a restricted set of Fibonaccis, (1, 3, 8, 21, 55…), you get this, where each vertical line represents a count of one:

Golden gaps from sum-counts of restricted Fibonacci numbers 1, 3, 8, 21, 55, 144…

That restricted Fibonacci graph is strangely attractive, because it has golden gaps (verb sap!).

Friday is Φiday

Friday, 23rd May, 2025 by Krilling for Company in Arithmetic, Fibonacci Sequence, Integer Sequences, φ, Mathematics, Phi and tagged algorithm for Fibonacci numbers, φ-day, φriday, recreational math, recreational mathematics, recreational maths | Leave a comment

The 11th, 12th and 23rd day of a month can be called a φ-day (pronounced fy-day). Why so? Because those numbers are consecutive entries in the famous Fibonacci sequence, which offers better and better approximations to a mathematical constant called φ = (√5 + 1) / 2 = 1.6180339887498948…:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, …

Each number after the second is the sum of the preceding two (so 11, 12, 23… could be the start of a similar sequence). When you divide fib(i) by fib(i-1), you get these approximations to φ:

2 = 2/1
1.5 = 3/2
1.6 = 5/3
1.6 = 8/5
1.625 = 13/8
1.6153846… = 21/13
1.619047619… = 34/21
1.6176470588235294117647… = 55/34
1.618… = 89/55
1.617977528… = 144/89
1.61805… = 233/144
1.618025751… = 377/233
1.618037135… = 610/377
1.618032786… = 987/610
1.618034447… = 1597/987
1.618033813… = 2584/1597
1.618034055… = 4181/2584
1.618033963… = 6765/4181
1.618033998… = 10946/6765
1.618033985… = 17711/10946

Today is the 23rd and not just a φ-day but a Friday (or φriday). So here’s one of the interesting results I’ve recently found while playing with the Fibonacci sequence. As any recreational mathematician kno, you can also find the Fibonacci sequence — and φ — with this little algorithm:

f = 0
LOOP
f = 1 / (f + 1)
print(f)
goto LOOP

The algorithm returns these values:

1, 1/2, 2/3, 3/5, 5/8, 8/13, 13/21, 21/34, 34/55, 55/89, 89/144, 144/233, 233/377, 377/610, 610/987, 987/1597, 1597/2584, 2584/4181, 4181/6765, 6765/10946, 10946/17711, …

I was playing with that algorithm and got an unexpected result with a simple adaptation of it:

f = 0
LOOP
f = 1 / (3 – f)
print(f)
goto LOOP

The values of f generated by this adapted algorithm are:

1/3, 3/8, 8/21, 21/55, 55/144, 144/377, 377/987, 987/2584, 2584/6765, 6765/17711, 17711/46368, 46368/121393, 121393/317811, 317811/832040, 832040/2178309, 2178309/5702887, 5702887/14930352, 14930352/39088169, 39088169/102334155, 102334155/267914296, …

The numerator and denominator in each fraction are next-but-one Fibonacci numbers, beautifully generated at each step:

3 – 0 = 3 → 1/3
3 – 1/3 = (3*3)/3 – 1/3 = 9/3 – 1/3 = (9-1) / 3 = 8 / 3 → 1/(8/3) = 3/8
3 – 3/8 = (3*8)/3 – 3/8 = 24/8 – 3/8 = (24-3) / 8 = 21/8 → 1/(21/8) = 8/21
3 – 8/21 = (3*21)/21 – 8/21 = 63/21 – 8/21 = (63-8)/21 = 55/21 → 1/(55/21) = 21/55
3 – 21/55 = (3*55)/55 – 21/55 = 165/55 – 21/55 = (165-21)/55 = 144/55 → 1/(144/55) = 55/144
3 – 55/144 = (3*144)/144 – 55/144 = (432-55)/144 = 377/144 → 1/(377/144) = 144/377
3 – 144/377 = (3*377)/377 – 144/377 = (1131-144)/377 = 987/377 → 1/(987/377) = 377/987
[…]

If you reverse numerator and denominator, the limit of the fraction is φ^2 = 2.6180339887498948… = φ+1:

3 = 3/1
2.6 = 8/3
2.625 = 21/8
2.6190476190476190476190476… = 55/21
2.6181818181818181818181818… = 144/55
2.6180555555555555555555555… = 377/144
2.6180371352785145888594164… = 987/377
2.6180344478216818642350557… = 2584/987
2.6180340557275541795665634… = 6765/2584
2.6180339985218033998521803… = 17711/6765
2.6180339901755970865563773… = 46368/17711
2.6180339889579020013802622… = 121393/46368
2.6180339887802426828565073… = 317811/121393
2.6180339887543225376088304… = 832040/317811
2.6180339887505408393827219… = 2178309/832040
2.6180339887499890970472967… = 5702887/2178309
2.6180339887499085989254214… = 14930352/5702887
2.6180339887498968544077192… = 39088169/14930352
2.6180339887498951409056791… = 102334155/39088169
2.6180339887498948909091006… = 267914296/102334155

Deep-Dive Dyadendricity

Monday, 19th May, 2025 by Krilling for Company in √2, Fractals, Mathematics and tagged animated gifs, H-tree fractal | Leave a comment

This simple equation helps you understand a fractal:

1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + … = 2 = Σ(1/2^k,k=0..∞)

Now here’s the construction of the H-tree fractal, in which the lines are divided in length by sqrt(2) = 1.41421356237… at each stage. Or multiplied by 0.70710678118… = &sqrt;0.5. This means that, after two divisions, the lines are 1/2 the size. So in the end they create a 1 x &sqrt;2 rectangle:

H-Tree fractal #1

H-Tree fractal #2

H-Tree fractal #3

H-Tree fractal #4

H-Tree fractal #5

H-Tree fractal #6

H-Tree fractal #7

H-Tree fractal #8

H-Tree fractal #9

H-Tree fractal #10

↓
[…]
↓

H-Tree fractal #13

H-Tree fractal #14

H-Tree fractal #15

Here’s an animation:

H-Tree fractal (animated at EZgif)

And here’s the H-tree in black-and-white:

H-Tree fractal #3

↓

H-Tree fractal #6

↓

H-Tree fractal #6

↓

H-Tree fractal #12

↓

H-Tree fractal #15

↓

H-Tree fractal (animated at EZgif)

Because the construction of the H-tree is governed by a string of directions — for example, left-right-right-left-left-left… or 211222… — you can perform tests on that string to create sub-fractals from the super-fractal. Like this:

count(1) = count(2) in string to step 12

count(1) = count(2) in string (omitting lines)

sum(string) = mul(string)

sum(string) > mul(string)

count(1) = 2 or count(2) = 2 after step 2

count(1) < count(2)

count(1) < 3 or count(2) < 3 after step 6

value of string after step 8 > value of string at step 1

value after step 8 > value at step 4

value after step 8 < value step 1

ispalindrome(string) to step 11

ispalindrome(string) to step 18

ispalindrome(string) to step 20

alternating 121… or 212… in string after step 9

ispolygonal(sum(string[i]-1),pol=10)

isprime(sum(string))

sum(string[i]-1) mod 13 = 0

sum(string[i]-1) mod 13 = 1

sum(string[i]-1) mod 16 = 0

sum(string[i]-1) mod 18 = 0

Jewel Tree

Saturday, 17th May, 2025 by Krilling for Company in Arithmetic, Mathematics and tagged Farey fractions, Farey tessellation, Farey tree, Johannes Kepler | Leave a comment

Dual tree for Farey tessellation and positive Farey tree (from ResearchGate)

• Die Geometrie besitzt zwei große Schätze: einer ist der Satz von Pythagoras, der andere die Teilung einer Strecke nach dem äußeren und mittleren Verhältnis. Den ersten dürfen wir mit einem Scheffel Gold vergleichen; den zweiten nennen wir ein kostbares Juwel. — Johannes Kepler

• “Geometry has two great treasures: one is the Theorem of Phythagoras, the other the division of a line in extreme and mean ratio. The first we can compare to a mass of gold; the other we may call a precious jewel.”

ResidUlam

Wednesday, 7th May, 2025 by Krilling for Company in Arithmetic, Mathematics and tagged OEIS, recreational math, recreational mathematics, recreational maths, ResidUlam spirals | Leave a comment

Seq’ and ye shall find. So what’s the next number in this sequence?

1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 1, 17, 9, 19, ?

It’s simple once you’ve spotted the rule. List the counting numbers. If a number is a multiple of 2, divide it by 2 until it’s no longer a multiple of 2 and it becomes what I call a 2-residue. Like this:

reduce(2,n) = 1, 2 → 1, 3, 4 → 2 → 1, 5, 6 → 3, 7, 8 → 4 → 2 → 1, 9, 10 → 5, 11, 12 → 6 → 3, 13, 14 → 7, 15, 16 → 8 → 4 → 2 → 1, 17, 18 → 9, 19, 20 → 10 → 5... — A000265 at the Online Encyclopedia of Integer Sequences (OEIS)

So the next number was 5. Now, what’s the next number in this sequence?

1, 2, 1, 4, 5, 2, 7, 8, 1, 10, 11, 4, 13, 14, 5, 16, 17, 2, 19, 20, ?

The rule now is: divide multiples of 3 by 3 until they’re no longer multiples of 3.

reduce(3,n) = 1, 2, 1, 4, 5, 2, 7, 8, 1, 10, 11, 4, 13, 14, 5, 16, 17, 2, 19, 20, 7, 22, 23, 8, 25, 26, 1, 28, 29, 10, 31, 32, 11, 34, 35, 4, 37, 38, 13, 40, 41, 14, 43, 44, 5, 46, 47, 16, 49, 50, 17, 52, 53, 2, 55, 56, 19, 58, 59, 20, 61, 62, 7, 64, 65, 22, 67, 68, 23, 70, 71, 8, 73, 74, 25, 76, ... — A038502 at OEIS

So the next number is 7, the 3-residue of 21. After looking at these sequences, I did what I usually did and tried them on an Ulam spiral. The sum of reduce(2,n) is this:

1, 2, 5, 6, 11, 14, 21, 22, 31, 36, 47, 50, 63, 70, 85, 86, 103, 112, 131, 136, 157, 168, 191, 194, 219, 232, 259, 266, 295, 310, 341, 342, 375, 392, 427, 436, 473, 492, 531, 536, 577, 598, 641, 652, 697, 720, 767, 770, 819, 844, 895, 908, 961, 988, 1043, 1050, 1107, 1136, 1195, 1210, 1271, 1302, 1365, 1366, 1431, 1464, 1531, 1548, 1617, 1652, 1723, 1732, 1805, 1842, 1917, 1936, 2013, 2052, 2131, 2136, 2217, 2258, 2341, 2362, 2447, 2490, 2577, 2588, 2677, 2722, 2813, 2836, 2929, 2976... — A135013 at OEIS

And on an Ulam spiral, the sequence looks like this:

Ulam-like spiral for sum(reduce(2,n)) = 1, 2, 5, 6, 11, 14, 21, 22, 31, 36, 47…

Here are more ResidUlam spirals (not all at the same resolution):

Spiral for sum(reduce(3,n))

Spiral for sum(reduce(4,n))

Spiral for sum(reduce(10,n))

Spiral for sum(reduce(11,n))

Spiral for sum(reduce(18,n))

Spiral for sum(reduce(28,n))

Spiral for sum(reduce(51,n))

N.B. THe 51-ResidUlam doesn’t look like that because the numbers are thinning, but because sum(reduce(51,n)) concentrates them in certain parts of the spiral. Compare sum(reduce(64,n)):

Spiral for sum(reduce(64,n))

Next, you can try reducing numbers with more than one multiple. For example, if you reduce the counting numbers by 2 and 3, you get this sequence:

reduce(2,3,n) = 1, 1, 1, 1, 5, 1, 7, 1, 1, 5, 11, 1, 13, 7, 5, 1, 17, 1, 19, 5, 7, 11, 23, 1, 25, 13, 1, 7, 29, 5, 31, 1, 11, 17, 35, 1, 37, 19, 13, 5, 41, 7, 43, 11, 5, 23, 47, 1, 49, 25, 17, 13, 53, 1, 55, 7, 19, 29, 59, 5, 61, 31, 7, 1, 65, 11, 67, 17, 23, 35, 71, 1, 73, 37, 25, 19, 77, 13, 79, 5, 1, ... — A065330 at OEIS sum(reduce(2,3,n)) = 1, 2, 3, 4, 9, 10, 17, 18, 19, 24, 35, 36, 49, 56, 61, 62, 79, 80, 99, 104, 111, 122, 145, 146, 171, 184, 185, 192, 221, 226, 257, ...

On an ResiduUlam spiral, sum(reduce(2,3,n)) looks like this at higher and higher resolution:

Spiral for sum(reduce(2,3,n)) #1

Spiral for sum(reduce(2,3,n)) #2

Spiral for sum(reduce(2,3,n)) #3

Spiral for sum(reduce(2,3,n)) #4

Now try another double-reducer:

reduce(6,3,n) = 1, 2, 1, 4, 5, 1, 7, 8, 1, 10, 11, 2, 13, 14, 5, 16, 17, 1, 19, 20, 7, 22, 23, 4, 25, 26, 1, 28, 29, 5, 31, 32, 11, 34, 35, 1, 37, 38, 13, 40, 41, 7, 43, 44, 5, 46, 47, 8, 49, 50, 17, 52, 53, 1, 55, 56, 19, 58, 59, 10, 61, 62, 7, 64, 65, 11, 67, 68, 23, ... sum(reduce(6,3,n)) = 1, 3, 4, 8, 13, 20, 28, 29, 39, 50, 52, 65, 79, ...

Note that it’s important to reduce by 6 before reducing by 3 (reducing by 3 first would mean no numbers to reduce by 6). Here’s the ResidUlam spiral:

Spiral for sum(reduce(6,3,n)) #1

Spiral for sum(reduce(6,3,n)) #2

Spiral for sum(reduce(6,3,n)) #3

Spiral for sum(reduce(6,3,n)) #4

And two more double-multiple ResidUlams:

Spiral for sum(reduce(7,3,n))

Spiral for sum(reduce(10,8,n))

Mathematicoynte

Tuesday, 29th Apr, 2025 by Krilling for Company in Arithmetic, Continued Fractions, Fractals, Mathematics and tagged denominators, fractions, numerators, rational fractions, recreational math, recreational mathematics, recreational maths, reduced fractions, sorting fractions | Leave a comment

Pre-previously, I looked at a fractal phallus. Now I want to look at a fractal fanny (in the older British sense). In fact, it’s a fractional fractal fanny. Take a look at these fractions:

1/10, 1/9, 1/8, 1/7, 1/6, 1/5, 2/10, 2/9, 1/4, 2/8, 2/7, 3/10, 1/3, 2/6, 3/9, 3/8, 2/5, 4/10, 3/7, 4/9, 1/2, 2/4, 3/6, 4/8, 5/10, 5/9, 4/7, 3/5, 6/10, 5/8, 2/3, 4/6, 6/9, 7/10, 5/7, 3/4, 6/8, 7/9, 4/5, 8/10, 5/6, 6/7, 7/8, 8/9, 9/10

They’re all the fractions for 1/2..(n-1)/n, n = 10, sorted by increasing size. But obviously some of them are the same: 1/2 = 2/4 = 3/6 = 5/10, 1/3 = 2/6 = 3/9, 1/4 = 2/8, and so on. If you remove the duplicates, you get this set of reduced fractions:

1/10, 1/9, 1/8, 1/7, 1/6, 1/5, 2/9, 1/4, 2/7, 3/10, 1/3, 3/8, 2/5, 3/7, 4/9, 1/2, 5/9, 4/7, 3/5, 5/8, 2/3, 7/10, 5/7, 3/4, 7/9, 4/5, 5/6, 6/7, 7/8, 8/9, 9/10

Now here are the reduced fractions for 1/2..(n-1)/n, n = 30:

1/30, 1/29, 1/28, 1/27, 1/26, 1/25, 1/24, 1/23, 1/22, 1/21, 1/20, 1/19, 1/18, 1/17, 1/16, 1/15, 2/29, 1/14, 2/27, 1/13, 2/25, 1/12, 2/23, 1/11, 2/21, 1/10, 3/29, 2/19, 3/28, 1/9, 3/26, 2/17, 3/25, 1/8, 3/23, 2/15, 3/22, 4/29, 1/7, 4/27, 3/20, 2/13, 3/19, 4/25, 1/6, 5/29, 4/23, 3/17, 5/28, 2/11, 5/27, 3/16, 4/21, 5/26, 1/5, 6/29, 5/24, 4/19, 3/14, 5/23, 2/9, 5/22, 3/13, 7/30, 4/17, 5/21, 6/25, 7/29, 1/4, 7/27, 6/23, 5/19, 4/15, 7/26, 3/11, 8/29, 5/18, 7/25, 2/7, 7/24, 5/17, 8/27, 3/10, 7/23, 4/13, 9/29, 5/16, 6/19, 7/22, 8/25, 9/28, 1/3, 10/29, 9/26, 8/23, 7/20, 6/17, 5/14, 9/25, 4/11, 11/30, 7/19, 10/27, 3/8, 11/29, 8/21, 5/13, 7/18, 9/23, 11/28, 2/5, 11/27, 9/22, 7/17, 12/29, 5/12, 8/19, 11/26, 3/7, 13/30, 10/23, 7/16, 11/25, 4/9, 13/29, 9/20, 5/11, 11/24, 6/13, 13/28, 7/15, 8/17, 9/19, 10/21, 11/23, 12/25, 13/27, 14/29, 1/2, 15/29, 14/27, 13/25, 12/23, 11/21, 10/19, 9/17, 8/15, 15/28, 7/13, 13/24, 6/11, 11/20, 16/29, 5/9, 14/25, 9/16, 13/23, 17/30, 4/7, 15/26, 11/19, 7/12, 17/29, 10/17, 13/22, 16/27, 3/5, 17/28, 14/23, 11/18, 8/13, 13/21, 18/29, 5/8, 17/27, 12/19, 19/30, 7/11, 16/25, 9/14, 11/17, 13/20, 15/23, 17/26, 19/29, 2/3, 19/28, 17/25, 15/22, 13/19, 11/16, 20/29, 9/13, 16/23, 7/10, 19/27, 12/17, 17/24, 5/7, 18/25, 13/18, 21/29, 8/11, 19/26, 11/15, 14/19, 17/23, 20/27, 3/4, 22/29, 19/25, 16/21, 13/17, 23/30, 10/13, 17/22, 7/9, 18/23, 11/14, 15/19, 19/24, 23/29, 4/5, 21/26, 17/21, 13/16, 22/27, 9/11, 23/28, 14/17, 19/23, 24/29, 5/6, 21/25, 16/19, 11/13, 17/20, 23/27, 6/7, 25/29, 19/22, 13/15, 20/23, 7/8, 22/25, 15/17, 23/26, 8/9, 25/28, 17/19, 26/29, 9/10, 19/21, 10/11, 21/23, 11/12, 23/25, 12/13, 25/27, 13/14, 27/29, 14/15, 15/16, 16/17, 17/18, 18/19, 19/20, 20/21, 21/22, 22/23, 23/24, 24/25, 25/26, 26/27, 27/28, 28/29, 29/30

Can you see the fractal fanny? Not unless you’re superhuman. But any normal human can see the fractal fanny when you turn those reduced and sorted fractions, a/b, into a graph, where y = b and x = n for a/b_n:

graph for b of reduced a/b = 1/2..29/30, sorted by size of a/b

(click for larger)

If you don’t reduce the fractions, you get this distorted coynte:

graph for b of all fractions 1/2..29/30, sorted by a/b

And you can use other variables for y, like the sum of the continued fraction of a/b:

graph for sum(contfrac(a/b)) of reduced fractions 1/2..29/30, sorted by a/b

graph for cfsum of all fractions 1/2..29/30, sorted by a/b

And the product of the continued fraction of a/b:

graph for prod(contfrac(a/b)) of reduced fractions 1/2..29/30, sorted by a/b

graph for cfmul of all fractions 1/2..29/30, sorted by a/b

And you can sort by the size of other variables, like the number of factors of b:

graph for a+b of all fractions 1/2..29/30, sorted by factornum(b)

And so on:

graph for a of reduced fractions 1/2..29/30, sorted by a/b

graph for a of all fractions 1/2..29/30, sorted by a/b

graph for a of all fractions 1/2..29/30, sorted by length(contfrac(a/b))

graph for a of all fractions 1/2..29/30, sorted by factornum(b)

graph for a of all fractions 1/2..29/30, sorted by gcd(a/b)

graph for a+b of all fractions 1/2..29/30, sorted by a/b

graph for a+b of reduced fractions 1/2..29/30, sorted by a/b

graph for a+b of all fractions 1/2..29/30, sorted by a+b

graph for a+b of all fractions 1/2..29/30, sorted by cflen(a/b)

graph for a+b of all fractions 1/2..29/30, sorted by gbd(a,b)

graph for b of all fractions 1/2..29/30, sorted by a+b

graph for b of all fractions 1/2..29/30, sorted by cflen(a/b)

graph for b of all fractions 1/2..29/30, sorted by factnum(b)

graph for b of all fractions 1/2..29/30, sorted by gcd(a,b)

graph for b-a of all fractions 1/2..29/30, sorted by a/b

graph for b-a of reduced fractions 1/2..29/30, sorted by a/b

graph for b-a of all fractions 1/2..29/30, sorted by a+b

graph for b-a of all fractions 1/2..29/30, sorted by factnum(b)

graph for cfmul of all fractions 1/2..29/30, sorted by a

graph for cfsum of all fractions 1/2..29/30, sorted by a

Previously Pre-Posted (Please Peruse)

• Phrallic Frolics — a look at fractal phalluses, a.k.a. phralluses

Points Pared

Wednesday, 23rd Apr, 2025 by Krilling for Company in Geometry, Hexagons, Mathematics, Squares, Triangles and tagged average distance, distance to center of polygon, distance to vertices, points in the plane, vertices of a hexagon, vertices of a square, vertices of a triangle | Leave a comment

There are an infinite number of points in the plane. And in part of the plane. So you have to pare points to create interesting shapes. And one way of paring them is by comparing them. The six red dots in the image below mark the three vertices of an equilateral triangle and the three mid-points of the sides. Now, test the other points in the surrounding plane and mark them in white if the average distance to (the centers of) any two of the red dots is equal to the average distance to (the centers of) the four other red dots:

Triangle + 1 side-point, sum(d1,d2)/2 = sum(d3,d4,d5,d6)/4

(click for larger)

Add a central red dot to the triangle and you get this pattern:

Triangle + 1 side-point + center, distfunc(2) = distfunc(5)

And so on:

v = 3 + sd = 2, distfunc(2) = distfunc(7)

v = 3 + sd = 2 + center, distfunc(2) = distfunc(7)

v = 3 + sd = 1, distfunc(3) = distfunc(3)

v = 3 + sd = 1 + center, distfunc(3) = distfunc(7)

v = 4 + center, distfunc(2) = distfunc(3)

v = 4 + sd = 1, distfunc(2) = distfunc(6)

v = 4 + sd = 1 + center, distfunc(2) = distfunc(7)

v = 4 + sd = 2, distfunc(2) = distfunc(10)

v = 4 + sd = 2, distfunc(2) = distfunc(10) (enlarged)

v = 4 + sd = 1, distfunc(3) = distfunc(5)

v = 4 + sd = 1, distfunc(4) = distfunc(4)

v = 5 + sd = 1, distfunc(2) = distfunc(8)

v = 5 + sd = 1, distfunc(2) = distfunc(8) (smaller scale)

v = 6 + sd = 1, distfunc(1) = distfunc(11)

v = 6 + sd = 1 + center, distfunc(1) = distfunc(12)

v = 6, distfunc(2) = distfunc(4)

v = 6 + center, distfunc(2) = distfunc(5)

v = 6, distfunc(3) = distfunc(3)

v = 6 + center, distfunc(3) = distfunc(4)

The Call of CFulhu

Thursday, 17th Apr, 2025 by Krilling for Company in Arithmetic, √2, Continued Fractions, Fibonacci Sequence, H.P. Lovecraft, Integer Sequences, φ, Mathematics, Phi, Square root of 2 and tagged "The Call of Cthulhu", fractions, HPL, recreational math, recreational mathematics, recreational maths | Leave a comment

“The most merciful thing in the world, I think, is the inability of the human mind to correlate all its contents.” So said HPL in “The Call of Cthulhu” (1926). But I’d still like to correlate the contents of mine a bit better. For example, I knew that φ, the golden ratio, is the most irrational of all numbers, in that it is the slowest to be approximated with rational fractions. And I also knew that continued fractions, or CFs, were a way of representing both rationals and irrationals as a string of numbers, like this:

contfrac(10/7) = [1; 2, 3]
10/7 = 1 + 1/(2 + 1/3)
10/7 = 1.428571428571…

contfrac(3/5) = [0; 1, 1, 2]
4/5 = 0 + 1/(1 + 1/(1 + 1/2))
4/5 = 0.8

contfrac(11/8) = [1; 2, 1, 2]
11/8 = 1 + 1/(2 + 1/(1 + 1/2))
11/8 = 1.375

contfrac(4/7) = [0; 1, 1, 3]
4/7 = 0 + 1/(1 + 1/(1 + 1/3))
4/7 = 0.57142857142…

contfrac(17/19) = [0; 1, 8, 2]
17/19 = 0 + 1/(1 + 1/(8 + 1/2))
17/19 = 0.8947368421052…

contfrac(8/25) = [0; 3, 8]
8/25 = 0 + 1/(3 + 1/8)
8/25 = 0.32

contfrac(√2) = [1; 2, 2, 2, 2, 2, 2, 2…] = [1; 2]

√2 = 1 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/2 + …))))))

√2 = 1.41421356237309504…

contfrac(φ) = [1; 1, 1, 1, 1, 1, 1, 1, 1…]

φ = 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/1 + …)))))))

φ = 1.6180339887498948…

But I didn’t correlate those two contents of my mind: the maximal irrationality of φ and the way continued fractions work.

That’s why I was surprised when I was looking at the continued fractions of 2..(n-1) / n for 3,4,5,6,7… That is, I was looking at the continued fractions of 2/3, 3/4, 2/5, 3/5, 4/5, 5/6, 2/7, 3/7… (skipping fractions like 2/4, 2/6, 3/6 etc, because they’re reducible: 2/4 = ½, 2/6 = 1/3, 3/6 = ½ etc). I wondered which fractions set successive records for the length of their continued fractions as one worked through ½, 2/3, 3/4, 2/5, 3/5, 4/5, 5/6, 2/7, 3/7… And because I hadn’t correlated the contents of my mind, I was surprised at the result. I shouldn’t have been, of course:

contfrac(1/2) = [0; 2] (cfl=1)
1/2 = 0 + 1/2
1/2 = 0.5

contfrac(2/3) = [0; 1, 2] (cfl=2)
2/3 = 0 + 1/(1 + 1/2)
2/3 = 0.666666666…

contfrac(3/5) = [0; 1, 1, 2] (cfl=3)
3/5 = 0 + 1/(1 + 1/(1 + 1/2))
3/5 = 0.6

contfrac(5/8) = [0; 1, 1, 1, 2] (cfl=4)
5/8 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/2)))
5/8 = 0.625

contfrac(8/13) = [0; 1, 1, 1, 1, 2] (cfl=5)
8/13 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2))))
8/13 = 0.615384615…

contfrac(13/21) = [0; 1, 1, 1, 1, 1, 2] (cfl=6)
13/21 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2)))))
13/21 = 0.619047619…

contfrac(21/34) = [0; 1, 1, 1, 1, 1, 1, 2] (cfl=7)
21/34 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2))))))
21/34 = 0.617647059…

contfrac(34/55) = [0; 1, 1, 1, 1, 1, 1, 1, 2] (cfl=8)
contfrac(55/89) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=9)
contfrac(89/144) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=10)
contfrac(144/233) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=11)
contfrac(233/377) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=12)
contfrac(377/610) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=13)
contfrac(610/987) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=14)
contfrac(987/1597) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=15)
contfrac(1597/2584) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=16)
contfrac(2584/4181) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=17)
contfrac(4181/6765) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=18)
[…]

Which n₁/n₂ set records for the length of their continued fractions (with n₂ > n₁)? It’s the successive Fibonacci fractions, fib(i)/fib(i+1), of course. I didn’t anticipate that answer because I didn’t understand φ and continued fractions properly. And I still don’t, because I’ve been surprised again today looking at palindromic CFs like these:

contfrac(2/5) = [0; 2, 2] (cfl=2)
2/5 = 0 + 1/(2 + 1/2)
2/5 = 0.4

contfrac(3/8) = [0; 2, 1, 2] (cfl=3)
3/8 = 0 + 1/(2 + 1/(1 + 1/2))
3/8 = 0.375

contfrac(3/10) = [0; 3, 3] (cfl=2)
3/10 = 0 + 1/(3 + 1/3)
3/10 = 0.3

contfrac(5/12) = [0; 2, 2, 2] (cfl=3)
5/12 = 0 + 1/(2 + 1/(2 + 1/2))
5/12 = 0.416666666…

contfrac(5/13) = [0; 2, 1, 1, 2] (cfl=4)
5/13 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/2)))
5/13 = 0.384615384…

contfrac(4/15) = [0; 3, 1, 3] (cfl=3)
4/15 = 0 + 1/(3 + 1/(1 + 1/3))
4/15 = 0.266666666…

contfrac(7/16) = [0; 2, 3, 2] (cfl=3)
7/16 = 0 + 1/(2 + 1/(3 + 1/2))
7/16 = 0.4375

contfrac(4/17) = [0; 4, 4] (cfl=2)
4/17 = 0 + 1/(4 + 1/4)
4/17 = 0.235294117…

Again, I wondered which of these fractions set successive records for the length of their palindromic continued fractions. Here’s the answer:

contfrac(1/2) = [0; 2] (cfl=1)
1/2 = 0 + 1/2
1/2 = 0.5

contfrac(2/5) = [0; 2, 2] (cfl=2)
2/5 = 0 + 1/(2 + 1/2)
2/5 = 0.4

contfrac(3/8) = [0; 2, 1, 2] (cfl=3)
3/8 = 0 + 1/(2 + 1/(1 + 1/2))
3/8 = 0.375

contfrac(5/13) = [0; 2, 1, 1, 2] (cfl=4)
5/13 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/2)))
5/13 = 0.384615384…

contfrac(8/21) = [0; 2, 1, 1, 1, 2] (cfl=5)
8/21 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/(1 + 1/2))))
8/21 = 0.380952380…

contfrac(13/34) = [0; 2, 1, 1, 1, 1, 2] (cfl=6)
13/34 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/(
1 + 1/(1 + 1/2)))))
13/34 = 0.382352941..

contfrac(21/55) = [0; 2, 1, 1, 1, 1, 1, 2] (cfl=7)
21/55 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2))))))
21/55 = 0.381818181…

contfrac(34/89) = [0; 2, 1, 1, 1, 1, 1, 1, 2] (cfl=8)
contfrac(55/144) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=9)
contfrac(89/233) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=10)
contfrac(144/377) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=11)
contfrac(233/610) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=12)
contfrac(377/987) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=13)
contfrac(610/1597) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=14)
contfrac(987/2584) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=15)
contfrac(1597/4181) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=16)
contfrac(2584/6765) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=17)
[…]

Now it’s the successive Fibonacci skip-one fractions, fib(i)/fib(i+2), that set records for the length of their palindromic continued fractions. But I think you’d have to be very good at maths not to be surprised by that result.

After that, I continued to be compelled by the Call of CFulhu and started to look at the CFs of Fibonacci skip-n fractions in general. That’s contfrac(fib(i)/fib(i+n)) for n = 1,2,3,… And I’ve found more interesting patterns, as I’ll describe in a follow-up post.

Message from Mater

Monday, 31st Mar, 2025 by Krilling for Company in Arithmetic, √2, Continued Fractions, Irrational numbers, π, φ, Mathematics, Square root of 2, Ulam Spirals and tagged animated gifs, continued fraction, Continued Fractions, irrational numbers, Mater Mathematica, rational numbers, Ulam Spirals, Ulamization, Ulamize | Leave a comment

As any recreational mathematician kno, the Ulam spiral shows the prime numbers on a spiral grid of integers. Here’s a Ulam spiral with 1 represented in blue and 2, 3, 5, 7… as white blocks spiralling anti-clockwise from the right of 1:

The Ulam spiral of prime numbers

Ulam spiral at higher resolution

I like the Ulam spiral and whenever I’m looking at new number sequences I like to Ulamize it, that is, display it on a spiral grid of integers. Sometimes the result looks good, sometimes it doesn’t. But I’ve always wondered something beforehand: will this be the spiral where I see a message appear? That is, will I see a message from Mater Mathematica, Mother Maths, the omniregnant goddess of mathematics? Is there an image or text embedded in some obscure number sequence, revealed when the sequence is Ulamized and proving that there’s divine intelligence and design behind the universe? Maybe the image of a pantocratic cat will appear. Or a text in Latin or Sanskrit or some other suitably century-sanctified language.

That’s what I wonder. I don’t wonder it seriously, of course, but I do wonder it. But until 22nd March 2025 I’d never seen any Ulam-ish spiral that looked remotely like a message. But 22nd May is the day I Ulamed some continued fractions. And I saw something that did look a little like a message. Like text, that is. But I might need to explain continued fractions first. What are they? They’re a fascinating and beautiful way of representing both rational and irrational numbers. The continued fractions for rational numbers look like this in expanded and compact format:

5/3 = 1 + 1/(1 + ½) = 1 + ⅔
5/3 = [1; 1, 2]

19/7 = 2 + 1/(1 + 1/(2 + ½)) = 2 + 4/7
19/7 = [2; 1, 2, 2]

2/3 = 0 + 1/(1 + 1/2)
2/3 = [0; 1, 2] (compare 5/3 above)

3/5 = 0 + 1/(1 + 1/(1 + 1/2))
3/5 = [0; 1, 1, 2]

5/7 = 0 + 1/(1 + 1/(2 + 1/2))
5/7 = [0; 1, 2, 2] (compare 19/7 above)

13/17 = 0 + 1/(1 + 1/(3 + 1/4))
13/17 = [0; 1, 3, 4]

30/67 = 0 + 1/(2 + 1/(4 + 1/(3 + ½)))
30/67 = [0; 2, 4, 3, 2]

The continued fractions of irrational numbers are different. Most importantly, they never end. For example, here are the infinite continued fractions for φ, √2 and π in expanded and compact format:

φ = 1 + (1/(1 + 1/(1 + 1/(1 + …)))φ = [1; 1]

√2 = 1 + (1/(2 + 1/(2 + 1/(2 + …)))
√2 = [1; 2]

π = 3 + 1/(7 + 1/(15 + 1/(1 + 1/(292 + 1/(1 + 1/(1 + 1/(1 + 1/(2 + 1/(1 + 1/(3 +…))))))))))
π = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3…]

As you can see, the continued fraction of π doesn’t fall into a predictable pattern like those for φ and √2. But I’ve already gone into continued fractions further than I need for this post, so let’s return to the continued fractions of rationals. I set up an Ulam spiral to show patterns based on the continued fractions for 1/1, ½, ⅓, ⅔, 1/4, 2/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 2/6, 3/6… (where the fractions are assigned to 1,2,3… and 2/4 = ½, 2/6 = ⅓ etc). For example, if the continued fraction contains a number higher than 5, you get this spiral:

Spiral for continued fractions containing at least number > 5

With tests for higher and higher numbers in the continued fractions, the spirals start to thin and apparent symbols start to appear in the arms of the spirals:

Spiral for contfrac > 10

Spiral for contfrac > 15

Spiral for contfrac > 20

Spiral for contfrac > 25

Spiral for contfrac > 30

Spiral for contfrac > 35

Spiral for contfrac > 40

Spirals for contfrac > 5..40 (animated at EZgif)

Here are some more of these spirals at increasing magnification:

Spiral for contfrac > 23 (#1)

Spiral for contfrac > 23 (#2)

Spiral for contfrac > 23 (#3)

Spiral for contfrac > 13

Spiral for contfrac > 15 (off-center)

Spiral for contfrac > 23 (off-center)

And here are some of the symbols picked out in blue:

Spiral for contfrac > 15 (blue symbols)

Spiral for contfrac > 23 (blue symbols)

But they’re not really symbols, of course. They’re quasi-symbols, artefacts of the Ulamization of a simple test on continued fractions. Still, they’re the closest I’ve got so far to a message from Mater Mathematica.

In 10 Words: Im-Precise

Tuesday, 25th Mar, 2025 by Krilling for Company in Base Behavior, Binary numbers, π, Mathematics, Pi and tagged digital precision, digits of pi in base 10, digits of pi in base 2, rational approximations of pi | Leave a comment

I was looking at the best rational approximations for π when I was puzzled for a moment or two by the way the precision of digits didn’t always improve:
22/7 → 3.1428571... = 22/7 (precision = 3 digits) 3.1415926... = π 333/106 → 3.141509433... = 333/106 (pr=5) 3.141592653... = π 355/113 → 3.14159292035... = 355/113 (pr=7) 3.14159265358... = π 103993/33102 → 3.14159265301190... (pr=10) 3.14159265358979... = π 104348/33215 → 3.14159265392142... (pr=10) 3.14159265358979... = π 208341/66317 → 3.14159265346743... (pr=10) 3.14159265358979... = π 312689/99532 → 3.14159265361893... (pr=10) 3.14159265358979... = π 833719/265381 → 3.1415926535810777... (pr=12) 3.1415926535897932... = π 1146408/364913 → 3.141592653591403... (pr=11) 3.141592653589793... = π 4272943/1360120 → 3.14159265358938917... (pr=13) 3.14159265358979323... = π 5419351/1725033 → 3.14159265358981538... (pr=13) 3.14159265358979323... = π 80143857/25510582 → 3.1415926535897926593... (pr=15) 3.1415926535897932384... = π 165707065/52746197 → 3.14159265358979340254... (pr=16) 3.14159265358979323846... = π 245850922/78256779 → 3.14159265358979316028... (pr=16) 3.14159265358979323846... = π 411557987/131002976 → 3.141592653589793257826... (pr=17) 3.141592653589793238462... = π 1068966896/340262731 → 3.1415926535897932353925... (pr=18) 3.1415926535897932384626... = π 2549491779/811528438 → 3.1415926535897932390140... (pr=18) 3.1415926535897932384626... = π 6167950454/1963319607 → 3.14159265358979323838637... (pr=19) 3.14159265358979323846264... = π 14885392687/4738167652 → 3.141592653589793238493875... (pr=20) 3.141592653589793238462643... = π

But it was my precision that was wrong, of course. I wasn’t thinking about digits precisely enough. One approximation can be closer to π with fewer precise digits than another (e.g. 3.14201… is closer to π than 3.14101…). The same applies in binary, but there the precision tends to increase much more obviously:
22/7 → 3.1428571... = 22/7 in base 10 (pr=3) 3.1415926... = π in base 10 11.0010010010010... = 22/7 in base 2 (pr=9) 11.0010010000111... = π in base 2 333/106 → 3.141509433... = 333/106 in b10 (pr=5) 3.141592653... = π in b10 11.001001000011100111... = 333/106 in b2 (pr=14) 11.001001000011111101... = π in b2 355/113 → 3.14159292035... (pr=7) 3.14159265358... = π 11.00100100001111110110111100... = 355/113 in b2 (pr=22) 11.00100100001111110110101010... = π in b2 103993/33102 → 3.14159265301190... (pr=10) 3.14159265358979... = π 11.001001000011111101101010100001100... (pr=29) 11.001001000011111101101010100010001... = π 104348/33215 → 3.14159265392142... (pr=10) 3.14159265358979... = π 11.001001000011111101101010100010011111... (pr=32) 11.001001000011111101101010100010001000... = π 208341/66317 → 3.14159265346743... (pr=10) 3.14159265358979... = π 11.001001000011111101101010100001111... (pr=29) 11.001001000011111101101010100010001... = π 312689/99532 → 3.14159265361893... (pr=10) 3.14159265358979... = π 11.001001000011111101101010100010001010010... (pr=35) 11.001001000011111101101010100010001000010... = π 833719/265381 → 3.1415926535810777... (pr=12) 3.1415926535897932... = π 11.0010010000111111011010101000100001111... (pr=33) 11.0010010000111111011010101000100010000... = π 1146408/364913 → 3.141592653591403... (pr=11) 3.141592653589793... = π 11.0010010000111111011010101000100010000111011... (pr=39) 11.0010010000111111011010101000100010000101101... = π 4272943/1360120 → 3.14159265358938917... (pr=13) 3.14159265358979323... = π 11.001001000011111101101010100010001000010100110... (pr=41) 11.001001000011111101101010100010001000010110100... = π 5419351/1725033 → 3.14159265358981538... (pr=13) 3.14159265358979323... = π 11.0010010000111111011010101000100010000101101010010... (pr=45) 11.0010010000111111011010101000100010000101101000110... = π 80143857/25510582 → 3.1415926535897926593... (pr=15) 3.1415926535897932384... = π 11.0010010000111111011010101000100010000101101000101101... (pr=48) 11.0010010000111111011010101000100010000101101000110000... = π 165707065/52746197 → 3.14159265358979340254... (pr=16) 3.14159265358979323846... = π 11.00100100001111110110101010001000100001011010001100010100... (pr=52) 11.00100100001111110110101010001000100001011010001100001000... = π 245850922/78256779 → 3.14159265358979316028... (pr=16) 3.14159265358979323846... = π 11.001001000011111101101010100010001000010110100011000000110... (pr=53) 11.001001000011111101101010100010001000010110100011000010001... = π 411557987/131002976 → 3.141592653589793257826... (pr=17) 3.141592653589793238462... = π 11.00100100001111110110101010001000100001011010001100001010001... (pr=55) 11.00100100001111110110101010001000100001011010001100001000110... = π 1068966896/340262731 → 3.1415926535897932353925... (pr=18) 3.1415926535897932384626... = π 11.00100100001111110110101010001000100001011010001100001000100110... (pr=58) 11.00100100001111110110101010001000100001011010001100001000110100... = π 2549491779/811528438 → 3.1415926535897932390140... (pr=18) 3.1415926535897932384626... = π 11.00100100001111110110101010001000100001011010001100001000110111010... (pr=61) 11.00100100001111110110101010001000100001011010001100001000110100110... = π 6167950454/1963319607 → 3.14159265358979323838637... (pr=19) 3.14159265358979323846264... = π 11.0010010000111111011010101000100010000101101000110000100011010001101... (pr=63) 11.0010010000111111011010101000100010000101101000110000100011010011000... = π 14885392687/4738167652 → 3.141592653589793238493875... (pr=20) 3.141592653589793238462643... = π 11.001001000011111101101010100010001000010110100011000010001101001110100... (pr=65) 11.001001000011111101101010100010001000010110100011000010001101001100010... = π

Post-Performative Post-Scriptum…

The title of this terato-toxic post is a maximal mash-up (wow) of two well-known toxico-teratic tropes:

• “There are 10 kinds of people in the world. Those who understand binary and those who don’t.”
• Sam Goldwyn’s malapropism: “In two words: im-possible!”

Overlord In Terms of Core Issues Around Maximal Engagement with Key Notions of the Über-Feral

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Category Archives: Mathematics

The Bellissima Curve

Friday is Φiday

Deep-Dive Dyadendricity

Jewel Tree

ResidUlam

Mathematicoynte

Points Pared

The Call of CFulhu

Message from Mater

In 10 Words: Im-Precise