Twi-Phi

Here’s a pentagon:

Stage #1

And here’s the pentagon with smaller pentagons on its vertices:

Stage #2

And here’s more of the same:

Stage #3

Stage #4

Stage #5

Stage #6

Stage #7

Stage #8

Animated fractal

At infinity, the smaller pentagons have reached out like arms to exactly fill the gaps between themselves without overlapping. But how much smaller is each set of smaller pentagons than its mother-pentagon when the gaps are exactly filled? Well, if the radius of the mother-pentagon is r, then the radius of each daughter-pentagon is r * 1/(φ^2) = r * 0·38196601125…

But what happens if the radius relationship of mother to daughter is r * 1/φ = r * 0·61803398874 = r * (φ-1)? Then you get this fractal:

Stage #1

Stage #2

Stage #3

Stage #4

Stage #5

Stage #6

Stage #7

Stage #8

Stage #9

Animated fractal

Mötley Vüe

Here’s the Fibonacci sequence, where each term (after the first two) is created by adding the two previous numbers:

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

In “Fib and Let Tri”, I described how my eye was caught by 55, which is a palindrome, reading the same backwards and forwards. “Were there any other Fibonacci palindromes?” I wondered. So I looked to see. Now my eye has been caught by 55 again, but for another reason. It should be easy to spot another interesting aspect to 55 when the Fibonacci numbers are set out like this:

fib(1) = 1
fib(2) = 1
fib(3) = 2
fib(4) = 3
fib(5) = 5
fib(6) = 8
fib(7) = 13
fib(8) = 21
fib(9) = 34
fib(10) = 55
fib(11) = 89
fib(12) = 144
fib(13) = 233
fib(14) = 377
fib(15) = 610
fib(16) = 987
fib(17) = 1597
fib(18) = 2584
fib(19) = 4181
fib(20) = 6765
[...]

55 is fib(10), the 10th Fibonacci number, and 5+5 = 10. That is, digsum(fib(10)) = 10. What other Fibonacci numbers work like that? I soon found some and confirmed my answer at the Online Encyclopedia of Integer Sequences:

1, 5, 10, 31, 35, 62, 72, 175, 180, 216, 251, 252, 360, 494, 504, 540, 946, 1188, 2222 — A020995 at OEIS

And that seems to be the lot, according to the OEIS. In base 10, at least, but why stop at base 10? When I looked at base 11, the numbers of digsum(fib(k)) = k didn’t stop coming, because I couldn’t take the Fibonacci numbers very high on my computer. But the OEIS gives a much longer list, starting like this:

1, 5, 13, 41, 53, 55, 60, 61, 90, 97, 169, 185, 193, 215, 265, 269, 353, 355, 385, 397, 437, 481, 493, 617, 629, 630, 653, 713, 750, 769, 780, 889, 905, 960, 1013, 1025, 1045, 1205, 1320, 1405, 1435, 1501, 1620, 1650, 1657, 1705, 1735, 1769, 1793, 1913, 1981, 2125, 2153, 2280, 2297, 2389, 2413, 2460, 2465, 2509, 2533, 2549, 2609, 2610, 2633, 2730, 2749, 2845, 2893, 2915, 3041, 3055, 3155, 3209, 3360, 3475, 3485, 3521, 3641, 3721, 3749, 3757, 3761, 3840, 3865, 3929, 3941, 4075, 4273, 4301, 4650, 4937, 5195, 5209, 5435, 5489, 5490, 5700, 5917, 6169, 6253, 6335, 6361, 6373, 6401, 6581, 6593, 6701, 6750, 6941, 7021, 7349, 7577, 7595, 7693, 7740, 7805, 7873, 8009, 8017, 8215, 8341, 8495, 8737, 8861, 8970, 8995, 9120, 9133, 9181, 9269, 9277, 9535, 9541, 9737, 9935, 9953, 10297, 10609, 10789, 10855, 11317, 11809, 12029, 12175... — A020995 at OEIS

The list ends with 1636597 = A18666[b11] and the OEIS says that 1636597 almost certainly completes the list. According to David C. Terr’s paper “On the Sums of Fibonacci Numbers” (pdf), published in the Fibonacci Quarterly in 1996, the estimated digit-sum for the k-th Fibonacci number in base b is given by the formula (b-1)/2 * k * log(b,φ), where log(b,φ) is the logarithm in base b of the golden ratio, 1·61803398874… Terr then notes that the simplified formula (b-1)/2 * log(b,φ) gives the estimated average ratio digsum(fib(k)) / k in base b. Here are the estimates for bases 2 to 20:

b02 = 0.3471209568153086...
b03 = 0.4380178794859424...
b04 = 0.5206814352229629...
b05 = 0.5979874356654401...
b06 = 0.6714235829697111...
b07 = 0.7418818776805580...
b08 = 0.8099488992357201...
b09 = 0.8760357589718848...
b10 = 0.9404443811249043...
b11 = 1.0034045909311624...
b12 = 1.0650963641043091...
b13 = 1.1256639207937723...
b14 = 1.1852250528196852...
b15 = 1.2438775226715552...
b16 = 1.3017035880574074...
b17 = 1.3587732842474014...
b18 = 1.4151468584732730...
b19 = 1.4708766105122322...
b20 = 1.5260083080264088...

In base 2, you can expect digsum(fib(k)) to be much smaller than k; in base 20, you can expect digsum(fib(k)) to be much larger. But as you can see, the estimate for base 11, 1.0034045909311624…, is very nearly 1. That’s why base 11 produces so many results for digsum(fib(k)) = k, because only a slight deviation from the estimate might create a perfect ratio of 1 for digsum(fib(k)) / k, i.e. digsum(fib(k)) = k. But in the end the results run out in base 11 too, because as k gets higher and fib(k) gets bigger, the estimate becomes more and more accurate and digsum(fib(k)) > k. With lower k, digsum(fib(k)) can easily fall below k or match k. That happens in other bases, but because their estimates are further from 1, results for digsum(fib(k)) = k run out much more quickly.

To see this base behavior represented visually, I’ve created Ulam-like spirals for k using three colors: blue for digsum(fib(k)) < k, yellow for digsum(fib(k)) > k, and red for digsum(fib(k)) = k (with the green square at the center representing fib(1) = 1). As you can see below, the spiral for base 11 immediately stands out. It’s motley, not dominated by blue or yellow like the other spirals:

Spiral for digsum(fib(k)) in base 9
(blue for digsum(fib(k)) < k, yellow for digsum(fib(k)) > k, red for digsum(fib(k)) = k, green for fib(1))

Spiral for digsum(fib(k)) in base 10

Spiral for digsum(fib(k)) in base 11 — a motley view of blue, yellow and red

Spiral for digsum(fib(k)) in base 12

Spiral for digsum(fib(k)) in base 13

Finally, here are spirals at higher and higher resolution for digsum(fib(k)) = k in base 11:

digsum(fib(k)) = k in base 11 (low resolution)
(green square is fib(1))

digsum(fib(k)) = k in base 11 (x2 resolution)

digsum(fib(k)) = k in base 11 (x4)

digsum(fib(k)) = k in base 11 (x8)

digsum(fib(k)) = k in base 11 (x16)

digsum(fib(k)) = k in base 11 (x32)

digsum(fib(k)) = k in base 11 (x64)

digsum(fib(k)) = k in base 11 (x128)

digsum(fib(k)) = k in base 11 (animated)

Fib and Let Tri

It’s a simple sequence with hidden depths:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155... — A000045 at OEIS

That’s the Fibonacci sequence, probably the most famous of all integer sequences after the integers themselves (1, 2, 3, 4, 5…) and the primes (2, 3, 5, 7, 11…). It has a very simple definition: if fib(fi) is the fi-th number in the Fibonacci sequence, then fib(fi) = fib(fi-1) + fib(fi-2). By definition, fib(1) = fib(2) = 1. After that, it’s easy to generate new numbers:

2 = fib(3) = fib(1) + fib(2) = 1 + 1
3 = fib(4) = fib(2) + fib(3) = 1 + 2
5 = fib(5) = fib(3) + fib(4) = 2 + 3
8 = fib(6) = fib(4) + fib(5) = 3 + 5
13 = fib(7) = fib(5) + fib(6) = 5 + 8
21 = fib(8) = fib(6) + fib(7) = 8 + 13
34 = fib(9) = fib(7) + fib(8) = 13 + 21
55 = fib(10) = fib(8) + fib(9) = 21 + 34
89 = fib(11) = fib(9) + fib(10) = 34 + 55
144 = fib(12) = fib(10) + fib(11) = 55 + 89
233 = fib(13) = fib(11) + fib(12) = 89 + 144
377 = fib(14) = fib(12) + fib(13) = 144 + 233
610 = fib(15) = fib(13) + fib(14) = 233 + 377
987 = fib(16) = fib(14) + fib(15) = 377 + 610
[...]

How to create the Fibonacci sequence is obvious. But it’s not obvious that fib(fi) / fib(fi-1) gives you ever-better approximations to a fascinating constant called φ, the golden ratio, which is 1.618033988749894…:

1/1 = 1
2/1 = 2
3/2 = 1.5
5/3 = 1.66666...
8/5 = 1.6
13/8 = 1.625
21/13 = 1.615384...
34/21 = 1.619047...
55/34 = 1.6176470588235294117647058823...
89/55 = 1.618181818...
144/89 = 1.617977528089887640...
233/144 = 1.6180555555...
377/233 = 1.618025751072961...
610/377 = 1.618037135278514...
987/610 = 1.618032786885245...
[...]

And that’s just the start of the hidden depths in the Fibonacci sequence. I stumbled across another interesting pattern for myself a few days ago. I was looking at the sequence and one of the numbers caught my eye:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597...

55 is a palindrome, reading the same forward and backwards. I wondered whether there were any other palindromes in the sequence (apart from the trivial single-digit palindromes 1, 1, 2, 3…). I couldn’t find any more. Nor can anyone else, apparently. But that’s in base 10. Other bases are more productive. For example, in bases 2, 3 and 4, you get this:

11 in b2 = 3
101 in b2 = 5
10101 in b2 = 21

22 in b3 = 8
111 in b3 = 13
22122 in b3 = 233

11 in b4 = 5
111 in b4 = 21
202 in b4 = 34
313 in b4 = 55

I decided to concentrate on tripals, or palindromes with three digits. I started looking at bases that set records for the greatest number of tripals. And there are some interesting patterns in the digits of the tripals in these bases (when a digit > 9, the digit is represented inside square brackets — see base-29 and higher). See how quickly you can spot the patterns:

Palindromic Fibonacci numbers in base-4

111 in b4 (fib=21, fi=8)
202 in b4 (fib=34, fi=9)
313 in b4 (fib=55, fi=10)

4 = 2^2 (pal=3)

Palindromic Fibonacci numbers in base-11

121 in b11 (fib=144, fi=12)
313 in b11 (fib=377, fi=14)
505 in b11 (fib=610, fi=15)
818 in b11 (fib=987, fi=16)

11 is prime (pal=4)

Palindromic Fibonacci numbers in base-29

151 in b29 (fib=987, fi=16)
323 in b29 (fib=2584, fi=18)
818 in b29 (fib=6765, fi=20)
[13]0[13] in b29 (fib=10946, fi=21)
[21]1[21] in b29 (fib=17711, fi=22)

29 is prime (pal=5)

Palindromic Fibonacci numbers in base-76

1[13]1 in b76 (fib=6765, fi=20)
353 in b76 (fib=17711, fi=22)
828 in b76 (fib=46368, fi=24)
[21]1[21] in b76 (fib=121393, fi=26)
[34]0[34] in b76 (fib=196418, fi=27)
[55]1[55] in b76 (fib=317811, fi=28)

76 = 2^2 * 19 (pal=6)

Palindromic Fibonacci numbers in base-199

1[34]1 in b199 (fib=46368, fi=24)
3[13]3 in b199 (fib=121393, fi=26)
858 in b199 (fib=317811, fi=28)
[21]2[21] in b199 (fib=832040, fi=30)
[55]1[55] in b199 (fib=2178309, fi=32)
[89]0[89] in b199 (fib=3524578, fi=33)
[144]1[144] in b199 (fib=5702887, fi=34)

199 is prime (pal=7)

Palindromic Fibonacci numbers in base-521

1[89]1 in b521 (fib=317811, fi=28)
3[34]3 in b521 (fib=832040, fi=30)
8[13]8 in b521 (fib=2178309, fi=32)
[21]5[21] in b521 (fib=5702887, fi=34)
[55]2[55] in b521 (fib=14930352, fi=36)
[144]1[144] in b521 (fib=39088169, fi=38)
[233]0[233] in b521 (fib=63245986, fi=39)
[377]1[377] in b521 (fib=102334155, fi=40)

521 is prime (pal=8)

Palindromic Fibonacci numbers in base-1364

1[233]1 in b1364 (fib=2178309, fi=32)
3[89]3 in b1364 (fib=5702887, fi=34)
8[34]8 in b1364 (fib=14930352, fi=36)
[21][13][21] in b1364 (fib=39088169, fi=38)
[55]5[55] in b1364 (fib=102334155, fi=40)
[144]2[144] in b1364 (fib=267914296, fi=42)
[377]1[377] in b1364 (fib=701408733, fi=44)
[610]0[610] in b1364 (fib=1134903170, fi=45)
[987]1[987] in b1364 (fib=1836311903, fi=46)

1364 = 2^2 * 11 * 31 (pal=9)

Two patterns are quickly obvious. Every digit in the tripals is a Fibonacci number. And the middle digit of one Fibonacci tripal, fib(fi), becomes fib(fi-2) in the next tripal, while fib(fi), the first and last digits (which are identical), becomes fib(fi+2) in the next tripal.

But what about the bases? If you’re an expert in the Fibonacci sequence, you’ll spot the pattern at work straight away. I’m not an expert, but I spotted it in the end. Here are the first few bases setting records for the numbers of Fibonacci tripals:

4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196...

These numbers come from the Lucas sequence, which is closely related to the Fibonacci sequence. But where fib(1) = fib(2) = 1, luc(1) = 1 and luc(2) = 3. After that, luc(li) = luc(li-2) + luc(li-1):

1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196... — A000204 at OEIS

It seems that every second number from 4 in the Lucas sequence supplies a base in which 1) the number of Fibonacci tripals sets a new record; 2) every digit of the Fibonacci tripals is itself a Fibonacci number.

But can I prove that this is always true? No. And do I understand why these patterns exist? No. My simple search for palindromes in the Fibonacci sequence soon took me far out of my mathematical depth. But it’s been fun to find huge bases like this in which every digit of every Fibonacci tripal is itself a Fibonacci number:

Palindromic Fibonacci numbers in base-817138163596

1[139583862445]1 in b817138163596 (fib=781774079430987230203437, fi=116)
3[53316291173]3 in b817138163596 (fib=2046711111473984623691759, fi=118)
8[20365011074]8 in b817138163596 (fib=5358359254990966640871840, fi=120)
[21][7778742049][21] in b817138163596 (fib=14028366653498915298923761, fi=122)
[55][2971215073][55] in b817138163596 (fib=36726740705505779255899443, fi=124)
[144][1134903170][144] in b817138163596 (fib=96151855463018422468774568, fi=126)
[377][433494437][377] in b817138163596 (fib=251728825683549488150424261, fi=128)
[987][165580141][987] in b817138163596 (fib=659034621587630041982498215, fi=130)
[2584][63245986][2584] in b817138163596 (fib=1725375039079340637797070384, fi=132)
[6765][24157817][6765] in b817138163596 (fib=4517090495650391871408712937, fi=134)
[17711][9227465][17711] in b817138163596 (fib=11825896447871834976429068427, fi=136)
[46368][3524578][46368] in b817138163596 (fib=30960598847965113057878492344, fi=138)
[121393][1346269][121393] in b817138163596 (fib=81055900096023504197206408605, fi=140)
[317811][514229][317811] in b817138163596 (fib=212207101440105399533740733471, fi=142)
[832040][196418][832040] in b817138163596 (fib=555565404224292694404015791808, fi=144)
[2178309][75025][2178309] in b817138163596 (fib=1454489111232772683678306641953, fi=146)
[5702887][28657][5702887] in b817138163596 (fib=3807901929474025356630904134051, fi=148)
[14930352][10946][14930352] in b817138163596 (fib=9969216677189303386214405760200, fi=150)
[39088169][4181][39088169] in b817138163596 (fib=26099748102093884802012313146549, fi=152)
[102334155][1597][102334155] in b817138163596 (fib=68330027629092351019822533679447, fi=154)
[267914296][610][267914296] in b817138163596 (fib=178890334785183168257455287891792, fi=156)
[701408733][233][701408733] in b817138163596 (fib=468340976726457153752543329995929, fi=158)
[1836311903][89][1836311903] in b817138163596 (fib=1226132595394188293000174702095995, fi=160)
[4807526976][34][4807526976] in b817138163596 (fib=3210056809456107725247980776292056, fi=162)
[12586269025][13][12586269025] in b817138163596 (fib=8404037832974134882743767626780173, fi=164)
[32951280099]5[32951280099] in b817138163596 (fib=22002056689466296922983322104048463, fi=166)
[86267571272]2[86267571272] in b817138163596 (fib=57602132235424755886206198685365216, fi=168)
[225851433717]1[225851433717] in b817138163596 (fib=150804340016807970735635273952047185, fi=170)
[365435296162]0[365435296162] in b817138163596 (fib=244006547798191185585064349218729154, fi=171)
[591286729879]1[591286729879] in b817138163596 (fib=394810887814999156320699623170776339, fi=172)

817138163596 = 2^2 * 229 * 9349 * 95419 (pal=30)

Performativizing the Polygonic #3

Pre-previously in my passionate portrayal of polygonic performativity, I showed how a single point jumping randomly (or quasi-randomly) towards the vertices of a polygon can create elaborate fractals. For example, if the point jumps 1/φth (= 0.6180339887…) of the way towards the vertices of a pentagon, it creates this fractal:

Point jumping 1/φth of the way to a randomly (or quasi-randomly) chosen vertex of a pentagon

But as you might expect, there are different routes to the same fractal. Suppose you take a pentagon and select a single vertex. Now, measure the distance to each vertex, v(1,i=1..5), of the original pentagon (including the selected vertex) and reduce it by 1/φ to find the position of a new vertex, v(2,i=1..5). If you do this for each vertex of the original pentagon, then to each vertex of the new pentagons, and so on, in the end you create the same fractal as the jumping point does:

Shrink pentagons by 1/φ, stage #1

Stage #2

Stage #3

Stage #4

Stage #5

Stage #6

Shrink by 1/φ (animated) (click for larger if blurred)

And here is the route to a centre-filled variant of the fractal:

Central pentagon, stage #1

Stage #2

Stage #3

Stage #4

Stage #5

Stage #6

Central pentagon (animated) (click for larger if blurred)

Using this shrink-the-polygon method, you can reach the same fractals by a third route. This time, use vertex v(1,i) of the original polygon as the centre of the new polygon with its vertices v(2,i=1..5). Creation of the fractal looks like this:

Pentagons over vertices, shrink by 1/φ, stage #1 (no pentagons over vertices)

Stage #2

Stage #3

Stage #4

Stage #4

Stage #5

Stage #7

Pentagons over vertices (animated) (click for larger if blurred)

And here is a third way of creating the centre-filled pentagonal fractal:

Pentagons over vertices and central pentagon, stage #1

Stage #2

Stage #3

Stage #4

Stage #5

Stage #6

Stage #7

Pentagons over vertices with central pentagon (animated) (click for larger if blurred)

And here is a fractal created when there are three pentagons to a side and the pentagons are shrunk by 1/φ^2 = 0.3819660112…:

Pentagon at vertex + pentagon at mid-point of side, shrink by 1/φ^2

Final stage

Pentagon at vertex + pentagon at mid-point of side (animated) (click for larger if blurred)

Pentagon at vertex + pentagon at mid-point of side + central pentagon, shrink by 1/φ^2 and c. 0.5, stage #1

Stage #2

Stage #3

Stage #4

Stage #5

Pentagon at vertex + mid-point + center (animated) (click for larger if blurred)

Previously pre-posted:

Performativizing the Polygonic #2

Suppose a café offers you free drinks for three days. You can have tea or coffee in any order and any number of times. If you want tea every day of the three, you can have it. So here’s a question: how many ways can you choose from two kinds of drink in three days? One simple way is to number each drink, tea = 1, coffee = 2, then count off the choices like this:

1: 111
2: 112
3: 121
4: 122
5: 211
6: 212
7: 221
8: 222

Choice #1 is 111, which means tea every day. Choice #6 is 212, which means coffee on day 1, tea on day 2 and coffee on day 3. Now look at the counting again and the way the numbers change: 111, 112, 121, 122, 211… It’s really base 2 using 1 and 2 rather than 0 and 1. That’s why there are 8 ways to choose two drinks over three days: 8 = 2^3. Next, note that you use the same number of 1s to count the choices as the number of 2s. There are twelve 1s and twelve 2s, because each number has a mirror: 111 has 222, 112 has 221, 121 has 212, and so on.

Now try the number of ways to choose from three kinds of drink (tea, coffee, orange juice) over two days:

11, 12, 13, 21, 22, 23, 31, 32, 33 (c=9)

There are 9 ways to choose, because 9 = 3^2. And each digit, 1, 2, 3, is used exactly six times when you write the choices. Now try the number of ways to choose from three kinds of drink over three days:

111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333 (c=27)

There are 27 ways and (by coincidence) each digit is used 27 times to write the choices. Now try three drinks over four days:

1111, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1212, 1213, 1221, 1222, 1223, 1231, 1232, 1233, 1311, 1312, 1313, 1321, 1322, 1323, 1331, 1332, 1333, 2111, 2112, 2113, 2121, 2122, 2123, 2131, 2132, 2133, 2211, 2212, 2213, 2221, 2222, 2223, 2231, 2232, 2233, 2311, 2312, 2313, 2321, 2322, 2323, 2331, 2332, 2333, 3111, 3112, 3113, 3121, 3122, 3123, 3131, 3132, 3133, 3211, 3212, 3213, 3221, 3222, 3223, 3231, 3232, 3233, 3311, 3312, 3313, 3321, 3322, 3323, 3331, 3332, 3333 (c=81)

There are 81 ways to choose and each digit is used 108 times. But the numbers don’t have represent choices of drink in a café. How many ways can a point inside an equilateral triangle jump four times half-way towards the vertices of the triangle? It’s the same as the way to choose from three drinks over four days. And because the point jumps toward each vertex in a symmetrical way the same number of times, you get a nice even pattern, like this:

vertices = 3, jump = 1/2

Every time the point jumps half-way towards a particular vertex, its position is marked in a unique colour. The fractal, also known as a Sierpiński triangle, actually represents all possible choices for an indefinite number of jumps. Here’s the same rule applied to a square. There are four vertices, so the point is tracing all possible ways to choose four vertices for an indefinite number of jumps:

v = 4, jump = 1/2

As you can see, it’s not an obvious fractal. But what if the point jumps two-thirds of the way to its target vertex and an extra target is added at the centre of the square? This attractive fractal appears:

v = 4 + central target, jump = 2/3

If the central target is removed and an extra target is added on each side, this fractal appears:

v = 4 + 4 midpoints, jump = 2/3

That fractal is known as a Sierpiński carpet. Now up to the pentagon. This fractal of endlessly nested contingent pentagons is created by a point jumping 1/φ = 0·6180339887… of the distance towards the five vertices:

v = 5, jump = 1/φ

With a central target in the pentagon, this fractal appears:

v = 5 + central, jump = 1/φ

The central red pattern fits exactly inside the five that surround it:

v = 5 + central, jump = 1/φ (closeup)

v = 5 + c, jump = 1/φ (animated)

For a fractal of endlessly nested contingent hexagons, the jump is 2/3:

v = 6, jump = 2/3

With a central target, you get a filled variation of the hexagonal fractal:

v = 6 + c, jump = 2/3

And for a fractal of endlessly nested contingent octagons, the jump is 1/√2 = 0·7071067811… = √½:

v = 8, jump = 1/√2

Previously pre-posted:

Square on a Three String

222 A.D. was the year in which the Emperor Heliogabalus was assassinated by his own soldiers. Exactly 1666 years later, the Anglo-Dutch classicist Sir Lawrence Alma-Tadema exhibited his painting The Roses of Heliogabalus (1888). I suggested in “Roses Are Golden” that Alma-Tadema must have chosen the year as deliberately as he chose the dimensions of his canvas, which, at 52″ x 84 1/8“, is an excellent approximation to the golden ratio.

But did Alma-Tadema know that lines at 0º and 222º divide a circle in the golden ratio? He could easily have done, just as he could easily have known that 222 precedes the 48th prime, 223. But it is highly unlikely that he knew that 223 yields a magic square whose columns, rows and diagonals all sum to 222. To create the square, simply list the 222 multiples of the reciprocal 1/223 in base 3, or ternary. The digits of the reciprocal repeat after exactly 222 digits and its multiples begin and end like this:

001/223 = 0.00001002102101021212111012022211122022... in base 3
002/223 = 0.00002011211202120201222101122200021121...
003/223 = 0.00010021021010212121110120222111220221...
004/223 = 0.00011100200112011110221210022100120020...
005/223 = 0.00012110002220110100102222122012012120...

[...]

218/223 = 0.22210112220002112122120000100210210102... in base 3
219/223 = 0.22211122022110211112001012200122102202...
220/223 = 0.22212201201212010101112102000111002001...
221/223 = 0.22220211011020102021000121100022201101...
222/223 = 0.22221220120121201010111210200011100200...

Each column, row and diagonal of ternary digits sums to 222. Here is the full n/223 square represented with 0s in grey, 1s in white and 2s in red:

(Click for larger)

It isn’t difficult to see that the white squares are mirror-symmetrical on a horizontal axis. Here is the symmetrical pattern rotated by 90º:

(Click for larger)

But why should the 1s be symmetrical? This isn’t something special to 1/223, because it happens with prime reciprocals like 1/7 too:

1/7 = 0.010212... in base 3
2/7 = 0.021201...
3/7 = 0.102120...
4/7 = 0.120102...
5/7 = 0.201021...
6/7 = 0.212010...

And you can notice something else: 0s mirror 2s and 2s mirror 0s. A related pattern appears in base 10:

1/7 = 0.142857...
2/7 = 0.285714...
3/7 = 0.428571...
4/7 = 0.571428...
5/7 = 0.714285...
6/7 = 0.857142...

The digit 1 in the decimal digits of n/7 corresponds to the digit 8 in the decimal digits of (7-n)/7; 4 corresponds to 5; 2 corresponds to 7; 8 corresponds to 1; 5 corresponds to 4; and 7 corresponds to 2. In short, if you’re given the digits d1 of n/7, you know the digits d2 of (n-7)/7 by the rule d2 = 9-d1.

Why does that happen? Examine these sums:

1/7 = 0.142857142857142857142857142857142857142857...
+6/7 = 0.857142857142857142857142857142857142857142...
7/7 = 0.999999999999999999999999999999999999999999... = 1.0

2/7 = 0.285714285714285714285714285714285714285714...
+5/7 = 0.714285714285714285714285714285714285714285...
7/7 = 0.999999999999999999999999999999999999999999... = 1.0

3/7 = 0.428571428571428571428571428571428571428571...
+4/7 = 0.571428571428571428571428571428571428571428...
7/7 = 0.999999999999999999999999999999999999999999... = 1.0

And here are the same sums in ternary (where the first seven integers are 1, 2, 10, 11, 12, 20, 21):

1/21 = 0.010212010212010212010212010212010212010212...
+20/21 = 0.212010212010212010212010212010212010212010...
21/21 = 0.222222222222222222222222222222222222222222... = 1.0

2/21 = 0.021201021201021201021201021201021201021201...
+12/21 = 0.201021201021201021201021201021201021201021...
21/21 = 0.222222222222222222222222222222222222222222... = 1.0

10/21 = 0.102120102120102120102120102120102120102120...
+11/21 = 0.120102120102120102120102120102120102120102...
21/21 = 0.222222222222222222222222222222222222222222... = 1.0

Accordingly, in base b with the prime p, the digits d1 of n/p correspond to the digits (p-n)/p by the rule d2 = (b-1)-d1. This explains why the 1s mirror themselves in ternary: 1 = 2-1 = (3-1)-1. In base 5, the 2s mirror themselves by the rule 2 = 4-2 = (5-1) – 2. In all odd bases, some digit will mirror itself; in all even bases, no digit will. The mirror-digit will be equal to (b-1)/2, which is always an integer when b is odd, but never an integer when b is even.

Here are some more examples of the symmetrical patterns found in odd bases:

Patterns of 1s in 1/19 in base 3

Patterns of 6s in 1/19 in base 13

Patterns of 7s in 1/19 in base 15

Elsewhere other-posted:

Roses Are Golden — more on The Roses of Heliogabalus (1888)
Three Is The Key — more on the 1/223 square

Breeding Bunnies

The Golden Ratio: The Story of Phi, the Extraordinary Number of Nature, Art and Beauty, Mario Livio (Headline Review 2003)

A good short popular guide to perhaps the most interesting, and certainly the most irrational, of all numbers: the golden ratio or phi (φ), which is approximately equal to 1·6180339887498948482… Prominent in mathematics since at least the ancient Greeks and Euclid, phi is found in many places in nature too, from pineapples and sunflowers to the flight of hawks. Livio catalogues its appearances in both maths and nature, looking closely at the Fibonacci sequence and rabbit-breeding, before going on to debunk mistaken claims that phi also appears a lot in art, music and poetry. Dalí certainly used it, but da Vinci, Debussy and Virgil almost certainly didn’t. Nor, almost certainly, did the builders of the Parthenon and pyramids. Finally, he examines what has famously been called (by the physicist Eugene Wiegner) the unreasonable effectiveness of mathematics: why is this human invention so good at describing the behaviour of the Universe? Livio quotes one of the best short answers I’ve seen:

Human logic was forced on us by the physical world and is therefore consistent with it. Mathematics derives from logic. That is why mathematics is consistent with the physical world. (ch. 9, “Is God a mathematician?”, pg. 252)

It’s not hard to recommend a book that quotes everyone from Johannes Kepler and William Blake to Lewis Carroll, Christopher Marlowe and Jef Raskin, “the creator of the Macintosh computer”, whose answer is given above. Recreational mathematicians should also find lots of ideas for further investigation, from fractal strings to the fascinating number patterns governed by Benford’s law. It isn’t just human beings who look after number one: as a leading figure, 1 turns up much more often in data from the real world, and in mathematical constructs like the Fibonacci sequence, than intuition would lead you to expect. If you’d like to learn more about that and about many other aspects of mathematics, hunt down a copy of this book.

Elsewhere other-posted:

Roses Are Golden – φ and floral homicide

Roses Are Golden

Sir Lawrence Alma-Tadema’s painting The Roses of Heliogabalus (1888) is based on an apocryphal episode in the sybaritic life of the Roman Emperor Elagabalus (204-222 A.D.), who is said to have suffocated guests with flowers at one of his feasts. The painting is in a private collection, but I saw it for real in an Alma-Tadema exhibition at the Walker Art Gallery in Liverpool sometime during the late 1990s. I wasn’t disappointed: it was a memorable meeting with a painting I’d been interested in for years. Roses is impressively large and impressively skilful. Close-up, the brush-strokes are obvious, obtrusive and hard to interpret as people and objects. It isn’t till you step back, far beyond the distance at which Alma-Tadema was painting, that the almost photographic realism becomes apparent. But you get more of the many details at close range, like the Latin inscription on a bowl below and slightly to the right of that scowling water-mask. Alas, I forgot to take a note of what the inscription was, though perhaps the memory is still locked away somewhere in my subconscious.

The Roses of Heliogabalus (1888)

Whatever it is, I feel sure it is significant, because Roses is rich with meaning. That’s a large part of why I’m interested in it. Yes, I like it a lot as art, but the women would have to be more attractive for it to be higher in the list of my favourite paintings. As it is, I think there are only four reasonably good-looking people in it: the man with the beard on the right; the flautist striding past the marble pillar on the left; the red-headed girl with a crown of white flowers; and Heliogabalus himself, crowned in roses and clutching a handful of grapes beside the overweight man who’s wearing a wreath and sardonically saluting one of the rose-pelted guests in the foreground. When I first wrote about Roses in a pub-zine whose name escapes me, I misidentified the overweight man as Heliogabalus himself, even though I noted that he seemed many years old than Heliogabalus, toppled as a teen tyrant, should have been. It was a bad mistake, but one that, with less knowledge and more excuse, many people must make when they look at Roses, because the overweight man and his sardonic salute are a natural focus for the eye. Once your eye has settled on and noted him, you naturally follow the direction of his gaze down to the man in the foreground, who’s gazing right back.

Something Like the Sun

And by following that gaze, you’ve performed a little ratio-ritual, just as Alma-Tadema intended you to do. Yes, Roses is full of meaning and much of that meaning is mathematical. I think the angle of the gaze is one of many references in Roses to the golden ratio, or φ (phi), a number that is supposed to have special aesthetic importance and has certainly been used by many artists and musicians to guide their work. A rectangle with sides in the proportions 8:13, for example, approximates the golden ratio pretty closely, but φ itself is impossible to represent physically, because it’s an irrational number with infinitely many decimal digits, like π or √2, the square root of two. π represents the ratio of a circle’s circumference to its diameter and √2 the ratio of a square’s diagonal to its side, but no earthly circle and no earthly square can ever capture these numbers with infinite precision. Similarly, no earthly rectangle can capture φ, but the rectangle of Roses is a good attempt, because it measures 52″ x 84 1/8". That extra eighth of an inch was my first clue to the painting’s mathematical meaningfulness. And sure enough, 52/84·125 = 416/673 = 0·61812…, which is a good approximation to φ’s never-ending 0·6180339887498948482045868343656…*

That deliberate choice of dimensions for the canvas led me to look for more instances of φ in the painting, though one of the most important and obvious might be called a meta-presence. The Roses of Heliogabalus is dated 1888, or 1666 years after the death of Heliogabalus in 222 AD. A radius at 222º divides a circle in the golden ratio, because 222/360 = 0·616… It’s very hard to believe Alma-Tadema didn’t intend this reference and I also think there’s something significant in 1888 itself, which equals 2 x 2 x 2 x 2 x 2 x 59 = 25 x 59. Recall that 416 is the expanded short side of Roses. This equals 25 x 13, while 673, the expanded long side, is the first prime number after 666. As one of the most technically skilled painters who ever lived, Alma-Tadema was certainly an exceptional implicit mathematician. But he clearly had explicit mathematical knowledge too and this painting is a phi-pie cooked by a master matho-chef. In short, when Roses is read, Roses turns out to be golden.

*φ is more usually represented as 1·6180339887498948482045868343656…, but it has the pecularity that 1/φ = φ-1, so the decimal digits don’t change and 0·6180339887498948482045868343656… is also legitimate.

Appendix I

I’ve looked at more of Alma-Tadema’s paintings to see if their dimensions approximate φ, √2, √3 or π, or their reciprocals. These were the results (ε = error, i.e. the difference between the constant and the ratio of the dimensions).

The Roman Wine Tasters (1861), 50" x 69 2/3": 150/209 = 0·717… ≈ 1/√2 (ε=0·02)
A Roman Scribe (1865), 21 1/2" x 15 1/2": 43/31 = 1·387… ≈ √2 (ε=0·027)
A Picture Gallery (1866), 16 1/8" x 23": 129/184 = 0·701… ≈ 1/√2 (ε=0·012)
A Roman Dance (1866), 16 1/8" x 22 1/8": 43/59 = 0·728… ≈ 1/√2 (ε=0·042)
In the Peristyle (1866), 23" x 16": 23/16 = 1·437… ≈ √2 (ε=0·023)
Proclaiming Emperor Claudius (1867), 18 1/2" x 26 1/3": 111/158 = 0·702… ≈ 1/√2 (ε=0·009)
Phidias and the Frieze of the Parthenon Athens (1868), 29 2/3" x 42 1/3": 89/127 = 0·7… ≈ 1/√2 (ε=0·012)
The Education of Children of Clovis (1868), 50" x 69 2/3": 150/209 = 0·717… ≈ 1/√2 (ε=0·02)
An Egyptian Juggler (1870), 31" x 19 1/4": 124/77 = 1·61… ≈ φ (ε=0·007)
A Roman Art Lover (1870), 29" x 40": 29/40 = 0·725… ≈ 1/√2 (ε=0·034)
Good Friends (1873), 4 1/2" x 7 1/4": 18/29 = 0·62… ≈ φ (ε=0·006)
Pleading (1876), 8 1/2" x 12 3/8": 68/99 = 0·686… ≈ 1/√2 (ε=0·041)
An Oleander (1882), 36 1/2" x 25 1/2": 73/51 = 1·431… ≈ √2 (ε=0·017)
Dolce Far Niente (1882), 9 1/4" x 6 1/2": 37/26 = 1·423… ≈ √2 (ε=0·008)
Anthony and Cleopatra (1884), 25 3/4" x 36 1/3": 309/436 = 0·708… ≈ 1/√2 (ε=0·003)
Rose of All Roses (1885), 15 1/4" x 9 1/4": 61/37 = 1·648… ≈ φ (ε=0·03)
The Roses of Heliogabalus (1888), 52" x 84 1/8": 416/673 = 0·618… ≈ φ (ε<0.001)
The Kiss (1891), 18" x 24 3/4": 8/11 = 0·727… ≈ 1/√2 (ε=0·039)
Unconscious Rivals (1893), 17 3/4" x 24 3/4": 71/99 = 0·717… ≈ 1/√2 (ε=0·019)
A Coign of Vantage (1895), 25 1/4" x 17 1/2": 101/70 = 1·442… ≈ √2 (ε=0·028)
A Difference of Opinion (1896), 15" x 9": 5/3 = 1·666… ≈ φ (ε=0·048)
Whispering Noon (1896), 22" x 15 1/2": 44/31 = 1·419… ≈ √2 (ε=0·005)
Her Eyes Are With Her Thoughts And Her Thoughts Are Far Away (1897), 9" x 15": 3/5 = 0·6… ≈ φ (ε=0·048)
The Baths of Caracalla (1899), 60" x 37 1/2": 8/5 = 1·6… ≈ φ (ε=0·018)
The Year’s at the Spring, All’s Right with the World (1902), 13 1/2" x 9 1/2": 27/19 = 1·421… ≈ √2 (ε=0·006)
Ask Me No More (1906), 31 1/2" x 45 1/2": 9/13 = 0·692… ≈ 1/√2 (ε=0·03)

Appendix II

The Roses of Heliogabalus is based on this section from Aelius Lampridius’ pseudonymous and largely apocryphal Vita Heliogabali, or Life of Heliogabalus, in the Historia Augusta (late fourth century):

XXI. 1 Canes iecineribus anserum pavit. Habuit leones et leopardos exarmatos in deliciis, quos edoctos per mansuetarios subito ad secundam et tertiam mensam iubebat accumbere ignorantibus cunctis, quod exarmati essent, ad pavorem ridiculum excitandum. 2 Misit et uvas Apamenas in praesepia equis suis et psittacis atque fasianis leones pavit et alia animalia. 3 Exhibuit et sumina apruna per dies decem tricena cottidie cum suis vulvis, pisum cum aureis, lentem cum cerauniis, fabam cum electris, orizam cum albis exhibens. 4 Albas praeterea in vicem piperis piscibus et tuberibus conspersit. 5 Oppressit in tricliniis versatilibus parasitos suos violis et floribus, sic ut animam aliqui efflaverint, cum erepere ad summum non possent. 6 Condito piscinas et solia temperavit et rosato atque absentato…

Historia Augusta: Vita Heliogabali

XXI. 1 He fed his dogs on goose-livers. He had pet lions and leopards, which had been rendered harmless and trained by tamers, and these he would suddenly order during the dessert and the after-dessert to get on the couches, thereby causing laughter and panic, for none knew that they were harmless. 2 He sent grapes from Apamea to his stables for the horses, and he fed parrots and pheasants to his lions and other beasts. 3 For ten days in a row, moreover, he served wild sows’ udders with the matrices, at a rate of thirty a day, serving, besides, peas with gold-pieces, lentils with onyx, beans with amber, and rice with pearls; 4 and he also sprinkled pearls on fish and used truffles instead of pepper. 5 In a banqueting-room with a reversible ceiling he once buried his parasites in violets and other flowers, so that some were actually smothered to death, being unable to crawl out to the top. 6 He flavoured his swimming-pools and bath-tubs with essence of spices or of roses or wormwood…

Augustan History: Life of Heliogabalus