The Overlord of the Über-Feral says: Welcome to my bijou bloguette. You can scroll down to sample more or simply:
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Gweel & Other Alterities – Incunabula’s new edition
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I identify as √-37 (the square root of -37) in bases 5 (on Mondays and Wednesdays), 17 (Tuesdays, Thursdays and alternate Saturdays) and 89 (Fridays, Sundays and alternate Saturdays).
My preferred pronouns are 6.08276i, 雲 and მსოფლიოს მეფე.
What could be duller than digits? They just sit there on the page or screen, mindlessly marking mathematics:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100…
But perhaps they become more interesting as images. Let’s display the final digit of the integers, or counting numbers, on a graph. Running left-right and up-down, the graph represents the final or rightmost digit of 1, 2, 3, … 10, 11, 12, 13, … 100, 101, 102, 103, …, 1000, 1001, 1002, 1003, …:
Rightmost single digit of the integers (click for larger)
No, that’s still dull: the graph just generates endlessly repeating triangles. After all, the final digits fall into a cycle: 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3… So do the final two digits: 1, 2, 3, 4, 5, […] 94, 95, 96, 97, 98, 99, 00, 01, 02, 03… Here they are as a graph:
Rightmost two digits of the integers
Now the triangles look like waves sweeping to shore. That’s a bit more interesting, but not much. So let’s try something different. The trailing digits of the integers generate triangles, so let’s see what the triangular numbers generate. The triangular numbers — 0, 1, 3, 6, 10, 15, 21… — are very simple to form. You just sum the integers: 1, 3 = 1 + 2, 6 = 1 + 2 + 3, 10 = 1 + 2 + 3 + 4, 15 = 1 + 2 + 3 + 4 + 5, 21 = 1 + 2 + 3 + 4 + 5 + 6, 28 = 1 + 2 + 3 + 4 + 5 + 6 + 7, 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8, 45 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9, 55 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10… Here are the final digits of the triangulars — 1, 3, 6, 0, 5, 1, 8, 6… — as a graph:
Final digit of triangular numbers in base 10 (click for larger)
Now something interesting has appeared. The final digits form a repeated palindromic pattern (counting 0 as the zero-th triangular number):
0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0, 1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, …
An Altar of Mathness created by the final digit of triangular numbers in base 10
And those palindromic digits create symmetric shapes that remind me of little altars — let’s call them “altars of mathness” in tribute to Morbid Angel’s genre-defining album Altars of Madness (1989). And what about the final two digits of the triangular numbers? Here’s the graph (adjusted so that 99 fits into the same space as 9):
Final two digits of triangulars in b10
Final two triangular digits in b10 (horizontal scale compressed)
The final two digits form palindromes too. And this time we don’t get just triangles, but curves too. But that’s in base 10. What happens with the trailing triangular digits in other bases? Well, here’s the final triangular digit creating more altars of mathness in different bases (note that the altars are more elaborate in even bases):
Final triangular digit in base 4
Final triangular digit in b5
Final triangular digit in b6
Final triangular digit in b7
Final triangular digit in b8
Final triangular digit in b9
Final triangular digit in b14
And here’s the graph for the final triangular digit in base 100:
Final triangular digit in b100
The graph for final single digit in b100 should look familiar, because it’s identical to the graph for final double triangular digits in b10:
Final two digits of triangulars in b10
That’s because two digits in b10 are equivalent in one digit in b100, four digits in b10 are equivalent to two digits in b100, and so on. But b100 can’t capture three digits in b10 (the graph is again adjusted so that 999 fits into the same space as 9 and 99 above):
Final three triangular digits in b10
If you compress the x-axis for that graph, you can see how long the symmetries are:
Final three triangular digits in b10 (x-axis / 2)
Final three triangular digits in b10 (x-axis / 4)
The final four digits of the triangulars in b10 create even longer symmetries:
Final quadruple triangular digits in b10
Final quadruple triangular digits in b10 (x-axis / 2)
Final quadruple triangular digits in b10 (x-axis / 8)
Note how, as the length of the final digits rises, you need to compress the x-axis more and more to see the symmetries. But integer sequences obviously don’t end with the counting numbers and triangulars. What about squares and powers of n? What about primes and Fibonacci numbers? Here’s the final two digits of the squares — 1, 4, 9, 16, 25, 49, 64, 81, 100, 121, 144, 169… — in b10:
Final two digits of the squares in b10
It’s reminiscent of the triangular numbers (so are the final-digit graphs for other polygonal numbers). So what about the powers of 2? That’s 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024… Here’s the graph for final single digits of 2^p in b10:
Final single digits of 2^p in b10
This time there’s repetition, but not symmetry. Here’s the graph for final double digits, or 2-digits, of 2^p in b10:
Final 2-dig of 2^p in b10
Now the graph looks a little like a range of eroded mountains. Now try dig-4, the final four digits of 2^p in b10:
Final 4-dig of 2^p in b10
The patterns are similar to those of dig-2 and don’t need compressing in the x-axis. This similarity and lack of need for compression are true of any number of final digits in 2^p. The final 10 digits look like this:
Final 10-dig of 2^p in b10
And the final 20 and 30 digits like this:
Final 20-dig of 2^p in b10
Final 30-dig of 2^p in b10
Powers don’t behave like polygonals: the finals are fractals. That is, the final digits create similar patterns at all scales: 1-dig, 2-dig, 10-dig, 100-dig, 1000-dig and so on. That’s true in other bases:
Final 5-dig of 3^p in b2
But a glimpse of b2 is all you’re going to get of other bases. There are other fish to fry — Fibonacci fish. The Fibonacci sequence, whose terms are equal to the sum of the previous two numbers (after seeding with “1, 1”), starts like this: 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… And what about the graphs for final fib-digits? As you’ll see, final Fib-digits are fractal too. Indeed, Fibonacci final-graphs look like 2-power final-graphs (in a way, Fibonacci numbers are powers of φ = 1.6180339887498948482…). The patterns are similar at all scales. And they remind me of the skyline of a ruined city in an Oriental tale, with collapsed domes and crumbling minarets:
Final 1-dig of Fibonacci numbers in b10
Final 2-fibdig in b10
Final 3-fibdig in b10
Final 4-fibdig in b10
Final 5-fibdig in b10
Final 10-fibdig in b10
Final 15-fibdig in b10
Final 20-fibdig in b10
Final 25-fibdig in b10
So final fibdigs are fractal. But final prime digits aren’t:
Final 1-digit of primes in b10
Final 1-digit of primes in b5
Final 2-digit of primes in b10
Primes aren’t final-digitally fractal like Fibonaccis and powers of 2. But there’s occasional symmetry in the prime fin-digs. I’ve marked some palindromic patterns in red and green:
Palindromic patterns in final 1-digits of the primes in b10 (click for larger)
The palindromic patterns, or pal-pats, in the primes look like the altars of mathness in the triangulars. They’re created by digital palindromes like these:
19, 23, 29 (c=3)
347, 349, 353, 359, 367 (c=5)
937, 941, 947, 953, 967, 971, 977 (c=7)
1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011 (c=9)
26423, 26431, 26437, 26449, 26459, 26479, 26489, 26497, 26501, 26513 (c=10)
Here are the first few pal-pats in the primes (note that 157, 163, 167 and 163, 167, 173 overlap):
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607…
And are there palindromes among the final 2-digits, 3-digits and higher n-digits of the primes in different bases? Yes, you can easily find some. But I haven’t put them on a graph yet:
Base 10 (2-dig)
58789, 58831, 58889 (c=3)
286873, 286927, 286973 (c=3)
360649, 360653, 360749 (c=3)
404851, 404941, 404951 (c=3)
590437, 590489, 590537 (c=3)
623071, 623107, 623171 (c=3)
651517, 651587, 651617 (c=3)
Base 6 (2-dig)
300335, 300401, 300441, 300501, 300535 (c=5) (23459 to 23531 in base 10)
1030255, 1030331, 1030351, 1030431, 1030455 (c=5) (50651 to 50723 in b10)
1140451, 1140501, 1140521, 1141001, 1141051 (c=5) (59791 to 59863 in b10)
1402451, 1402545, 1403031, 1403045, 1403051 (c=5) (78367 to 78439 in b10)
1435431, 1435451, 1435505, 1435551, 1440031 (c=5) (82891 to 82963)
2400505, 2401001, 2401015, 2401101, 2401105 (c=5) (124601 to 124673)
2442235, 2442311, 2442351, 2442411, 2442435 (c=5) (130127 to 130199)
2444215, 2444225, 2444311, 2444325, 2444415 (c=5) (130547 to 130619)
2533105, 2533121, 2533215, 2533221, 2533305 (c=5) (136769 to 136841)
Base 4 (3-dig)
20013013, 20013133, 20020013 (c=3) (33223 to 33287 in base 10)
21031111, 21031303, 21032111 (c=3) (37717 to 37781)
22310011, 22310333, 22311011 (c=3) (44293 to 44357)
33030121, 33031001, 33031121 (c=3) (62233 to 62297)
102031333, 102032131, 102032333 (c=3) (74623 to 74687)
110013121, 110013311, 110020121 (c=3) (82393 to 82457)
Base 3 (3-dig)
112121020012, 112121021211, 112121022021, 112121100211, 112121101012 (c=5) (287393 to 287501 in base 10)
202002212002, 202002212101, 202002220001, 202002221101, 202010000002 (c=5) (395741 to 395849)
1001012111212, 1001012112202, 1001012121022, 1001012121202, 1001012122212 (c=5) (555143 to 555251)
1010112112012, 1010112112201, 1010112120222, 1010112121201, 1010112200012 (c=5) (601079 to 601187)
1011202211211, 1011202212212, 1011202220022, 1011202221212, 1011202222211 (c=5) (625369 to 625477)
Base 2 (5-dig)
101110001111101, 101110010000111, 101110010001001, 101110010100111, 101110010111101 (c=5) (23677 to 23741 in base 10)
10000001000111101, 10000001001011001, 10000001001110001, 10000001001111001, 10000001001111101 (c=5) (66109 to 66173)
10111100110011011, 10111100110011111, 10111100110111001, 10111100110111111, 10111100111011011 (c=5) (96667 to 96731)
11010000111110001, 11010001000001101, 11010001000011001, 11010001000101101, 11010001000110001 (c=5) (106993 to 107057)
And I conjecture that you’ll get palindromes for any number of final digits in all bases. And can these palindromes be of arbitrary length? Again, I conjecture so. There are infinitely many primes and very rare patterns can occur infinitely often in an infinite set of numbers.
Post-Performative Post-Scriptum
Here’s Dan Seagrave’s classic cover for Morbid Angel’s Altars of Madness (1989):
• Morbid Angel — official website
• Dan Seagrave — official website
Let’s look at a simple arithmetical rule and a simple arithmetical fact. And the complexity they can create. First the rule. Subtracting a negative number is the same as adding the positive form of that number:
7 – +2 = 5
7 – -2 = 7 + 2 = 9-10 – +4 = -14
-10 – -4 = -10 + 4 = -6
Now the simple arithmetical fact: The reciprocal of positive x, namely 1/x, is less than 1 when x > 1, identical to x when x = 1, and greater than 1 when 0 < x < 1. Negative x, -x, works in the opposite direction:
1/5 = 0.2; 1/-5 = -0.2
1/4 = 0.25; 1/-4 = -0.25
1/3 = 0.333333…; 1/-3 = -0.333333…
1/2 = 0.5; 1/-2 = -0.51/1 = 1; 1/-1 = -1
1/0.5 = 2; 1/-0.5 = -2
1/0.25 = 4; 1/-0.25 = -4
1/0.333333.. = 3; 1/-0.333333.. = -3
1/0.2 = 5; 1/-0.2 = -5
Now, the simple arithmetical rule and the simple arithmetical fact explain the wildly different behaviour of these two nearly identical formulae:
Formula #1: x = x + 1/x
Formula #2: x = x – 1/x
If you seed x = x + 1/x with 2, this is what happens:
2 = x
2.5 = 2 + 1/2 = 2 + 0.5
2.9 = 2.5 + 1/2.5 = 2.5 + 0.4
3.244827586206896551724137931… = 2.9 + 1/2.9 = 2.9 + 0.3448275862…
3.553010370478947561288431236…
3.834461842815967366750790750…
4.095254632258778985771918456…
4.339439692724345181049239663…
4.569884190357676650018985962…
4.788708116379690742064597208…
4.997532704493448986664559639…
5.197631445038131469095668466…
5.390026771750770995914851381…
5.575554607204394029915651664…
5.754908962142979073283550015…
5.928673657045750549124213874…
6.097345447373015508408978797…
6.261351244425377152997703626…
6.421061179383957004641284553…
6.576798676981813718180627345…
The value of x steadily (but more and more slowly) increases. But when you seed the other formula, x = x – 1/x, with 2, this is what happens:
+2
+1.5 = 2 – 1/2 = 2 – 0.5
+0.8333333… = 1.5 – 1/1.5 = 1.5 – 0.666666…
-0.3666666… = 0.8333333… – 1/0.8333333… = 0.8333333… – 1.2
+2.3606060606… = -0.3666666… – 1/-0.3666666… = -0.3666666… – -2.72727272… = -0.3666666… + 2.72727272…
+1.936986034932119656124790913…
+1.420720051612810742016492942…
+0.716851616121389735975863550…
-0.678137217705362317788764881…
+0.796490591963802485322149292…
-0.459017018658980935029501857…
+1.719551442531198550688634398…
+1.138004432499332885157841729…
+0.259273233005005595158072588…
-3.597661740227243739940039228…
-3.319703423907923593779727545…
-3.018471695555874174383708009…
-2.687178213005645221877765061…
-2.315040631969854351245993463…
-1.883082770759830608578236571…
-1.352038668223383718148747858…
-0.612414851610188982350276645…
+1.020465208974159220420492697…
+0.040519992610273807119693182…
-24.63865528804984441050796942…
-24.59806865747650234381633987…
-24.55741505926418687092326558…
-24.51669416101057552476382150…
-24.47590562755526483917345018…
-24.43504912094763238695385804…
The value of x swings between positive and negative in an irregular, non-periodic way, alternating between slow deterministic decay and instantaneous jumps to sometimes large positive or negative values. The deterministic decays explains why, as we’ll see, there are beautiful regular curves — parabolic curves — amid the irregularity. When the function creates a positive number x > 1, it nibbles away at x until x x > -1, x becomes positive at the next step and the process continues. Represented as a graph, x = x – 1/x looks like this when seeded with 2 — note the parabolic curves:
x[i] = x[i-1] – 1/x[i-1], x[1] = 2 (click for larger)
A shark-fin, some red sails and Sydney Opera House (images StockCake + Para-Sailing World Championship + Wikipedia)
When x > 0, its value is represented in white; when x < 0, its value is represented in red. The curves created remind of me of shark-fins or sails or Sydney Opera House. So you could say the graph contains red sails in the subset, i.e. the set of values of x that are sub-zero. Here are some variations on the formula:
x = x – (1/4)/x, x[1] = 2
x = x – (4/3)/x, x[1] = 2
x = x – (4/5)/x, x[1] = 2
Now try this formula, x = 1 – 1/x. When it’s seeded with 2, it behaves like this:
2
0.5 = 1 – 1/2 = 1 – 0.5
-1 = 1 – 1/0.5 = 1 – 2
2 = 1 – -1/-1 = 1 – -1 = 1 + 1
1/2
-1
2
[…]
The values cycles through 2, 0.5, -1, 2, 0.5… for ever. So try varying the formula. This is what happens with x = 0.1 – 1.7/x, seeded with 2:
+2
-0.75
+2.366666666666666666666666666…
-0.618309859154929577464788732…
+2.849430523917995444191343963…
-0.496610440482852346310656327…
+3.523206323143542441364433927…
-0.382515028663775083373274222…
+4.544269826308639632084352464…
-0.274097504104620631734792447…
+6.302172491695235560374503419…
-0.169748249867834571832745868…
+10.11483079397645866692882142…
-0.068070038404634195480677189…
+25.07427708053510247498559239…
When you look at the graph of x = 0.1 – 1.7/x, you’ll see it’s also cycling, just in a more complicated way:
x = 0.1 – 1.7/x, x[1] = 2 (click for larger)
And here’s how different seeds can change the graph:
x = 2/3 – 1/x, x[1] = 2/3
x = 2/3 – 1/x, x[1] = 3/2
This graph reminds me of vertebrae:
x = 2/5 – 1/x, x[1] = 2
And this graph reminds of a bone:
x = 9/7 – 1/x, x[1] = 2
As Lucretius nearly said: Mathematica Moles et Machina Mundi — Mathematics is the Mass and Body of the World.
Elsewhere Other-Accessible…
• Moto-Motto — what Lucretius did say
A word inherited from Germanic.
Cognate with Old Frisian finger, Old Saxon fingar (Middle Low German finger), Old Dutch fingar (Middle Dutch, Dutch vinger), Old High German fingar (Middle High German vinger, German Finger), Old Icelandic fingr, Old Swedish finger (Swedish finger), Old Danish fingær (Danish finger), Gothic figgrs.
Further etymology uncertain, perhaps < a suffixed form of the Indo-European base of five adj., cognate with Old Frisian fīf (West Frisian fiif), Old Saxon fīf (Middle Low German vīf), Old Dutch fīf (Middle Dutch, Dutch vijf), Old High German fimf, finf, funf (Middle High German vünf, German fünf (although this presents semantic difficulties with regard to the function of the suffix), or perhaps < a suffixed form of the Indo-European base of fang v., Old English fón, reduplicated strong verb corresponding to Old Frisian fâ, Old Saxon fâhan, Old High German fâhan (Middle High German vâhen, modern German (poet) fahen), Old Norse fá (although this presents phonological difficulties).
Compare fist n., Old English fýst strong feminine corresponds to Old Frisian fêst, Middle Low German fûst (Dutch vuist), Old High German fûst (Middle High German vûst, modern German Faust) < West Germanic *fûsti.
A word inherited from Germanic.
Old English fýst strong feminine corresponds to Old Frisian fêst, Middle Low German fûst (Dutch vuist), Old High German fûst (Middle High German vûst, modern German Faust) < West Germanic *fûsti.
By some scholars this is referred to an Old Germanic form *fûhsti-z, *funhsti-z < pre-Germanic *pṇqstis (whence Old Church Slavonic pęstĭ of same meaning), < ablaut-variant of *penqe, five adj. & n., cognate with Old Frisian fīf (West Frisian fiif), Old Saxon fīf (Middle Low German vīf), Old Dutch fīf (Middle Dutch, Dutch vijf), Old High German fimf, finf, funf (Middle High German vünf).
• from the Oxford English Dictionary
Here’s a sequence of decreasing numbers. Which number comes next?
612 → 600 → 594 → 414 → 398 → 182 → ?
It’s 166, because the numbers decrease by the product of their digits higher than 0:
612 – 6*2 = 612 – 12 = 600 → 600 – 6 = 594 → 594 – 5*9*4 = 594 – 180 = 414 → 414 – 4*4 = 414 – 16 = 398 → 398 – 3*9*8 = 398 – 216 = 166
Eventually the sequence will reached 0 and stop. If you want to see how this function looks on a graph, here it is:
x = n <= 3722, f(i) -= digmul(f(i)) → 0 (click for larger)
The graph represents n on the x-axis, with the red circles marking n = 100 and n = 1000. The sequence of falling digit-products is on the y-axis, but the graph has a special feature there. The y-axis is compressed according to the size of n, so that n = 1000 falls to 0 with n -= digmul(n) in the same height as n = 100. Here’s a graph for the same function in base 7:
x = n <= 3722 in base 7, f(i) -= digmul(f(i)) → 0
Now the red circles represent 7^2 = 49, 7^3 = 343, 7^4 = 2401, i.e. 100b7, 1000b7, 10000b7. And you can try other functions for n = n – func(n) = n -= func(n). Here’s a graph for n -= hailstep(n), where hailstep(n) returns the number of steps in the Collatz sequence for n:
x = n <= 3722 in base 7, f(i) -= hailstep(f(i)) → f(i) < 2
You form a Collatz sequence by starting with a whole number and finding the next number according to two rules:
1. If n(i) is divisible by 2, n(i+1) = n(i) / 2
2. If n(i) is not divisible by 2, n(i+1) = n(i) * 3 + 1
So the Collatz sequence for n = 10 looks like this:
10 → 10 / 2 = 5 → 5 * 3 + 1 = 16 → 16 / 2 = 8 → 8 / 2 = 4 → 4 / 2 = 2 → 2 / 2 = 1.
When you reach 1, you stop. So that’s six steps for n = 10. But does every n reach 1 in the end? It’s a very simple question about a very simple function. But nobody knows and nobody can prove that either all numbers do or at least one number doesn’t. The German mathematician Lothar Collatz (1910-90) conjectured that all numbers do reach 1. But it can take a surprisingly long time, even with small n. This is the Collatz sequence for n = 27:
27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 395, 1186, 593, 1780, 890, 445, 1336, 668, 334, 167, 502, 251, 754, 377, 1132, 566, 283, 850, 425, 1276, 638, 319, 958, 479, 1438, 719, 2158, 1079, 3238, 1619, 4858, 2429, 7288, 3644, 1822, 911, 2734, 1367, 4102, 2051, 6154, 3077, 9232, 4616, 2308, 1154, 577, 1732, 866, 433, 1300, 650, 325, 976, 488, 244, 122, 61, 184, 92, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1
Now some more functions for the y-compressed fall-bands, as I call them. If you use the sum of the factors * powers, you get this:
x = n <= 7422, f(i) -= factpowsum(f(i)) → f(i) < 2
The factpowsum(n) is the sum of the factors multiplied by their powers. For example, 37692 = 2^2 * 3^3 * 349, so factpowsum(4188) = 2*2 + 3*3 + 349*1 = 362. Here’s factpowsum for more n:
x = n <= 14822, f(i) -= factpowsum(f(i)) → f(i) < 2
You can also use the very simple function f(i) -= 1, that is, compress the numbers from n to 1 into the y-gap. But if you do that, you’ll get a completely filled screen:
x = n <= 3722, f(i) -= 1 → f(i) = 0
So you can adjust the color of a pixel according to how many times it’s written to:
x = n <= 1862, f(i) -= 1 → f(i) = 0 (color-adjust)
The patterns in the colors are artifacts of the limited resolution of the screen, so I call these patterns strartifacts = strata + artifacts. Here’s another example:
x = n <= 3722, f(i) -= 1 → f(i) = 0 (color-adjust)
Or adjust the greytone of the pixel:
x = n <= 3722, f(i) -= 1 → f(i) = 0 (greytone-adjust)
And so on (in all cases, you can click for a larger image):
x = n <= 1862, f(i) -= blockmul(f(i)) (multiply run-lengths of same digits) → f(i) < 2
x is triangular(n) = 3 to 1734453, y is 1 < triangular numbers <= n
x = n <= , f(i) -= 1 → f(i) < 2
x = n <= 7442, f(i) -= 1 → f(i) < 2
x = n <= 1862, f(i) -= leaddig(f(i)) → f(i) < 2 # 1
x = n <= , f(i) -= leaddig(f(i)) → f(i) < 2 # 2
x = n <= , f(i) -= trailingdigit(f(i)) + 1 → f(i) < 2
x = n <= , f(i) -= trailingdigit(f(i) in base 5) + 3 → f(i) < 2
for triangular(n) = 3 to 6928503, f(i) -= primes → f(i) < 2
x = n <= 1862, f(i) -= blockmul(f(i) in base 5) (multiply run-lengths of same digits) → f(i) < 2
x = n <= 1862, f(i) -= blockmul(f(i) in base 2) → f(i) < 2
x = n <= 1862, f(i) -= digsum(f(i)) → f(i) < 2
x = n <= 3722, f(i) -= func(x = 1/4 → x < 0, x(1) = 4) → f(i) < 2
x = n <= 932, f(i) -= func(x -= 3/x → x < 0, x(1) = 6) → f(i) < 2
What is VDSP? Inter alia, it’s the complicated consonant-cluster you get when you carefully pronounce the phrase “sieved spiral”. And here is a sieved spiral:
An Ulam Spiral of primes represented on a square grid
The pattern above is called an Ulam spiral (OO-lam) after its inventor, the Polish-Jewish mathematician Stanisław Ulam (1909-84). The white squares represent the prime numbers as you spiral counter-clockwise on a square grid — the little boot or reversed-L in the middle is the only time that filled squares are in direct contact, because it includes square #2, the only even prime. #2 is the heel of the boot, with #3 as the shaft and #11 as the toe.
But the Ulam spiral could also be called a sieved spiral, because you can build it by using the Sieve of Erastosthenes, whose invention is attributed to the Greek scholar Eratosthenes of Cyrene (c. 276–c. 194 BC). Create a list of whole numbers skipping 1. Then choose the first number on the list, which is 2. Cross out every higher number that’s divisible by 2. Then choose the next number that isn’t crossed out. It’ll be 3. Cross out every higher number that’s divisible by 3. Then choose the next available number, 5, and cross out all higher numbers divisible by 5. When you’ve crossed out everything you can, you’ll be left with just prime numbers. Here’s an animation of the Sieve from Wikipedia:
Animated Sieve of Erastosthenes from Wikipedia
Now we can sieve-and-let-spi’, as it were. First create a square grid with white squares. Choose the square in the middle as #1 and fill it it. Then choose white square to the right of #1 and call it #2. Then spiral outwards counter-clockwise filling with black all squares whose count is divisible by 2. Then do that for squares #3, #5, #7, #11 and so on. In the end, the only white squares on the grid will be the primes. And you’ll have a sieved Ulam spiral:
Sieving a spiral — creating the Ulam spiral using the Sieve of Eratosthenes
You can also sieve and let spi’ in reverse, blacking the squares using primes from higher to lowest. With this method, the sieved spiral looks like this:
Sieving a spiral — creating the Ulam spiral using the Sieve of Eratosthenes (higher primes first)
Today isn’t Friday, let alone Phriday. But it is Phiday, that is, it’s a date when the digits of the day of the month, the 23rd, reproduce two successive terms in the famous Fibonacci sequence:
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,…
The rule for the Fibonacci sequence is very simple. If n(i) represents the i-th term of the sequence, n(i) = n(i-1) + n(i-2). So 8, the sixth term, equals the sum of 3 and 5, the fourth and fifth terms, respectively. Dividing successive terms in the sequence, n(i)/n(i-1), gives you closer and closer approximations to a famous constant called the golden ratio or phi or φ, which equals 1.6180339887498948482045868343656381177203091798… So the Phidays in a month are the 11th, the 12th and the 23rd (except in other bases, where you can get 112 in base 3 = 14 in base 10 and 123 in b4 = 27 in b10). Obviously, you can try variants on the Fibonacci sequence. Here’s one I hadn’t tried before, summing the reciprocals of the two previous terms (1/x is the reciprocal of x) and seeded with (1,1):
f(i) = 1/f(i-1) + 1/f(i-2)
A good mathematician might not be surprised by what happens when you apply that function, but I was. Here’s the sequence in action:
1/1 + 1/1 = 1 + 1 = 2
1/2 + 1/1 = 1/2 + 1 = 3/2 = 1.5
1/2 + 1/(3/2) = 1/2 + 2/3 = 7/6 = 1.1666666666666666…
1/(3/2) + 1/(7/6) = 2/3 + 6/7 = 32/21 = 1.5238095238095…
1/(7/6) + 1/(32/21) = 6/7 + 21/32 = 339/224 = 1.51339285714285…
1/(32/21) + 1/(339/224) = 14287/10848 = 1.3170169616519174…
1/(339/224) + 1/(14287/10848) = 6877760/4843293 = 1.4200586254…
1/(14287/10848) = 1/(6877760/4843293) = 143806067571/98262557120 = 1.4634879427713391…
And here’s the sequence as a list:
1, 1, 2, 1.5, 1.16666666…, 1.52380952…, 1.51339286…, 1.31701696…, 1.42005863…, 1.46348794…, 1.38749538…, 1.40402222…, 1.43296256…, 1.41009438…, 1.40702733…, 1.41989064…, 1.41499784…, 1.41099445…, 1.41543487…, 1.41521666…, 1.41310224…, 1.41426846…, 1.41474221…, 1.41392189…, 1.4140952…, 1.41441861…, 1.41417024…, 1.41413272…, 1.41427565…, 1.41422294…, 1.41417783…, 1.41422674…, 1.41422484…, 1.41420133…, 1.41421404…, 1.41421944…, 1.41421039…, 1.41421221…, 1.41421583…, 1.41421311…, 1.41421266…, 1.41421424…, 1.41421367…, 1.41421317…, 1.4142137…, 1.41421369…, 1.41421343…, 1.41421357…, 1.41421363…, 1.41421353…, 1.41421355…, 1.41421359…
The successive terms themselves (not the division of two successive terms) get closer and closer to another famous constant, 1.4142135623… = sqrt(2) = √2 = the square root of 2. That is, 2 = 1.4142135623…^2 = √2 * √2 = 2.
Now try this faux-Fibonacci function, which sums the reciprocals of three previous terms and is seeded with (1,1,1):
f(i) = 1/f(i-1) + 1/f(i-2) + 1/f(i-3)
1, 1, 1, 3, 2.33333333…, 1.76190476…, 1.32947233…, 1.74831712…, 1.8917243…, 1.85277499…, 1.64032782…, 1.67798344…, 1.74531862…, 1.77854896…, 1.73117082…, 1.71286113…, 1.72371834…, 1.74160342…, 1.73814321…, 1.7296513…, 1.72766133…, 1.73229495…, 1.73423726…, 1.73270842…, 1.73102242…, 1.73144679…, 1.7323761…, 1.73248681…, 1.73199851…, 1.73181453…, 1.73200171…, 1.73216337…, 1.73210842…, 1.73201045…, 1.73200754…, 1.73205948…, 1.73207579…, 1.73205401…, 1.73203852…, 1.73204551…, 1.7320556…, 1.73205507…, 1.73204955…, 1.73204821…, 1.73205067…, 1.73205214…, 1.73205128…, 1.73205025…, 1.73205039…, 1.73205097…, 1.73205108…, 1.7320508…, 1.73205066…
This time the terms are approximating √3 = 1.7320508075688772…
The pattern should be becoming clear. Here’s a faux-Fibonacci function summing the reciprocals of four previous terms and seeded with (1,1,1,1):
f(i) = 1/f(i-1) + 1/f(i-2) + 1/f(i-3) + 1/f(i-4) = f(i) = sum(j=1,4,1/f(i-j)) → 1, 1, 1, 1, 4, 3.25…, 2.55769231…, 1.94866975…, 1.46184034…, 1.89590957…, 2.11566858…, 2.19735496…, 2.13927698…, 1.9226554…, 1.91531809…, 1.96476075…, 2.01863597…, 2.04657233…, 2.0150802…, 1.98923188…, 1.98297066…, 1.99188053…, 2.00529681…, 2.00771794…, 2.00308927…, 1.99802424…, 1.99648053…, 1.99868265…, 2.00093427…, 2.00147194…, 2.0006098…, 1.99957598…, 1.99935245…, 1.99974785…, 2.00017861…, 2.00028636…, 2.00010876…, 1.99991963…, 1.99987668…, 1.99995216…, 2.0000357…, 2.00005396…, 2.00002038…, 1.99998445…, 1.99997638…, 1.99999121…, 2.0000069…, 2.00001027…, 2.00000381…, 1.99999695…, 1.99999552…, 1.99999836…, 2.00000134…, 2.00000196…
It’s approximating √4 = 2. Finally, a faux-Fibonacci function summing five previous terms and seeded with (1,1,1,1,1):
f(i) = sum(j=1,5,1/f(i-j)) → 1, 1, 1, 1, 1, 5, 4.2, 3.43809524…, 2.72895396…, 2.09539474…, 1.57263179…, 2.00850854…, 2.26829518…, 2.41829618…, 2.46536969…, 2.39375132…, 2.1756289…, 2.13738424…, 2.1643861…, 2.2128966…, 2.25917432…, 2.28405957…, 2.26223931…, 2.2364176…, 2.22153651…, 2.21977901…, 2.22763471…, 2.23872439…, 2.24336745…, 2.24198222…, 2.23787719…, 2.23423394…, 2.23290801…, 2.23407155…, 2.23592634…, 2.2371344…, 2.23728276…, 2.23667283…, 2.235919…, 2.23554916…, 2.23562462…, 2.23592649…, 2.23619761…, 2.23629263…, 2.2362179…, 2.23608413…, 2.23599221…, 2.23597907…, 2.23602277…, 2.23607674…, 2.23610497…, 2.2361008…, 2.23607908…, 2.23605908…, 2.23605182…
f(i) = sum(j=1,5,1/f(i-j)) → √5 = 2.2360679774997896964…
What’s going on? Why does a faux-Fibonacci function summing the reciprocals of n previous terms approximate √n? Simple. It’s because √n/n = 1/√n, so:
√2 = 1/√2 + 1/√2
√3 = 1/√3 + 1/√3 + 1/√3
√4 = 1/√4 + 1/√4 + 1/√4 + 1/√4
√5 = 1/√5 + 1/√5 + 1/√5 + 1/√5 + 1/√5
√6 = 1/√6 + 1/√6 + 1/√6 + 1/√6 + 1/√6 + 1/√6
√7 = 1/√7 + 1/√7 + 1/√7 + 1/√7 + 1/√7 + 1/√7 + 1/√7
Now try a variant of this variant on the standard Fibonacci function. Rather than summing the reciprocals of n previous terms, that is, adding all the reciprocals, you can try adding some, subtracting others and leaving others out. When I played with add-subtract-ignore, I had another surprise. Consider the faux-Fibonacci function using four terms where you add f(i-4), subtract f(i-3) and f(i-2), and neither add nor subtract f(i-1). If it’s seeded with (1,1,1,1), it behaves like this (note that subtracting a negative number is the same as adding its positive form):
f(i) = 1/f(i-4) – 1/f(i-3) – 1/f(i-2) = f(i) = -1/f(i-2) – 1/f(i-3) + 1/f(i-4)
1, 1, 1, 1, -1, -1, 1, 3, -1, -2.33333333…, 1.66666667…, 1.76190476…, -1.17142857…, -1.596139…, 0.886090969…, 2.04773796…, -1.35569898…, -2.24340789…, 1.37783544…, 1.67172102…, -1.01765253…, -1.76971243…, 1.11024382…, 2.14590316…, -1.31829317…, -1.93177087…, 1.19325541…, 1.7422206…, -1.07894042…, -1.92968341…, 1.19089866…, 2.01903508…, -1.24831753…, -1.85320783…, 1.14549415…, 1.83596921…, -1.13445903…, -1.95726202…, 1.20979165…, 1.93706664…, -1.19714822…, -1.85375091…, 1.14566257…, 1.89100976…, -1.16872901…, -1.94112216…, 1.19966969…, 1.89961427…, -1.17402719…, -1.87515137…, 1.15890915…, 1.91148197…, -1.18135916…, -1.91932498…, 1.18620874…, 1.89065342…, -1.16848806…, -1.89295612…, 1.16991105…, 1.91299871…, -1.18229835…, -1.90577961…, 1.17783653…, 1.89326935…, -1.1701048…, -1.90192076…, 1.17545169…, 1.90859542…, -1.17957683…, -1.90046656…, 1.17455293…, 1.89789374…, -1.17296284…, -1.90447424…, 1.17702981…, 1.90452112…, -1.17705878…, -1.89974183…, 1.17410502…, 1.90102893…, -1.17490049…, -1.9041308…, 1.17681755…, 1.90234084…, -1.1757113…, -1.90059155…, 1.17463018…, 1.90236908…, -1.17572875…, -1.90314411…, 1.17620774…, 1.90164296…, -1.17527998…, -1.9014973…, 1.17518996…, 1.90262351…, -1.17588599…, -1.90241767…, 1.17575878…, 1.90165956…, -1.17529024…, -1.902018…, 1.17551177…, 1.90246751…
The function cycles through approximations of four constants consisting of successive pairs that are identical except for their sign (positive and negative). When you square those constants, you get this (multiplying two negative numbers is the same as multiplying their positive forms):
+1.1755705045849462583374119093…^2 = 1.3819660112501051517954131656…
+1.9021130325903071442328786668…^2 = 3.6180339887498948482045868343…
-1.1755705045849462583374119093…^2 = 1.3819660112501051517954131656…
-1.9021130325903071442328786668…^2 = 3.6180339887498948482045868343…
I was surprised to see that φ had appeared:
3.6180339887498948482045868344… = 2 + φ = 1 + φ^2
1.3819660112501051517954131656… = 1 + (φ-1)^2 = 1 + (1/φ)^2
Elsewhere Other-Accessible…
• Friday is Φday — a first look at Phiday
Here’s an interesting formula:
fr(1) = 1/2; mx = 3
fr(i) = fr(i-1) + 1/fr(i-1)
if fr(i) > mx, fr(i) = fr(i) – mx
Early terms look like this:
0.5, 2.5, 2.9, 3.244827586…, 4.329334628…, 2.081590666…, 2.561992513…, 2.952313716…, 3.291031107…, 3.727089920…, 2.102435627…, 2.578074447…, 2.965960841…, 3.303119709…, 3.602146368…, 2.262872154…, 2.704788415…, 3.074503101…, 13.49676325…, 10.59203071…, 7.723747777…, 4.935444092…, 2.452121378…, 2.859931533…, 3.209590254…, 4.980804482…, 2.485649867…, 2.887959143…, 3.234224430…, 4.503633905…, 2.168689406…
Can you see any patterns emerging? I’d guess not. And I’d guess a thousand more terms wouldn’t help you see any better. It’s hard for humans to see patterns in a jumble of numbers. Our eyes don’t work as well on numbers as on shapes. That’s why you can make that formula focal to the flesh, as it were, by plotting the numbers on a graph. Or part of the numbers, anyway. Suppose you take the fractional parts of each pair of terms and use them to map (x,y) on a FractL (my name for a graph whose arms run from 0 to 1). For example, the terms 4.935444092… and 2.452121378… would yield x = 0.935444092… and y = 0.452121378… (or vice versa). The resultant graph makes the formula focal to the flesh. And it’s replete with patterns:
fr(i)+=1/fr(i-1); if fr(i)>3, fr(i)-=3; x = frac(fr(i)), y = frac(fr(i+1))
I can’t explain the patterns and they may arise from limited precision in the decimal digits. But I like them however they arise. The graph doesn’t change when mx = 4 (although it creates the lines in a different order):
if fr(i)>4, fr(i)-=4
But it does change when mx = 4/3. The lines almost vanish, except for a tiny comet-like mark towards the upper right-hand corner:
if fr(i)>4/3, fr(i)-=4/3
When mx = 7/2, the graph of mx = 3|4 is back in a slightly different form:
if fr(i)>7/2, fr(i)-=7/2
And again with 7/3:
if fr(i)>7/3, fr(i)-=7/3
There’s a big change with 7/4, Most of the lines disappear:
if fr(i)>7/4, fr(i)-=7/4
And only the main lines appear with 9/5:
if fr(i)>9/5, fr(i)-=9/5
And so on till you try fr -= 2/f, as noted below:
if fr(i)>11/5, fr(i)-=11/5
if fr(i)>11/6, fr(i)-=11/6
if fr(i)>15/8, fr(i)-=15/8
if fr(i)>29/15, fr(i)-=29/15
Now try fr += 2/fr and fr += 3/fr. This is what happens:
fr += 2/fr; if fr(i)>3, fr(i)-=3
fr += 2/fr; if fr(i)>8/3, fr(i)-=8/3
fr += 2/fr; if fr(i)>11/4, fr(i)-=11/4
fr += 3/fr; if fr(i)>6, fr(i)-=6
And what about these graphs?
They’re created by seeding a sum, s, with a fraction, then adding more fractions < 1 whose numerators = 1,2,3… and whose denominators are the prime numbers 1, s -= 1. When s > 1, s -= 1. Then you take the fractional parts of s(i) and s(i+1) and graph (x,y) as above.
Post-Performative Post-Scriptum
The title of this post refers to Morbid Angel’s Formulas Fatal to the Flesh (1998). I’ve never heard it, but I like Morbid Angel’s alphabetically alliterative album-titles.
An attractive and intriguing image by Kristina Armitage for an article on Zermelo-Fraenkel set theory at Quanta Magazine (click for larger image).
Post-Performative Post-Scriptum
The title of this post is a pun on Charles Baudelaire’s Les Fleurs du mal / Flowers of Evil (1857). If you’re already familiar with the book, you might have missed the “ZF” of ZFleurs.
Satanas
To Jorge Santayana
ECCE! Princeps infernorum,
Rex veneficus amorum
Vilium et mortiferorum,
Ecce! regnat Lucifer:
Animis qui dominatur,
Quibus coelum spoliatur;
Qui malignus bona fatur,
Cor corrumpens suaviter.
Fructus profert; inest cinis:
Profert flores plenos spinis:
Vitae eius mors est finis:
Crux est eius requies.
Qualis illic apparebit
Cruciatus, et manebit!
Quantas ista quot habebit
Mors amaritudines!
Iuventutis quam formosa
Floret inter rosas rosa!
Venit autem vitiosa
Species infamiae:
Veniunt crudeles visus,
Voces simulati risus;
Et inutilis fit nisus
Flebilis laetitae.
Quanto vitium splendescit,
Tanto anima nigrescit;
Tanto tandem cor marcescit,
Per peccata dulcia.
Gaudens mundi Princeps mali
Utitur veneno tali,
Voluptate Avernali;
0 mellita vitia!
Gaudet Princeps huius mundi
Videns animam confundi;
Cordis amat moribundi
Aspectare proelium.
Vana tentat, vana quaerens,
Cor anhelum, frustra moerens;
Angit animae inhaerens
Flamma cor miserrimum.
Gaudet Rector tenebrarum
Immolare cor amarum;
Satiare furiarum
Rex sorores avidas.
Vae! non stabit in aeternum
Regnum, ait Rex, infernum:
Sed, dum veniat Supernum,
Dabo vobis victimas.
• Lionel Johnson (1867-1902), text courtesy Laudator Temporis Acti
Today’s a Phiday Friday or Φiday Friday or Φriday, so let’s have some more Fibonacci Fun. Here is the famous Fibonacci sequence, where each number (after seeding with “0, 1”) is formed by adding the previous two numbers:
0, 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 the Online Encyclopedia of Integer Sequences (OEIS)
It’s obvious that the numbers get bigger for ever and that no number repeats except 1. But what happens to the final digit of the Fibonacci numbers, as underlined here?:
0, 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, …
If you think about it, you’ll realize that the final digit has to repeat. Look for the “0, 1, 1” re-appearing:
0, 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, 1, 6, 7, 3, 0, 3, 3, 6, 9, 5, 4, 9, 3, 2, 5, 7, 2, 9, 1, 0, 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, … — A003893 at the OEIS, “a(n) = Fibonacci(n) mod 10”
As you’ll see, all the numbers 0 to 9 appear in that sequence. But what about the final two digits of the Fibonacci sequence? Do all the numbers 0 to 99 appear before the sequence repeats?
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 44, 33, 77, 10, 87, 97, 84, 81, 65, 46, 11, 57, 68, 25, 93, 18, 11, 29, 40, 69, 9, 78, 87, 65, 52, 17, 69, 86, 55, 41, 96, 37, 33, 70, 3, 73, 76, 49, 25, 74, 99, 73, 72, 45, 17, 62, 79, 41, 20, 61, 81, 42, 23, 65, 88, 53, 41, 94, 35, 29, 64, … — A105471 at the OEIS, “a(n) = Fibonacci(n) mod 100”
And what about the the final three digits and the numbers 0 to 999?
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 597, 584, 181, 765, 946, 711, 657, 368, 25, 393, 418, 811, 229, 40, 269, 309, 578, 887, 465, 352, 817, 169, 986, 155, 141, 296, 437, 733, 170, 903, 73, 976, 49, 25, 74, 99, 173, 272 … — A248740 at the OEIS, “a(n) = Fibonacci(n) mod 1000”
In fact, some numbers do go missing as the final block of digits gets longer. But all that is based on the representation of the Fibonacci numbers in base 10. What about other bases? I had a look at that question and came up with some interesting patterns when I represented the final-block numbers on an Ulam-like spiral, where numbers are represented as squares on a spiral rotating counter-clockwise. This is the spiral for powers of 2 (the red square marks the center of the spiral and the number 1):
Spiral of final Fib-digits modulo 2^p
Fib-spiral mod 2^p (smaller scale)
Fib-spiral mod 2^p (smaller scale still)
What fraction of numbers are missing from the spiral? Watch this space. In the meantime, here’s the Fib-spiral for powers of 3:
Fib-spiral mod 3^p
It’s completely filled, because no numbers are missing (the red square marks “1” at the center of the spiral). What about powers of 4? Well, we’ve already seen that Fib-spiral, because all powers of 4 are also powers of 2:
Fib-spiral mod 4^p
The Fib-spiral for powers of 5 is the same as the Fib-spiral for powers of 3: it’s completely filled again. But the Fib-spiral for powers of 6 is interesting:
Fib-spiral mod 6^p
Fib-spiral mod 6^p (smaller scale)
Fib-spiral mod 6^p (smaller scale still)
And here are more Fib-spirals and more interesting patterns:
Fib-spiral mod 7^p
Fib-spiral mod 10^p — identical to the Fib-spiral for 2^p
Fib-spiral mod 11^p
Fib-spiral mod 11^p (smaller scale)
Fib-spiral mod 13^p
Fib-spiral mod 13^p (smaller scale)
Fib-spiral mod 17^p
Fib-spiral mod 17^p (smaller scale)
Fib-spiral mod 19^p
Fib-spiral mod 19^p (smaller scale)
Fib-spiral mod 41^p
Fib-spiral mod 41^p (smaller scale)
Fib-spiral mod 47^p
Elsewhere Other-Accessible…
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