Category Archives: Fibonacci Sequence

Altars of Mathness

by in Album Art, Arithmetic, Base Behavior, Fibonacci Sequence, Fractals, Integer Sequences, Mathematics, Polygonal numbers, Powers, Prime numbers, Triangular Numbers and tagged , , , , , , , , , , , | Leave a comment

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

Elsewhere Other-Accessible…

• Formulas Focal to the Flesh — a pre-previous post paronomasizing the title of a Morbid-Angel album…

Faux-Fib Funcs for Phiday

by in Arithmetic, √2, Fibonacci Sequence, Fractions, Irrational numbers, φ, Mathematics, Phi, Reciprocals, Square root of 2, Square Roots and tagged , , , , , , , , , | Leave a comment

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

Fibonacci Friday Factors

by in Arithmetic, Base Behavior, Fibonacci Sequence, Integer Sequences, Mathematics, Ulam Spirals and tagged , , , , , , , , , | Leave a comment

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…

• Friday is Φday — a first look at Phiday on Friday

Sequence Unfurls…

by in Arithmetic, Egyptian Fractions, Fibonacci Sequence, Integer Sequences, Mathematics and tagged , , , , | Leave a comment

The Fibonacci sequence is beautiful like clockwork. There’s a perfectly clear, rigorously defined mechanism ticking out an entirely predictable result for ever:

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)

And there’s a formula to calculate any term in the sequence without calculating all the terms that precede it:

Binet’s formula for Fn, the n-th Fibonacci number

But I also like sequences that you might call definitely arbitrary. That is, there’s a perfectly clear, rigorously defined mechanism, but the results seem arbitrary — not predictable at all:

6, 15, 5, 22, 6, 3, 30, 9, 7, 2, 45, 15, 6, 5, 1, 36, 14, 6, 5, 3, 1, 62, 22, 16, 6, 5, 3, 2, 69, 21, 15, 4, 9, 5, 2, 1, 84, 30, 15, 9, 6, 7, 2, 2, 1, 56, 22, 13, 7, 3, 5, 2, 0, 0, 0, 142, 45, 22, 15, 12, 6, 9, 5, 3, 1, 2, 53, 17, 8, 4, 5, 1, 6, 3, 1, 1, 1, 0, 124, 36, 27, 14, 18, 6, 6, 5, 2, 3, 1, 1, 0, … A349083 at OEIS

What’s the formula there? That sequence is defined at the OEIS as “The number of three-term Egyptian fractions of rational numbers x/y, 0 < x/y < 1, ordered as below. The sequence is the number of (p,q,r) such that x/y = 1/p + 1/q + 1/r where p, q, and r are integers with p < q < r.” For example: “The sixth rational number is 3/4 [and] 3/4 = 1/2 + 1/5 + 1/20 = 1/2 + 1/6 + 1/12 = 1/3 + 1/4 + 1/5, so a(6)=3.”

Phascinating Phibonacci Phact Phor Phiday

by in Arithmetic, Fibonacci Sequence, Integer Sequences, φ, Mathematics, Phi and tagged , , , , , , , , | Leave a comment

Phiday falls on the 11th, 12th and 23rd of each month, because 11, 12 and 23 represent entries 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, 165580141, 267914296, 433494437, 701408733, 1134903170, 1836311903, 2971215073, 4807526976, 7778742049, 12586269025, 20365011074, 32951280099, 53316291173, 86267571272, 139583862445, 225851433717, 365435296162, 591286729879, 956722026041, 1548008755920, 2504730781961, 4052739537881, 6557470319842, 10610209857723, 17167680177565, 27777890035288, 44945570212853, 72723460248141, 117669030460994, 190392490709135, 308061521170129, 498454011879264, 806515533049393, 1304969544928657, …

Successive entries in the Fibonacci sequence provide better and better approximations to the golden ratio or φ = 1.61803398874989484820458683…

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 23rd June, so here’s a Fascinating Fibonacci Fact for Phiday. First, list the rational fractions < 1 in simplified form and mark the Fibonacci fractions:

1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5, 4/5, 1/6, 5/6, 1/7, 2/7, 3/7, 4/7, 5/7, 6/7, 1/8, 3/8, 5/8, 7/8, 1/9, 2/9, 4/9, 5/9, 7/9, 8/9, 1/10, 3/10, 7/10, 9/10, 1/11, 2/11, 3/11, 4/11, 5/11, 6/11, 7/11, 8/11, 9/11, 10/11, 1/12, 5/12, 7/12, 11/12, 1/13, 2/13, 3/13, 4/13, 5/13, 6/13, 7/13, 8/13, 9/13, 10/13, 11/13, 12/13, 1/14, 3/14, 5/14, 9/14, 11/14, 13/14, 1/15, 2/15, 4/15, 7/15, 8/15, 11/15, 13/15, 14/15, 1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16, 1/17, 2/17, 3/17, 4/17, 5/17, 6/17, 7/17, 8/17, 9/17, 10/17, 11/17, 12/17, 13/17, 14/17, 15/17, 16/17, 1/18, 5/18, 7/18, 11/18, 13/18, 17/18, 1/19, 2/19, 3/19, 4/19, 5/19, 6/19, 7/19, 8/19, 9/19, 10/19, 11/19, 12/19, 13/19, 14/19, 15/19, 16/19, 17/19, 18/19, 1/20, 3/20, 7/20, 9/20, 11/20, 13/20, 17/20, 19/20, 1/21, 2/21, 4/21, 5/21, 8/21, 10/21, 11/21, 13/21, 16/21, 17/21, 19/21, 20/21, 1/22, 3/22, 5/22, 7/22, 9/22, 13/22, 15/22, 17/22, 19/22, 21/22, 1/23, 2/23, 3/23, 4/23, 5/23, 6/23, 7/23, 8/23, 9/23, 10/23, 11/23, 12/23, 13/23, 14/23, 15/23, 16/23, 17/23, 18/23, 19/23, 20/23, 21/23, 22/23…

Next, record the positions in the fraction list of the FibFracs, i.e. pos(fibonacci(i)/fibonacci(i+1)) = pos(fibfrac(i)):

1, 3, 8, 20, 53, 135, 353, 924, 2422, 6311, 16529, 43229, 113066, 296173, 775286, 2029661, 5313844, 13911391, 36419909, 95348490, 249624578, 653521015, 1710943906, 4479312193, 11726939926, 30701521655, 80377560978, 210431191133, 550915866198, 1442316294349, 3776032465954, 9885782372588, 25881314454327, 67758160822605, 177393168080718, 464421339906882, 1215870841639593, …

What do you get when you divide pos(fibfrac(i+1)) by pos(fibfrac(i))?

pos(1/2) = 1
pos(2/3) = 3 (3/1 = 3)
pos(3/5) = 8 (8/3 = 2.6…)
pos(5/8) = 20 (20/8 = 2.5)
pos(8/13) = 53 (53/20 = 2.65)
pos(13/21) = 135 (2.5471698113207…)
pos(21/34) = 353 (2.6148…)
pos(34/55) = 924 (2.617563739376770538243626062…)
pos(55/89) = 2422 (2.621…)
pos(89/144) = 6311 (2.605697770437654830718414533…)
pos(144/233) = 16529 (2.619077800665504674378070037…)
pos(233/377) = 43229 (2.615342730957710690301893642…)
pos(377/610) = 113066 (2.615512734506928219482291980…)
pos(610/987) = 296173 (2.619470044045071020465922559…)
pos(987/1597) = 775286 (2.617679531895209894217231145…)
pos(1597/2584) = 2029661 (2.617951310871084993150914630…)
pos(2584/4181) = 5313844 (2.618094351716863062353762525…)
pos(4181/6765) = 13911391 (2.617952465296309037299551888…)
pos(6765/10946) = 36419909 (2.617991903182075753603647543…)
pos(10946/17711) = 95348490 (2.618032076906068051954770123…)
pos(17711/28657) = 249624578 (2.618023400265699016313735016…)
pos(28657/46368) = 653521015 (2.618015502463863954934758067…)
pos(46368/75025) = 1710943906 (2.618039614227248683043497844…)
pos(75025/121393) = 4479312193 (2.618035680358535377956453004…)
pos(121393/196418) = 11726939926 (2.618022459860278821159630657…)
pos(196418/317811) = 30701521655 (2.618033506501651708043379296…)
pos(317811/514229) = 80377560978 (2.618031831816708695313688353…)
pos(514229/832040) = 210431191133 (2.618034045479393794998913484…)
pos(832040/1346269) = 550915866198 (2.618033302153394031845776103…)
pos(1346269/2178309) = 1442316294349 (2.618033683260502304564996035…)
pos(2178309/3524578) = 3776032465954 (2.618033562227999267671331082…)
pos(3524578/5702887) = 9885782372588 (2.618034262608066669117450079…)
pos(5702887/9227465) = 25881314454327 (2.618034008728793003503058474…)
pos(9227465/14930352) = 67758160822605 (2.618033985181798482654668954…)
pos(14930352/24157817) = 177393168080718 (2.618033989221521810752093192…)
pos(24157817/39088169) = 464421339906882 (2.618033969017113072183685603…)
pos(39088169/63245986) = 1215870841639593 (2.618033964338027806153843993…)
[…]

In other words, pos(fibfrac(i+1)) / pos(fibfrac(i)) → φ^2 = 2.61803398874989484820458683… = φ + 1

Previously Pre-Posted (Please Peruse)

• Friday is Φiday

The Bellissima Curve

by in Arithmetic, Fibonacci Sequence, Geometry, Mathematics and tagged , , , , , , | 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

by in Arithmetic, Fibonacci Sequence, Integer Sequences, φ, Mathematics, Phi and tagged , , , , , , , | Leave a comment

The 11th, 12th and 23rd day of a month can be called a Phiday or 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 Phiday but a Friday or Phriday (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

The Call of CFulhu

by in Arithmetic, √2, Continued Fractions, Fibonacci Sequence, H.P. Lovecraft, Integer Sequences, φ, Mathematics, Phi, Square root of 2 and tagged , , , , , | 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.

Fract-L Geometry

by in Arithmetic, Fibonacci Sequence, Geometry, Mathematics, Triangular Numbers and tagged , , , , , , , | Leave a comment

Suppose you set up an L, i.e. a vertical and horizontal line, representing the x,y coordinates between 0 and 1. Next, find the fractional pairs x = 1/2, 1/3, 2/3, 1/4, 2/4…, y = 1/2, 1/3, 2/3, 1/4, 2/4… and mark the point (x,y). That is, find the point, say, 1/5 of the way along the x-line, then the points 1/5, 2/5, 3/5 and 4/5 along the y-line, marking the points (1/5, 1/5), (1/5, 2/5), (1/5, 3/5), (1/5, 4/5). Then find (2/5, 1/5), (2/5, 2/5), (2/5, 3/5), (2/5, 4/5) and so on. Some interesting patterns appear in what I call a Frac-L (pronounced “frackle”) or Fract-L:

Frac-L for 1/2 to 21/22

Frac-L for 1/2 to 48/49

Frac-L for 1/2 to 75/76

Frac-L for 1/2 to 102/103

Frac-L for 1/2 to 102/103 (animated)

If the (x,y) point is first red, then becomes different colors as it is repeatedly found, you get these patterns:

Frac-L for 1/2 to 48/49 (color)

Frac-L for 1/2 to 75/79 (color)

Frac-L for 1/2 to 102/103 (color) (animated)

Now try polygonal numbers. The triangular numbers are 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78…, so you’re finding the fractional pairs, say, (1/21, 1/21), (1/21, 3/21, (1/21, 6/21), (1/21, 10/21), (1/21, 15/21), then (3/21, 1/21), (3/21, 3/21, (3/21, 6/21), (3/21, 10/21), (3/21, 15/21), and so on:

Frac-L for triangular fractions

The frac-L for square numbers (1, 4, 9, 16, 25, 36, 49, 64, 81, 100…) is almost identical:

Frac-L for square fractions, e.g. (1/16, 1/16), (1/16, 4/16), (1/16, 9/16)…

So is the frac-L for pentagonal numbers (1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, 247, 287, 330…):

Frac-L for pentagonal fractions, e.g. (1/35, 5/35), (1/35, 12/35), (1/35,22/35)…

Here are frac-Ls for tetrahedral and square-pyramidal numbers:

Frac-L for tetrahedral fractions

Frac-L for square pyramidal fractions

But what about prime numbers (skipping 2)? Here the fractional pairs are, say, (1/17, 1/17), (1/17, 3/17), (1/17, 5/17), (1/17, 7/17), (1/17, 11/17), (1/17, 13/17), then (3/17, 1/17), (3/17, 3/17), (3/17, 5/17), (3/17, 7/17), (3/17, 11/17), (3/17, 13/17), and so on:

Frac-L for 1/3 to 73/79 (prime fractions)

Frac-L for 1/3 to 223/227

Frac-L for 1/3 to 307/331

Frac-L for 1/3 to 307/331 (animated)

Frac-L for 1/3 to 73/79 (color) (prime fractions)

Frac-L for 1/3 to 223/227 (color)

Frac-L for 1/3 to 307/331 (color)

Frac-L for 1/3 to 307/331 (color) (animated)

And finally (for now), a frac-L for Fibonnaci numbers, where the fractional pairs are, say, (1/13, /13), (1/13, 2/13), (1/13, 3/13), (1/13, 5/13), (1/13, 8/13), then (2/13, /13), (2/13, 2/13), (2/13, 3/13), (2/13, 5/13), (2/13, 8/13), and so on:

Frac-L for Fibonacci fractions to 14930352/2178309 = fibonacci(36)/fibonacci(37)