Altars of Mathness

Tuesday, 7th Jul, 2026 by Krilling for Company in Album Art, Arithmetic, Base Behavior, Fibonacci Sequence, Fractals, Integer Sequences, Mathematics, Polygonal numbers, Powers, Prime numbers, Triangular Numbers and tagged Altars of Madness, altars of mathness, Dan Seagrave, final digits, Morbid Angel, palindromic patterns, patterns in trailing digits, recreational math, recreational mathematics, recreational maths, repetition of trailing digits, trailing digits | 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…

Strartifacts

Monday, 29th Jun, 2026 by Krilling for Company in Arithmetic, Base Behavior, Geometry, Integer Sequences, Mathematics and tagged Collatz sequence, digit sum, digit-mul, graph of falling numbers, integers, product of digits, recreational math, recreational mathematics, recreational maths, screen resolution, strartifacts, sum of digits | Leave a comment

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. 100_b7, 1000_b7, 10000_b7. 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

Fibonacci Friday Factors

Friday, 12th Jun, 2026 by Krilling for Company in Arithmetic, Base Behavior, Fibonacci Sequence, Integer Sequences, Mathematics, Ulam Spirals and tagged Fibonacci numbers, Friday, modulo, OEIS, Online Encyclopedia of Integer Sequences, Phiday, Phriday, recreational math, recreational mathematics, recreational maths | 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

The Power of Cow

Monday, 11th May, 2026 by Krilling for Company in Arithmetic, Integer Sequences, Mathematics and tagged Fibonacci-like sequences, Narayana's cow sequence, Narayana's cows sequence, OEIS, Online Encyclopedia of Integer Sequences | Leave a comment

1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201, 39865, 58425, 85626, 125491, 183916, 269542, 395033, 578949, 848491, 1243524, 1822473, 2670964, 3914488, 5736961, 8407925, …

• Narayana’s cows sequence: a(0) = a(1) = a(2) = 1; thereafter a(n) = a(n-1) + a(n-3). […] Number of digits in A061582. — A000930 at the Online Encyclopedia of Integer Sequences

1, 3, 9, 27, 621, 1863, 324189, 961232427, 2718369612621, 6213249182718361863, 1863961227324621324918324189, 32418927183662196121863961227324961232427, 961232427621324918186327183632418927183662196122718369612621, …

• a(1) = 1, a(n) = number obtained by replacing each digit of a(n-1) with three times its value. — A061582 at the OEIS

Moniliform Maths

Monday, 27th Apr, 2026 by Krilling for Company in Arithmetic, Integer Sequences, Mathematics, Powers and tagged OEIS, Online Encyclopedia of Integer Sequences, sums of fractions | Leave a comment

2 = 1/2 + 2/4 + 3/8 + 4/16 + 5/32…

sum(n^p / 2ⁿ)

2 = prime = sum(n / 2ⁿ)
6 = 2·3 = sum(n² / 2ⁿ)
26 = 2·13 = sum(n³ / 2ⁿ)
150 = 2·3·5² = sum(n⁴ / 2ⁿ)
1082 = 2·541 = sum(n⁵ / 2ⁿ)
9366 = 2·3·7·223 = sum(n⁶ / 2ⁿ)
94586 = 2·47293 = sum(n⁷ / 2ⁿ)
1091670 = 2·3·5·36389 = sum(n⁸ / 2ⁿ)
14174522 = 2·7087261 = sum(n⁹ / 2ⁿ)
204495126 = 2·3·11·41·75571 = sum(n¹⁰ / 2ⁿ)

• A000629 Number of necklaces of partitions of n+1 labeled beads.

1, 2, 6, 26, 150, 1082, 9366, 94586, 1091670, 14174522, 204495126, 3245265146, 56183135190, 1053716696762, 21282685940886, 460566381955706, 10631309363962710, 260741534058271802, 6771069326513690646, 185603174638656822266, 5355375592488768406230

• moniliform ← French moniliforme (1800 or earlier) ← classical Latin monīle necklace

sum(n^p / 3ⁿ)

3/4 = prime / (2²) = sum(n / 3ⁿ)
3/2 = prime / prime = sum(n² / 3ⁿ)
33/8 = (3·11) / (2³) = sum(n³ / 3ⁿ)
15 = 3·5 = sum(n⁴ / 3ⁿ)
273/4 = (3·7·13) / (2²) = sum(n⁵ / 3ⁿ)
1491/4 = (3·7·71) / (2²) = sum(n⁶ / 3ⁿ)
38001/16 = (3·53·239) / (2⁴) = sum(n⁷ / 3ⁿ)
17295 = 3·5·1153 = sum(n⁸ / 3ⁿ)
566733/4 = (3·188911) / (2²) = sum(n⁹ / 3ⁿ)
2579313/2 = (3·11·47·1663) / prime = sum(n¹⁰ / 3ⁿ)

sum(n^p / 4ⁿ)

4/9 = (2²) / (3²) = sum(n / 4ⁿ)
20/27 = (2²·5) / (3³) = sum(n² / 4ⁿ)
44/27 = (2²·11) / (3³) = sum(n³ / 4ⁿ)
380/81 = (2²·5·19) / (3⁴) = sum(n⁴ / 4ⁿ)
4108/243 = (2²·13·79) / (3⁵) = sum(n⁵ / 4ⁿ)
17780/243 = (2²·5·7·127) / (3⁵) = sum(n⁶ / 4ⁿ)
269348/729 = (2²·17²·233) / (3⁶) = sum(n⁷ / 4ⁿ)
4663060/2187 = (2²·5·107·2179) / (3⁷) = sum(n⁸ / 4ⁿ)
10091044/729 = (2²·2522761) / (3⁶) = sum(n⁹ / 4ⁿ)
218374420/2187 = (2²·5·11·23·103·419) / (3⁷) = sum(n¹⁰ / 4ⁿ)

sum(n^p / 5ⁿ)

5/16 = prime / (2⁴) = sum(n / 5ⁿ)
15/32 = (3·5) / (2⁵) = sum(n² / 5ⁿ)
115/128 = (5·23) / (2⁷) = sum(n³ / 5ⁿ)
285/128 = (3·5·19) / (2⁷) = sum(n⁴ / 5ⁿ)
3535/512 = (5·7·101) / (2⁹) = sum(n⁵ / 5ⁿ)
26355/1024 = (3·5·7·251) / (2¹⁰) = sum(n⁶ / 5ⁿ)
458555/4096 = (5·91711) / (2¹²) = sum(n⁷ / 5ⁿ)
1139685/2048 = (3·5·75979) / (2¹¹) = sum(n⁸ / 5ⁿ)
25492435/8192 = (5·17·443·677) / (2¹³) = sum(n⁹ / 5ⁿ)
316786305/16384 = (3·5·11·1919917) / (2¹⁴) = sum(n¹⁰ / 5ⁿ)

sum(n^p / 6ⁿ)

6/25 = (2·3) / (5²) = sum(n / 6ⁿ)
42/125 = (2·3·7) / (5³) = sum(n² / 6ⁿ)
366/625 = (2·3·61) / (5⁴) = sum(n³ / 6ⁿ)
4074/3125 = (2·3·7·97) / (5⁵) = sum(n⁴ / 6ⁿ)
11334/3125 = (2·3·1889) / (5⁵) = sum(n⁵ / 6ⁿ)
189714/15625 = (2·3·7·4517) / (5⁶) = sum(n⁶ / 6ⁿ)
3706518/78125 = (2·3·181·3413) / (5⁷) = sum(n⁷ / 6ⁿ)
82749954/390625 = (2·3·7·1970237) / (5⁸) = sum(n⁸ / 6ⁿ)
2078250726/1953125 = (2·3·31·1061·10531) / (5⁹) = sum(n⁹ / 6ⁿ)
11598884682/1953125 = (2·3·7·11·23²·47459) / (5⁹) = sum(n¹⁰ / 6ⁿ)

sum(n^p / 7ⁿ)

7/36 = prime / (2²·3²) = sum(n / 7ⁿ)
7/27 = prime / (3³) = sum(n² / 7ⁿ)
91/216 = (7·13) / (2³·3³) = sum(n³ / 7ⁿ)
70/81 = (2·5·7) / (3⁴) = sum(n⁴ / 7ⁿ)
2149/972 = (7·307) / (2²·3⁵) = sum(n⁵ / 7ⁿ)
3311/486 = (7·11·43) / (2·3⁵) = sum(n⁶ / 7ⁿ)
285929/11664 = (7·40847) / (2⁴·3⁶) = sum(n⁷ / 7ⁿ)
220430/2187 = (2·5·7·47·67) / (3⁷) = sum(n⁸ / 7ⁿ)
1359337/2916 = (7·17·11423) / (2²·3⁶) = sum(n⁹ / 7ⁿ)
5239157/2187 = (7·11·68041) / (3⁷) = sum(n¹⁰ / 7ⁿ)

sum(n^p / 8ⁿ)

8/49 = (2³) / (7²) = sum(n / 8ⁿ)
72/343 = (2³·3²) / (7³) = sum(n² / 8ⁿ)
776/2401 = (2³·97) / (7⁴) = sum(n³ / 8ⁿ)
10440/16807 = (2³·3²·5·29) / (7⁵) = sum(n⁴ / 8ⁿ)
174728/117649 = (2³·21841) / (7⁶) = sum(n⁵ / 8ⁿ)
3525192/823543 = (2³·3²·11·4451) / (7⁷) = sum(n⁶ / 8ⁿ)
11870648/823543 = (2³·41·36191) / (7⁷) = sum(n⁷ / 8ⁿ)
319735800/5764801 = (2³·3²·5²·19·9349) / (7⁸) = sum(n⁸ / 8ⁿ)
9686934584/40353607 = (2³·1210866823) / (7⁹) = sum(n⁹ / 8ⁿ)
326084753016/282475249 = (2³·3²·11·16273·25301) / (7¹⁰) = sum(n¹⁰ / 8ⁿ)

sum(n^p / 9ⁿ)

9/64 = (3²) / (2⁶) = sum(n / 9ⁿ)
45/256 = (3²·5) / (2⁸) = sum(n² / 9ⁿ)
531/2048 = (3²·59) / (2¹¹) = sum(n³ / 9ⁿ)
1935/4096 = (3²·5·43) / (2¹²) = sum(n⁴ / 9ⁿ)
34983/32768 = (3²·13²·23) / (2¹⁵) = sum(n⁵ / 9ⁿ)
381465/131072 = (3²·5·7²·173) / (2¹⁷) = sum(n⁶ / 9ⁿ)
9725787/1048576 = (3²·67·127²) / (2²⁰) = sum(n⁷ / 9ⁿ)
35420535/1048576 = (3²·5·787123) / (2²⁰) = sum(n⁸ / 9ⁿ)
1160703963/8388608 = (3²·47·409·6709) / (2²³) = sum(n⁹ / 9ⁿ)
21129845715/33554432 = (3²·5·11·2213·19289) / (2²⁵) = sum(n¹⁰ / 9ⁿ)

sum(n^p / 10ⁿ)

10/81 = (2·5) / (3⁴) = sum(n / 10ⁿ)
110/729 = (2·5·11) / (3⁶) = sum(n² / 10ⁿ)
470/2187 = (2·5·47) / (3⁷) = sum(n³ / 10ⁿ)
7370/19683 = (2·5·11·67) / (3⁹) = sum(n⁴ / 10ⁿ)
142870/177147 = (2·5·7·13·157) / (3¹¹) = sum(n⁵ / 10ⁿ)
1114190/531441 = (2·5·7·11·1447) / (3¹²) = sum(n⁶ / 10ⁿ)
30495890/4782969 = (2·5·3049589) / (3¹⁴) = sum(n⁷ / 10ⁿ)
953934190/43046721 = (2·5·11·569·15241) / (3¹⁶) = sum(n⁸ / 10ⁿ)
3728765410/43046721 = (2·5·372876541) / (3¹⁶) = sum(n⁹ / 10ⁿ)
145739620510/387420489 = (2·5·11·1324905641) / (3¹⁸) = sum(n¹⁰ / 10ⁿ)

Sequence Unfurls…

Tuesday, 31st Mar, 2026 by Krilling for Company in Arithmetic, Egyptian Fractions, Fibonacci Sequence, Integer Sequences, Mathematics and tagged Egyptian Fractions, Fibonacci Sequence, OEIS, Online Encyclopedia of Integer Sequences, three-term Egyptian fractions | 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 F_n, 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.”

Primal Pellicles

Sunday, 19th Oct, 2025 by Krilling for Company in Arithmetic, Base Behavior, Integer Sequences, Mathematics, Prime numbers and tagged Integer Sequences, OEIS, prime factorization, prime factors, recreational math, recreational mathematics, recreational maths, VFNs, visible factor numbers | Leave a comment

Numbers have thin skins. And they’re easily replaced. Take 71624133. Here it is permuting its pellicles:

71624133 in base 10 = 100010001001110010111000101 in base 2 = 11222202212211200 in b3 = 10101032113011 in b4 = 121313433013 in b5 = 11035053113 in b6 = 1526536500 in b7 = 421162705 in b8 = 158685750 in b9 = 374802A9 in b11 = 1BBA1199 in b12 = 11AB9B59 in b13 = 9726137 in b14 = 644BE73 in b15 = F3855B7 in b16

But if digits are the skin of 71624133, what are its bones? Well, you could say the skeleton of a number, something that doesn’t change from base to base, is its prime factorization:

71624133 = 3² × 7² × 162413

But the primes themselves are numbers, so they’re wearing pellicles too. And it turns out that, in base 10, the pellicles of the prime factors of 71624133 match the pellicle of 71624133 itself:

71624133 = 3².7².162413

Here’s a list of primal pellicles in base 10:

735 = 3.5.7²
3792 = 2⁴.3.79
1341275 = 5².13.4127
13115375 = 5³.7.13.1153
22940075 = 5².229.4007
29373375 = 3.5³.29.37.73
71624133 = 3².7².162413
311997175 = 5².7.17².31.199
319953792 = 2⁷.3.53.79.199
1019127375 = 3².5³.7.127.1019
1147983375 = 3.5³.7.11.83.479
1734009275 = 5².173.400927
5581625072 = 2⁴.5581.62507
7350032375 = 5³.7.23.73.5003
17370159615 = 3⁴.5.17.59.61.701
33061224492 = 2².3³.306122449
103375535837 = 7².37.103.553583
171167303912 = 2³.11.17².6730391
319383665913 = 3.13³.19.383.6659
533671737975 = 3⁴.5².17.53.367.797
2118067737975 = 3².5².7.79.211.80677
3111368374257 = 3.11².13².683.74257
3216177757191 = 3.7³.191.757.21617
3740437158475 = 5².37.4043715847
3977292332775 = 3.5².29².233.277.977
4417149692375 = 5³.7.23.4969.44171
7459655393232 = 2⁴.3².7².23.45965539
7699132721175 = 3.5².7².27211.76991
7973529228735 = 3.5.7.97².2287.3529
10771673522535 = 3⁴.5.67.71.107.52253

You can find them at the Online Encyclopedia of Integer Sequences under A121342, “Composite numbers that are a concatenation of their distinct prime divisors in some order.” But what about pairs of primal pellicles, that is, pairs of numbers where the prime factors of each form the pellicle of the other?

35 = 5.7 ↔ 75 = 3.5²
1275 = 3.5².17 ↔ 3175 = 5².127
131715 = 3².5.2927 ↔ 329275 = 5².13171
3199767 = 3.359.2971 ↔ 35932971 = 3.19.67.97²
14931092 = 2².11.61.5563 ↔ 116155632 = 2⁴.3.109.149²

And here are a few primal pellicles I’ve found in other bases:

Primal Pellicles in Base 2

1111011011110 = 10.11¹⁰.110110111 in b2 = 7902 = 2.3².439 in b10
1110001100110111 = 11¹⁰.10111.100011001 in b2 = 58167 = 3².23.281 in b10
1111011011011110 = 10.11¹⁰.110110110111 in b2 = 63198 = 2.3².3511 in b10
11101001100001101 = 11¹⁰.101.101001100001 in b2 = 119565 = 3².5.2657 in b10
1111011011011011110 = 10.11¹⁰.110110110110111 in b2 = 505566 = 2.3².28087 in b10
1111011111101111011 = 11¹⁰.1011.10111.11011111 in b2 = 507771 = 3².11.23.223 in b10

Primal Pellicles in Base 3

121022 = 2¹⁰.12.102 in b3 = 440 = 2³.5.11 in b10
212212 = 2².21.212 in b3 = 644 = 2².7.23 in b10
20110112 = 2¹⁰.201.1011 in b3 = 4712 = 2³.19.31 in b10
21110110 = 10.21².1101 in b3 = 5439 = 3.7².37 in b10
121111101 = 12².111.1101 in b3 = 12025 = 5².13.37 in b10
222112121 = 2².21.221121 in b3 = 19348 = 2².7.691 in b10
2202122021 = 2².2021.22021 in b3 = 54412 = 2².61.223 in b10
120212201221 = 2.12².21.201.1202 in b3 = 312550 = 2.5².7.19.47 in b10

Primal Pellicles in Base 7

2525 = 2.5².25 in b7 = 950 = 2.5².19 in b10
3210 = 2.3⁴.10 in b7 = 1134 = 2.3⁴.7 in b10
5252 = 2.5².52 in b7 = 1850 = 2.5².37 in b10
332616 = 3³.16.326 in b7 = 58617 = 3³.13.167 in b10
336045 = 3².5.3604 in b7 = 59715 = 3².5.1327 in b10
2251635 = 2².3.5.16.25² in b7 = 281580 = 2².3.5.13.19² in b10

Primal Pellicles in Base 11

253 = 2².3.5² in b11 = 300 = 2².3.5² in b10
732 = 2.3².7² in b11 = 882 = 2.3².7² in b10
2123 = 2³.3³.12 in b11 = 2808 = 2³.3³.13 in b10
3432 = 2⁵.3.43 in b11 = 4512 = 2⁵.3.47 in b10
3710 = 3².7².10 in b11 = 4851 = 3².7².11 in b10
72252 = 2³.7².225 in b11 = 105448 = 2³.7².269 in b10

Primal Pellicles in Base 15

275 = 2⁴.5.7 in b15 = 560 = 2⁴.5.7 in b10
2D5 = 2.5².D in b15 = 650 = 2.5².13 in b10
2CD5 = 2.5².CD in b15 = 9650 = 2.5².193 in b10
7BE3 = 3.7².BE in b15 = 26313 = 3.7².179 in b10
21285 = 2⁴.5².128 in b15 = 105200 = 2⁴.5².263 in b10

Lime Time

Tuesday, 7th Oct, 2025 by Krilling for Company in Arithmetic, Circles, Fractals, Geometry, Integer Sequences, Mathematics and tagged Calkin-Wilf tree, fractal geometry, gryke fractal, limestone fractal, pavement fractal, perspective, rational fractions, simplest fractions, Stern-Brocot sequence, stretching polygons into circles, turning polygons into circles | Leave a comment

What do you get if you list every successive pair of entries in this sequence?

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, … — A038568 at the Online Encyclopedia of Integer Sequence

You get the rational fractions ordered by denominator in their simplest form: 1/2, 1/3, 2/3, 1/4, 3/4, 1/5, 2/5, 3/5… There are no pairs like 2/4 and 5/35, because those can be simplified: 2/4 → 1/2; 15/35 → 3/7. You can get the same set of rational fractions by listing every successive pair in this sequence, the Stern-Brocot sequence:

1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4, 1, 5, 4, 7, 3, 8, 5, 7, 2, 7, 5, 8, 3, 7, 4, 5, 1, 6, 5, 9, 4, 11, 7, 10, 3, 11, 8, 13, 5, 12, 7, 9, 2, 9, 7, 12, 5, 13, 8, 11, 3, 10, 7, 11, 4, 9, 5, 6, 1, 7, 6, 11, 5, 14, 9, 13, 4, 15, 11, 18, 7, 17, 10, 13, 3, 14, 11, 19, 8, 21, 13, 18, 5, 17, 12, 19, … — A002487 at the OEIS

But the fractions don’t come ordered by denominator this time. In fact, they seem to come at random: 1/2, 1/3, 2/3, 1/4, 3/5, 2/5, 3/4, 1/5, 4/7, 3/8, 5/7, 2/7, 5/8… But they’re not random at all. There’s a complicated way of generating them and a simple way. An amazingly simple way, I think:

Moshe Newman proved that the fraction a(n+1)/a(n+2) can be generated from the previous fraction a(n)/a(n+1) = x by 1/(2*floor(x) + 1 – x). The successor function f(x) = 1/(floor(x) + 1 – frac(x)) can also be used. — A002487, “Stern-Brocot Sequence”, at the OEIS

In another form, the Stern-Brocot sequence is generated by what’s called the Calkin-Wilf Tree. Now suppose you use the Stern-Brocot sequence to supply the x co-ordinate of an L-graph whose arms run from 0 to 1. And you use the Calkin-Wilf Tree to supply the y co-ordinate of the L-tree. What do you get? As I described in “I Like Gryke”, you get this fractal:

Limestone fractal

I call it a limestone fractal or pavement fractal or gryke fractal, because it reminds me of the fissured patterns you see in the limestone pavements of the Yorkshire Dales:

Fissured limestone pavement, Yorkshire Dales (Wikipedia)

But what happens when you plot the (x,y) of the Stern-Brocot sequence and the Calkin-Wilf Tree on a circle instead? You get an interestingly distorted limestone fractal:

Limestone fractal on circle

You can also plot the (x,y) around the perimeter of a polygon, then stretch the polygon into a circle. Here’s a square:

Limestone fractal on square

⇓

Limestone square stretched to circle

And here are a pentagon, hexagon, heptagon and octagon — note the interesting perspective effects:

Limestone fractal on pentagon

⇓

Limestone pentagon stretched to circle

Limestone fractal on hexagon

⇓

Limestone hexagon stretched to circle

Limestone fractal on heptagon

⇓

Limestone heptagon stretched to circle

Limestone fractal on octagon

⇓

Limestone octagon stretched to circle

And finally, here are animations of limestone polygons stretching to circles:

Limestone square stretched to circle (animated at EZgif)

Limestone pentagon to circle (animated)

Limestone hexagon to circle (animated)

Limestone heptagon to circle (animated)

Limestone octagon to circle (animated)

Previously Pre-Posted (Please Peruse)

• I Like Gryke — a first look at the limestone fractal

Summult-Time Hues

Saturday, 27th Sep, 2025 by Krilling for Company in Arithmetic, Integer Sequences, Mathematics and tagged OEIS, recreational math, recreational mathematics, recreational maths, sums of consecutive integers | Leave a comment

sum(3,6) = 3 * 6 = 18
• 3 * 2.3 = 2.3^2
sum(15,35) = 15 * 35 = 525
• 3.5 * 5.7 = 3.5^2.7
sum(85,204) = 85 * 204 = 17340
• 5.17 * 2^2.3.17 = 2^2.3.5.17^2
sum(493,1189) = 493 * 1189 = 586177
• 17.29 * 29.41 = 17.29^2.41
sum(2871,6930) = 2871 * 6930 = 19896030
• 3^2.11.29 * 2.3^2.5.7.11 = 2.3^4.5.7.11^2.29
sum(16731,40391) = 16731 * 40391 = 675781821
• 3^2.11.13^2 * 13^2.239 = 3^2.11.13^4.239
[…]

Elsewhere Other-Accessible

1, 18, 525, 17340, 586177, 19896030, 675781821, 22956120408, 779829016225, 26491211221770, 899921240562957, 30570830315362260, 1038508305678375841, 35278711540581704598, 1198437683944896688125, 40711602541832856049200, 1382996048733983114022337 — A011906 at the Online Encyclopedia of Integer Sequences

Partitional Pulchritude

Wednesday, 17th Sep, 2025 by Krilling for Company in Arithmetic, Geometry, Integer Sequences, Mathematics, Partitions and tagged animated gifs, count of partition lengths, partition lengths, recreational math, recreational mathematics, recreational maths | Leave a comment

If you want a good example of how, in math, something very simple can quickly get very deep, just look at partitions. Here are the partitions of 1 to 5, that is, the ways 1 to 5 can be expressed as a sum of integers smaller than or equal to themselves:

1 = 1

numbpart(1) = 1

2 = 2
1 + 1 = 2

numbpart(2) = 2

3 = 3
1 + 2 = 3
1 + 1 + 1 = 3

numbpart(3) = 3

4 = 4
1 + 3 = 4
2 + 2 = 4
1 + 1 + 2 = 4
1 + 1 + 1 + 1 = 4

numbpart(4) = 5

5 = 5
1 + 4 = 5
2 + 3 = 5
1 + 1 + 3 = 5
1 + 2 + 2 = 5
1 + 1 + 1 + 2 = 5
1 + 1 + 1 + 1 + 1 = 5

numbpart(5) = 7

It’s very easy to understand the concept of partitions, but very difficult to understand how partitions behave. For example, here is numbpart(n), the count of partitions for 1, 2, 3,…

1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, 6842, 8349, 10143, 12310, 14883, 17977, 21637, 26015, 31185, 37338, 44583, 53174, 63261, 75175, 89134, 105558, 124754, 147273, 173525, 204226, … A000041 at the Online Encyclopedia of Integer Sequences, “a(n) is the number of partitions of n (the partition numbers)”

What’s the formula for numbpart(n)? That’s a tricky question. And what’s the formula for the curves produced by counting the various lengths of partitions(n)? That’s another tricky question, but one thing is easy to see. As n gets bigger, the graph of countlen(partitions(n)) acquires a strange, lopsided beauty. Here are the partitions of 8, with the count of how many partitions of a particular length there are:

8 = 8 (1 partition of length 1)
1 + 7 = 8
2 + 6 = 8
3 + 5 = 8
4 + 4 = 8 (4 partitions of length 2)
1 + 1 + 6 = 8
1 + 2 + 5 = 8
1 + 3 + 4 = 8
2 + 2 + 4 = 8
2 + 3 + 3 = 8 (5 of length 3)
1 + 1 + 1 + 5 = 8
1 + 1 + 2 + 4 = 8
1 + 1 + 3 + 3 = 8
1 + 2 + 2 + 3 = 8
2 + 2 + 2 + 2 = 8 (5 of length 4)
1 + 1 + 1 + 1 + 4 = 8
1 + 1 + 1 + 2 + 3 = 8
1 + 1 + 2 + 2 + 2 = 8 (3 of length 5)
1 + 1 + 1 + 1 + 1 + 3 = 8
1 + 1 + 1 + 1 + 2 + 2 = 8 (2 of length 6)
1 + 1 + 1 + 1 + 1 + 1 + 2 = 8 (1 of length 7)
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 8 (1 of length 8)

When counts like that are shown as a graph, the graphs look like this (maximum counts are normalized to the same height):

graph of countlen(partitions(2))

countlen(partitions(3))

countlen(partitions(4))

countlen(partitions(5))

countlen(partitions(6))

countlen(partitions(7))

countlen(partitions(8))

countlen(partitions(9))

countlen(partitions(10))

countlen(partitions(15))

countlen(partitions(20))

countlen(partitions(30))

countlen(partitions(40))

countlen(partitions(50))

countlen(partitions(60))

countlen(partitions(70))

countlen(partitions(80))

countlen(partitions(90))

countlen(partitions(100))

Animated gif of partlen graphs (courtesy EZgif)

The graphs have a long, low right tail because the counts rise to great heights very quick, then fall away again, as you can see with partitions(100):

1 = count(partitions(10),len=1)
50 = count(partitions(10),len=2)
833 = count(partitions(10),len=3)
7153 = count(partitions(10),len=4)
38225 = count(partitions(10),len=5)
143247 = count(partitions(10),len=6)

[…]

10643083 = count(partitions(10),len=16)
11022546 = count(partitions(10),len=17)
11087828 = count(partitions(10),len=18)
10885999 = count(partitions(10),len=19)
10474462 = count(partitions(10),len=20)

[…]

30 = count(partitions(10),len=91)
22 = count(partitions(10),len=92)
15 = count(partitions(10),len=93)
11 = count(partitions(10),len=94)
7 = count(partitions(10),len=95)
5 = count(partitions(10),len=96)
3 = count(partitions(10),len=97)
2 = count(partitions(10),len=98)
1 = count(partitions(10),len=99)
1 = count(partitions(10),len=100)

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Category Archives: Integer Sequences