
Koch snowflake infinitely magnified (from Wikipedia)

Koch snowflake infinitely magnified (from Wikipedia)

A spinning Sierpiński tetrahedron or tetrix
15,527,402,881 = 3534 = 304 + 1204 + 2724 + 3154 — from David Wells’ Penguin Dictionary of Curious and Interesting Numbers (1986), entry for “15,527,402,881”
In “First Whirled Warp”, I looked at the paths traced by the midpoint of two points moving at varying speeds around the perimeter of a circle or polygon. Now I wanted to look at the midpoint of two points moving on the perimeter of a star. Suppose the star looks like this:

Four-pointed star
If the two points start at the same vertex and one point is moving 1/2 as fast as the other, the midpoint traces a shape like the head of a fox:

Fox-head from midpoint of two points moving in speed-ratio 1/2 : 1 (or 1 : 2)
If one point is moving 1/3 as fast (or 3x faster), the trace looks like this:
Midpoint of two points moving in speed-ratio 1/3 : 1
And if the points are moving -1/3 : 1, that is, in opposite directions (one clockwise, one widdershins):

Speed-ratio -1/3 : 1
And you can adjust all pixels outward so that the outer vertices of the star lie on the perimeter of a circle:

Speed-ratio -1/3 : 1 (circular)
Here are more traces created by the midpoint of two points moving around the perimeter of a four-pointed star:

Speed-ratio 1/5 : 1
Speed-ratio 3/5 : 1
↓
Speed-ratio 3/5 : 1 (circular)
Speed-ratio -3/7 : 1/3
↓
Speed-ratio -3/7 : 1/3 (circular)
Speed-ratio 7/3 : 6/7
↓
Speed-ratio 7/3 : 6/7 (circular)
Speed-ratio -7/3 : 6/7
↓
Speed-ratio -7/3 : 6/7 (circular)
If the star is adjusted like this:

Variant on four-pointed star
You can get mid-traces like this:
Speed-ratio -1/7 : 1 (adjusted star)
↓
Speed-ratio -1/7 : 1 (adjusted star) (circular)
Speed-ratio -4/5 : 1 (3p star)
↓
Speed-ratio -4/5 : 1 (3p star) (circular)
And some five-pointed stars:
Speed-ratio 2/7 : 1 (5p star)
↓
Speed-ratio 2/7 : 1 (5p star) (circular)
Speed-ratio -7/5 : 3/7 (5p star)
↓
Speed-ratio -7/5 : 3/7 (5p star) (circular)
Previously Pre-Posted
• First Whirled Warp — an earlier look at points performativizing on perimeters

An ox-eye daisy (Leucanthemum sp.) with a harlequin ladybird, Harmonia axyridis, sitting at its center
Peri-Performative Post-Scriptum…
The title of this incendiary intervention refers to
1) The Fibonacci sequence present in the beautiful interlocking curves at the heart of the
2) daisy, whose name comes from Anglo-Saxon dæges ēage, meaning “day’s eye”.
3) The eye-like appearance of the daisy, with the ladybird like a slightly off-centered pupil
These are the odd numbers:
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59...
If you add the odd numbers, 1+3+5+7…, you get the square numbers:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900...
And if you add the square numbers, 1+4+9+16…, you get what are called the square pyramidal numbers:
1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455...
There’s not a circle in sight, so you wouldn’t expect to find π amid the pyramids. But it’s there all the same. You can get π from this formula using the square pyramidal numbers:
π from a formula using square pyramidal numbers (Wikipedia)
Here are the approximations getting nearer and near to π:
3.1415926535897932384... = π
3.1666666666666666666... = sqpyra2pi(i=1) / 6 + 3
1 = sqpyra(1)3.1415926535897932384... = π
3.1452380952380952380... = sqpyra2pi(i=3) / 6 + 3
14 = sqpyra(3)3.1415926535897932384... = π
3.1412548236077647842... = sqpyra2pi(i=8) / 6 + 3
204 = sqpyra(8)3.1415926535897932384... = π
3.1415189855952756236... = sqpyra2pi(i=14) / 6 + 3
1,015 = sqpyra(14)3.1415926535897932384... = π
3.1415990074057163751... = sqpyra2pi(i=33) / 6 + 3
12,529 = sqpyra(33)3.1415926535897932384... = π
3.1415920110950124679... = sqpyra2pi(i=72) / 6 + 3
127,020 = sqpyra(72)3.1415926535897932384... = π
3.1415926017980070553... = sqpyra2pi(i=168) / 6 + 3
1,594,684 = sqpyra(168)3.1415926535897932384... = π
3.1415926599504002195... = sqpyra2pi(i=339) / 6 + 3
13,043,590 = sqpyra(339)3.1415926535897932384... = π
3.1415926530042565359... = sqpyra2pi(i=752) / 6 + 3
142,035,880 = sqpyra(752)3.1415926535897932384... = π
3.1415926535000384883... = sqpyra2pi(i=1406) / 6 + 3
927,465,791 = sqpyra(1406)3.1415926535897932384... = π
3.1415926535800054618... = sqpyra2pi(i=2944) / 6 + 3
8,509,683,520 = sqpyra(2944)3.1415926535897932384... = π
3.1415926535890006043... = sqpyra2pi(i=6806) / 6 + 3
105,111,513,491 = sqpyra(6806)3.1415926535897932384... = π
3.1415926535897000092... = sqpyra2pi(i=13892) / 6 + 3
893,758,038,910 = sqpyra(13892)3.1415926535897932384... = π
3.1415926535897999990... = sqpyra2pi(i=33315) / 6 + 3
12,325,874,793,790 = sqpyra(33315)3.1415926535897932384... = π
3.1415926535897939999... = sqpyra2pi(i=68985) / 6 + 3
109,433,980,000,485 = sqpyra(68985)3.1415926535897932384... = π
3.1415926535897932999... = sqpyra2pi(i=159563) / 6 + 3
1,354,189,390,757,594 = sqpyra(159563)3.1415926535897932384... = π
3.1415926535897932300... = sqpyra2pi(i=309132) / 6 + 3
9,847,199,658,130,890 = sqpyra(309132)3.1415926535897932384... = π
3.1415926535897932389... = sqpyra2pi(i=774865) / 6 + 3
155,080,688,289,901,465 = sqpyra(774865)3.1415926535897932384... = π
3.1415926535897932384... = sqpyra2pi(i=1586190) / 6 + 3
1,330,285,259,163,175,415 = sqpyra(1586190)
10 can be represented in exactly 10 ways as a sum of distinct integers:
10 = 1 + 2 + 3 + 4
10 = 2 + 3 + 5
10 = 1 + 4 + 5
10 = 1 + 3 + 6
10 = 4 + 6 (c=5)
10 = 1 + 2 + 7
10 = 3 + 7
10 = 2 + 8
10 = 1 + 9
10 = 10 (c=10)
But there’s something unsatisfying about including 10 as a sum of itself. It’s much more satisfying that 76 can be represented in exactly 76 ways as a sum of distinct primes:
76 = 2 + 3 + 7 + 11 + 13 + 17 + 23
76 = 5 + 7 + 11 + 13 + 17 + 23
76 = 2 + 3 + 5 + 11 + 13 + 19 + 23
76 = 3 + 7 + 11 + 13 + 19 + 23
76 = 2 + 3 + 5 + 7 + 17 + 19 + 23 (c=5)
76 = 2 + 3 + 5 + 7 + 13 + 17 + 29
76 = 2 + 3 + 5 + 7 + 11 + 19 + 29
76 = 3 + 5 + 7 + 13 + 19 + 29
76 = 11 + 17 + 19 + 29
76 = 11 + 13 + 23 + 29 (c=10)
76 = 2 + 5 + 17 + 23 + 29
76 = 7 + 17 + 23 + 29
76 = 2 + 3 + 19 + 23 + 29
76 = 5 + 19 + 23 + 29
76 = 2 + 3 + 5 + 7 + 11 + 17 + 31 (c=15)
76 = 3 + 5 + 7 + 13 + 17 + 31
76 = 3 + 5 + 7 + 11 + 19 + 31
76 = 2 + 11 + 13 + 19 + 31
76 = 2 + 7 + 17 + 19 + 31
76 = 2 + 7 + 13 + 23 + 31 (c=20)
76 = 2 + 3 + 17 + 23 + 31
76 = 5 + 17 + 23 + 31
76 = 3 + 19 + 23 + 31
76 = 2 + 3 + 11 + 29 + 31
76 = 5 + 11 + 29 + 31 (c=25)
76 = 3 + 13 + 29 + 31
76 = 3 + 5 + 7 + 11 + 13 + 37
76 = 2 + 7 + 13 + 17 + 37
76 = 2 + 7 + 11 + 19 + 37
76 = 2 + 5 + 13 + 19 + 37 (c=30)
76 = 7 + 13 + 19 + 37
76 = 3 + 17 + 19 + 37
76 = 2 + 3 + 11 + 23 + 37
76 = 5 + 11 + 23 + 37
76 = 3 + 13 + 23 + 37 (c=35)
76 = 2 + 3 + 5 + 29 + 37
76 = 3 + 7 + 29 + 37
76 = 3 + 5 + 31 + 37
76 = 2 + 5 + 11 + 17 + 41
76 = 7 + 11 + 17 + 41 (c=40)
76 = 2 + 3 + 13 + 17 + 41
76 = 5 + 13 + 17 + 41
76 = 2 + 3 + 11 + 19 + 41
76 = 5 + 11 + 19 + 41
76 = 3 + 13 + 19 + 41 (c=45)
76 = 2 + 3 + 7 + 23 + 41
76 = 5 + 7 + 23 + 41
76 = 2 + 7 + 11 + 13 + 43
76 = 2 + 3 + 11 + 17 + 43
76 = 5 + 11 + 17 + 43 (c=50)
76 = 3 + 13 + 17 + 43
76 = 2 + 5 + 7 + 19 + 43
76 = 3 + 11 + 19 + 43
76 = 2 + 3 + 5 + 23 + 43
76 = 3 + 7 + 23 + 43 (c=55)
76 = 2 + 31 + 43
76 = 2 + 3 + 11 + 13 + 47
76 = 5 + 11 + 13 + 47
76 = 2 + 3 + 7 + 17 + 47
76 = 5 + 7 + 17 + 47 (c=60)
76 = 2 + 3 + 5 + 19 + 47
76 = 3 + 7 + 19 + 47
76 = 29 + 47
76 = 2 + 3 + 7 + 11 + 53
76 = 5 + 7 + 11 + 53 (c=65)
76 = 2 + 3 + 5 + 13 + 53
76 = 3 + 7 + 13 + 53
76 = 23 + 53
76 = 2 + 3 + 5 + 7 + 59
76 = 17 + 59 (c=70)
76 = 3 + 5 + 7 + 61
76 = 2 + 13 + 61
76 = 2 + 7 + 67
76 = 2 + 3 + 71
76 = 5 + 71 (c=75)
76 = 3 + 73
Here is the sequence of triangular numbers, created by summing consecutive integers from 1 (i.e., 1+2+3+4+5…):
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431, 1485, 1540, 1596, 1653, 1711, 1770, 1830, 1891, 1953, 2016, 2080, 2145, 2211, 2278, 2346, 2415, 2485, 2556, 2628, 2701, 2775, 2850, 2926, 3003, 3081, 3160, 3240, 3321, 3403, 3486, 3570, 3655, 3741, 3828, 3916, 4005, 4095, 4186, 4278, 4371, 4465, 4560, 4656, 4753, 4851, 4950, 5050, 5151, 5253, 5356, 5460, 5565, 5671, 5778, 5886, 5995...
And here is a sequence of truncated triangulars, created by summing consecutive integers from 15 (i.e., 15+16+17+18+19…):
15, 31, 48, 66, 85, 105, 126, 148, 171, 195, 220, 246, 273, 301, 330, 360, 391, 423, 456, 490, 525, 561, 598, 636, 675, 715, 756, 798, 841, 885, 930, 976, 1023, 1071, 1120, 1170, 1221, 1273, 1326, 1380, 1435, 1491, 1548, 1606, 1665, 1725, 1786, 1848, 1911, 1975, 2040, 2106, 2173, 2241, 2310, 2380, 2451, 2523, 2596, 2670, 2745, 2821, 2898, 2976, 3055, 3135, 3216, 3298, 3381, 3465, 3550, 3636, 3723, 3811, 3900, 3990, 4081, 4173, 4266, 4360, 4455, 4551, 4648, 4746, 4845, 4945, 5046, 5148, 5251, 5355, 5460, 5566, 5673, 5781...
It’s obvious that the sequences are different at each successive step: 1 ≠ 15, 3 ≠ 31, 6 ≠ 48, 10 ≠ 66, 21 ≠ 85, and so on. But seven numbers occur in both sequences: 15, 66, 105, 171, 561, 1326 and 5460. And that’s it — 7 is the 14-th entry in A309507 at the Encyclopedia of Integer Sequences:
0, 1, 1, 1, 3, 3, 1, 2, 5, 3, 3, 3, 3, 7, 3, 1, 5, 5, 3, 7, 7, 3, 3, 5, 5, 7, 7, 3, 7, 7, 1, 3, 7, 7, 11, 5, 3, 7, 7, 3, 7, 7, 3, 11, 11, 3, 3, 5, 8, 11, 7, 3, 7, 15, 7, 7, 7, 3, 7, 7, 3, 11, 5, 3, 15, 7, 3, 7, 15, 7, 5, 5, 3, 11, 11, 7, 15, 7, 3, 9, 9, 3, 7 — A309507
I decided to take create graphs of shared numbers in compared sequences like this. In the 135×135 grid below, the brightness of the squares corresponds to the count of shared numbers in the sequence-pair sum(x..x+n) and sum(y..y+n), where x and y are the coordinates of each individual square. I think the grid looks like a city of skyscrapers bisected by a highway:
Count of shared numbers in sequence-pairs sum(x..x+n) and sum(y..y+n)
Note that the bright white diagonal in the grid corresponds to the sequence-pairs where x = y. Because the sequences are identical in each pair, the count of shared numbers is infinite. The grid is symmetrically reflected along the diagonal because, for example, the sequence-pair for x=12, y=43, where sum(12..12+n) is compared with sum(43..43+n), corresponds to the sequence pair for x=43, y=12, where sum(43..43+n) is compared with sum(12..12+n). The scale of brightness runs from 0 (black) to 255 (full white) and increases by 32 for each shared number in the sequence. Obviously, then, the brightness can’t increase indefinitely and some maximally bright squares will represent sequence-pairs that have different counts of shared pairs.
Now try altering the size of the step in brightness. You get grids in which the width of the central strip increases (smaller step) or decreases (bigger step). Here are grids for steps for 1, 2, 4, 8, 16, 32 and 64 (I’ve removed the bright x=y diagonal for the first few grids, because it’s too prominent against duller shades):
Brightness-step = 1
Brightness-step = 2
Brightness-step = 4
Brightness-step = 8
Brightness-step = 16
Brightness-step = 32
Brightness-step = 63
Brightness-step = 1, 2, 4, 8, 16, 32, 63 (animated)
12 is an interesting number in a lot of ways. Here’s one way I haven’t seen mentioned before:
12 = 3^1 * 2^2
The digits of 12 represent the powers of the primes in its factorization, if primes are represented from right-to-left, like this: …7, 5, 3, 2. But I couldn’t find any more numbers like that in base 10, so I tried a power flip, from right-left to left-right. If the digits from left-to-right represent the primes in the order 2, 3, 5, 7…, then this number is has prime-power digits too:
81312000 = 2^8 * 3^1 * 5^3 * 7^1 * 11^2 * 13^0 * 17^0 * 19^0
Or, more simply, given that n^0 = 1:
81312000 = 2^8 * 3^1 * 5^3 * 7^1 * 11^2
I haven’t found any more left-to-right prime-power digital numbers in base 10, but there are more in other bases. Base 5 yields at least three (I’ve ignored numbers with just two digits in a particular base):
110 in b2 = 2^1 * 3^1 (n=6)
130 in b6 = 2^1 * 3^3 (n=54)
1010 in b2 = 2^1 * 3^0 * 5^1 (n=10)
101 in b3 = 2^1 * 3^0 * 5^1 (n=10)
202 in b7 = 2^2 * 3^0 * 5^2 (n=100)
3020 in b4 = 2^3 * 3^0 * 5^2 (n=200)
330 in b8 = 2^3 * 3^3 (n=216)
13310 in b14 = 2^1 * 3^3 * 5^3 * 7^1 (n=47250)
3032000 in b5 = 2^3 * 3^0 * 5^3 * 7^2 (n=49000)
21302000 in b5 = 2^2 * 3^1 * 5^3 * 7^0 * 11^2 (n=181500)
7810000 in b9 = 2^7 * 3^8 * 5^1 (n=4199040)
81312000 in b10 = 2^8 * 3^1 * 5^3 * 7^1 * 11^2
Post-Performative Post-Scriptum
When I searched for 81312000 at the Online Encyclopedia of Integer Sequences, I discovered that these are Meertens numbers, defined at A246532 as the “base n Godel encoding of x [namely,] 2^d(1) * 3^d(2) * … * prime(k)^d(k), where d(1)d(2)…d(k) is the base n representation of x.”