Category Archives: Mathematics

Formulas Focal to the Flesh

by in Arithmetic, Fractions, Geometry, Mathematics and tagged , , , , , | Leave a comment

Here’s an interesting formula:

fr(1) = 1/2; mx = 3
fr(i) = fr(i-1) + 1/fr(i-1)
if fr(i) > mx, fr(i) = fr(i) – mx

Early terms look like this:

0.5, 2.5, 2.9, 3.244827586…, 4.329334628…, 2.081590666…, 2.561992513…, 2.952313716…, 3.291031107…, 3.727089920…, 2.102435627…, 2.578074447…, 2.965960841…, 3.303119709…, 3.602146368…, 2.262872154…, 2.704788415…, 3.074503101…, 13.49676325…, 10.59203071…, 7.723747777…, 4.935444092…, 2.452121378…, 2.859931533…, 3.209590254…, 4.980804482…, 2.485649867…, 2.887959143…, 3.234224430…, 4.503633905…, 2.168689406…

Can you see any patterns emerging? I’d guess not. And I’d guess a thousand more terms wouldn’t help you see any better. It’s hard for humans to see patterns in a jumble of numbers. Our eyes don’t work as well on numbers as on shapes. That’s why you can make that formula focal to the flesh, as it were, by plotting the numbers on a graph. Or part of the numbers, anyway. Suppose you take the fractional parts of each pair of terms and use them to map (x,y) on a FractL (my name for a graph whose arms run from 0 to 1). For example, the terms 4.935444092… and 2.452121378… would yield x = 0.935444092… and y = 0.452121378… (or vice versa). The resultant graph makes the formula focal to the flesh. And it’s replete with patterns:

fr(i)+=1/fr(i-1); if fr(i)>3, fr(i)-=3; x = frac(fr(i)), y = frac(fr(i+1))

I can’t explain the patterns and they may arise from limited precision in the decimal digits. But I like them however they arise. The graph doesn’t change when mx = 4 (although it creates the lines in a different order):

if fr(i)>4, fr(i)-=4

But it does change when mx = 4/3. The lines almost vanish, except for a tiny comet-like mark towards the upper right-hand corner:

if fr(i)>4/3, fr(i)-=4/3

When mx = 7/2, the graph of mx = 3|4 is back in a slightly different form:

if fr(i)>7/2, fr(i)-=7/2

And again with 7/3:

if fr(i)>7/3, fr(i)-=7/3

There’s a big change with 7/4, Most of the lines disappear:

if fr(i)>7/4, fr(i)-=7/4

And only the main lines appear with 9/5:

if fr(i)>9/5, fr(i)-=9/5

And so on till you try fr -= 2/f, as noted below:

if fr(i)>11/5, fr(i)-=11/5

if fr(i)>11/6, fr(i)-=11/6

if fr(i)>15/8, fr(i)-=15/8

if fr(i)>29/15, fr(i)-=29/15

Now try fr += 2/fr and fr += 3/fr. This is what happens:

fr += 2/fr; if fr(i)>3, fr(i)-=3

fr += 2/fr; if fr(i)>8/3, fr(i)-=8/3

fr += 2/fr; if fr(i)>11/4, fr(i)-=11/4

fr += 3/fr; if fr(i)>6, fr(i)-=6

And what about these graphs?

They’re created by seeding a sum, s, with a fraction, then adding more fractions < 1 whose numerators = 1,2,3… and whose denominators are the prime numbers 1, s -= 1. When s > 1, s -= 1. Then you take the fractional parts of s(i) and s(i+1) and graph (x,y) as above.

Post-Performative Post-Scriptum

The title of this post refers to Morbid Angel’s Formulas Fatal to the Flesh (1998). I’ve never heard it, but I like Morbid Angel’s alphabetically alliterative album-titles.

Les ZFleurs des Mathématiques

by in Mathematics, Art and tagged , , , , , | Leave a comment


An attractive and intriguing image by Kristina Armitage for an article on Zermelo-Fraenkel set theory at Quanta Magazine (click for larger image).

Post-Performative Post-Scriptum

The title of this post is a pun on Charles Baudelaire’s Les Fleurs du mal / Flowers of Evil (1857). If you’re already familiar with the book, you might have missed the “ZF” of ZFleurs.

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

Run Fun

by in Arithmetic, Base Behavior, Mathematics and tagged , | Leave a comment

I was looking at numbers where the digit-runs of n in one base, b1, represented the digits of n in another base, b2. To understand digit-runs, take the number 775555111 (in whatever base). The function digitrun(775555111) = 2,4,3, because reading 775555111 left-to-right you see two 7s, four 5s, three 1s. Here’s one pattern I found:

10000011111 in base 2 = 1055 in base 10. The digit runs of 10000011111 are 1,5,5 and 15530 = 105510 = 100000111112.
10000022222 in base 3 = 59291 in base 10. The digit runs of 10000022222 are again 1,5,5 and 155241 = 5929110 = 100000222223.
1000009999910 → 1,5,5 and 15599998 = 1000009999910

The pattern isn’t difficult to understand — what are 30+2 and 99998+2? — and it’s easy to prove that it applies to all standard bases. But I wondered how AI would go about the proof. Here’s the result:

For base b ≥ 2, let the largest digit be d = b − 1. Then

100000dddddb

has digit runs

1, 5, 5

so the run-length encoding is always

155.

The value of 155 is interpreted in base

b5 − 2,

because

155b5−2 = (b5−2)2 + 5(b5−2) + 5.

Thus

100000dddddb = 155b5−2.

Base b n = 100000(b−1)(b−1)(b−1)(b−1)(b−1)b Run Lengths Compressed Form Base-10 Value of n
2 100000111112 1,5,5 15530 1,055
3 100000222223 1,5,5 155241 59,291
4 100000333334 1,5,5 1551022 1,049,599
5 100000444445 1,5,5 1553123 9,765,624
6 100000555556 1,5,5 1557774 60,466,175
7 100000666667 1,5,5 15516805 282,475,248
8 100000777778 1,5,5 15532766 1,074,266,111
9 100000888889 1,5,5 15559047 3,486,270,320
10 1000009999910 1,5,5 15599998 10,000,099,999
11 100000AAAAA11 1,5,5 155161049 26,094,908,050
12 100000BBBBB12 1,5,5 155248830 62,191,972,031
13 100000CCCCC13 1,5,5 155371291 136,858,491,292
14 100000DDDDD14 1,5,5 155537822 281,793,968,543
15 100000EEEEE15 1,5,5 155759373 547,918,611,874
16 100000FFFFF16 1,5,5 1551048574 1,099,512,627,199
17 100000GGGGG17 1,5,5 1551419855 2,141,423,647,200
18 100000HHHHH18 1,5,5 1551889566 4,058,363,974,751
19 100000IIIII19 1,5,5 1552476097 7,502,570,010,432
20 100000JJJJJ20 1,5,5 1553199998 13,120,000,000,000

(Here A = 10, B = 11, …, J = 19.)

Proof for All b ≥ 2

The number

100000(b−1)(b−1)(b−1)(b−1)(b−1)b

has value

b10 + (b−1)(b4 + b3 + b2 + b + 1).

Since

b4 + b3 + b2 + b + 1 = (b5 − 1)/(b − 1),

this becomes

b10 + (b5 − 1) = b10 + b5 − 1.

Now let

B = b5 − 2.

Then

155B = B2 + 5B + 5.

Substituting B = b5 − 2:

(B2 + 5B + 5) = (b5−2)2 + 5(b5−2) + 5

= b10−4b5+4+5b5−10+5

= b10 + b5−1.

Hence

b10 + b5 − 1 = (b5 − 2)2+5(b5 − 2)+5

for every b ≥ 2.

Therefore,

100000(b−1)5b = 155b5−2.

This is exactly the pattern observed in bases 2 to 20, and it holds for all bases.

Partition Fission

by in Arithmetic, Mathematics, Partitions and tagged , , , , | Leave a comment

Simple but seductive. That’s how I’d describe partitions. Except that they’re not so simple. There are hidden depths in the task of finding how many ways an integer can be expressed as the sum of smaller integers (and as the sum of itself). Here are the partitions of 4, for example:

4 = 1+3 = 2+2 = 1+1+2 = 1+1+1+1
4 (1) = 1+3 (2) = 2+2 (3) = 1+1+2 (4) = 1+1+1+1 (5), ∴ partcount(4) = 5

There are five partitions of 4, because an integer is counted as its own partition. Accordingly, partcount(4) = 5. But partcount(n) doesn’t return values in a predictable way:

partcount(1) = 1 ← 1
partcount(2) = 2 ← 2 = 1+1
partcount(3) = 3 ← 3 = 1+2 = 1+1+1
partcount(4) = 5 ← 4 = 1+3 = 2+2 = 1+1+2 = 1+1+1+1
partcount(5) = 7 ← 5 = 1+4 = 2+3 = 1+1+3 = 1+2+2 = 1+1+1+2 = 1+1+1+1+1
partcount(6) = 11 ← 6 = 1+5 = 2+4 = 3+3 = 1+1+4 = 1+2+3 = 2+2+2 = 1+1+1+3 = 1+1+2+2 = 1+1+1+1+2 = 1+1+1+1+1+1
partcount(7) = 15 ← 7 = 1+6 = 2+5 = 3+4 = 1+1+5 = 1+2+4 = 1+3+3 = 2+2+3 = 1+1+1+4 = 1+1+2+3 = 1+2+2+2 = 1+1+1+1+3 = 1+1+1+2+2 = 1+1+1+1+1+2 = 1+1+1+1+1+1+1
partcount(8) = 22 ← 8 = 1+7 = 2+6 = 3+5 = 4+4 = 1+1+6 = 1+2+5 = 1+3+4 = 2+2+4 = 2+3+3 = 1+1+1+5 = 1+1+2+4 = 1+1+3+3 = 1+2+2+3 = 2+2+2+2 = 1+1+1+1+4 = 1+1+1+2+3 = 1+1+2+2+2 = 1+1+1+1+1+3 = 1+1+1+1+2+2 = 1+1+1+1+1+1+2 = 1+1+1+1+1+1+1+1

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 — the number of partitions of n, A000041 at the Online Encyclopedia of Integer Sequences

And there are fractals — self-similarities at smaller and smaller scales — hidden in that simple arithmetic. Take the partitions of 8:

8 = 1+7 = 2+6 = 3+5 = 4+4 = 1+1+6 = 1+2+5 = 1+3+4 = 2+2+4 = 2+3+3 = 1+1+1+5 = 1+1+2+4 = 1+1+3+3 = 1+2+2+3 = 2+2+2+2 = 1+1+1+1+4 = 1+1+1+2+3 = 1+1+2+2+2 = 1+1+1+1+1+3 = 1+1+1+1+2+2 = 1+1+1+1+1+1+2 = 1+1+1+1+1+1+1+1 (c=22)

By definition, the sum of each partition of 8 is the same: 8. But the products — the result of multiplying the numbers of a partition — rise and fall wildly, hitting a maximum of 18 and a minimum of 1:

1.7 = 7
2.6 = 12
3.5 = 15
4.4 = 16
1.1.6 = 6
1.2.5 = 10
1.3.4 = 12
2.2.4 = 16
2.3.3 = 18
1.1.1.5 = 5
1.1.2.4 = 8
1.1.3.3 = 9
1.2.2.3 = 12
2.2.2.2 = 16
1.1.1.1.4 = 4
1.1.1.2.3 = 6
1.1.2.2.2 = 8
1.1.1.1.1.3 = 3
1.1.1.1.2.2 = 4
1.1.1.1.1.1.2 = 2
1.1.1.1.1.1.1.1 = 1

It’s interesting to ask when partitions(n) yield the biggest product (the answer is here). It’s also interesting to create graphs of prod(part(n)), the products of the partitions of n. You’ll see something I call partition fission. The graphs start to fissure into what look like fins or sails, and then each fin or sail starts to fissure too:

Graph for multiples of partitions(8) (partcount(8) = 22)

Graph for prod(part(9)) (partcount = 30)

prod(part(10)) (partcount = 42)

prod(part(11)) (partcount = 56)

prod(part(12)) (partcount = 77)

prod(part(13)) (partcount = 101)

prod(part(14)) (partcount = 135)

prod(part(15)) (partcount = 176)

prod(part(16)) (partcount = 231)

Those graphs are all on the same scale. The two graphs below have been adjusted to capture many more partitions and show the fractality coming into full flower:

prod(part(20)) (partcount = 627)

prod(part(28)) (partcount = 3718)

Finally, here’s an animated gif of the graphs for the partition-products of 8 to 16:

Animated gif for prod(part(8..16)) (animation at EZgif) (click for larger image)

The Bird Dimension

by in Art, Geometry, Mathematics, Optical Illusions, Post-Weird, Surrealism and tagged , , , , , , , | 2 Comments

M.C. Escher, Another World / Andere Wereld (1947)

This is almost my favorite image by Escher. But I’d like a frEscher perspective in it: I think the bird should be looking in the other direction, out into the impossibly overlapping universes, not into the cupola and at the viewer.

The Power of Cow

by in Arithmetic, Integer Sequences, Mathematics and tagged , , , , | 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

Third Whirled Warp

by in Mathematics, Geometry, Triangles, Squares and tagged , , | Leave a comment

Here’s a regular hexagon inside a regular triangle, that is, an equilateral triangle:

Regular hexagon inside regular triangle

Imagine that two points are moving around the perimeter of each polygon, with the hex-point moving half as fast as the tri-point (after adjustment for the incommensurate relative lengths of the perimeters). If you trace the midpoint of the twin spinning points, you get this shape:

v3v6, 1 : 1/2, pol

And if you adjust the midpoint path as though the triangle had been stretched into a circle, you get this shape:

v3v6, 1 : 1/2, circ, pol

Here’s the same when the ratio of speeds is 1/2 to 1/3, that is, 1 to 2/3:

v3v6, 1/2 : 1/3, circ, pol

Without the polygons, it looks like this:

v3v6, 1/2 : 1/3, circ

When the ratio of speeds if -1/3 to 2/3, that is, the tri-point is moving counter-clockwise around the triangle, you get this shape:

v3v6, -1/3 : 2/3, pol

When it’s stretched into a circle, you get this:

v3v6, -1/3 : 2/3, circ, pol

It looks like a moustache:

v3v6, -1/3 : 2/3, circ

Here are more midpoint shapes created with a hexagon inside a triangle:

v3v6, 2/2 : 3/3, circ

v3v6, -1/2 : 3/4, circ

v3v6, 1/4 : 1/5, circ

v3v6, -1/4 : 3/4, circ

v3v6, -1/4 : 4/5, circ

v3v6, 2/3 : 3/4, circ

v3v6, 2/3 : 3/5, circ

v3v6, 3/4 : 4/5, circ

v3v6, 3/4 : 4/5, circ

Now try aligning the nested hexagon like this, so that the sides of the hexagon coincide with the middle third of the sides of the triangle:

v3v6, side alignment

With two points moving in a ratio of 1/3 to 1/4, you get this midpoint shape:

v3v6, sided, 1/3 : 1/4, pol

Here it is without the polygons:

v3v6, sided, 1/3 : 1/4

Now try a regular octagon inside a square:

v4v8, 1/2 : 1/3, circ, pol

v4v8, 1/2 : 1/3, circ

v4v8, -1/3 : 3/4, circ

v4v8, 2/3 : 3/5, circ

Now place a triangle inside a hexagon:

v6v3, 1 : 1/4, pol

If you stretch the midpoint path according to perimeter of the triangle, you get this:

v6v3, 1 : 1/4, circ, pol

v6v3, 1 : 1/4, circ

The three stretching shapes remind me of hands in Egyptian art, like this image of King Tutankhamun and Queen Ankhesenamun:

Detail from the Golden Throne of Tutankhamnun

Here are more midpoint paths:

v6v3, 1 : -1/4, circ

v6v3, 1 : 1/2, circ

v6v3, 1 : 1/3, circ

v6v3, -1 : 1/3, circ

v6v3, -1 : 1/4, circ

v6v3, 1 : 1/5, circ

v6v3, 2/3 : 1/4, circ

Now try a square inside an octagon:

v8v4, 2/3 : 1/4, circ, pol

v8v4, 2/3 : 1/4, circ

v8v4, 2/5 : 1/6, circ

v8v4, 2/5 : 3/7, circ

v8v4, 4/5 : 3/7, circ

Elsewhere Other-Accessible…

• First Whirled Warp — an earlier look at this kind of geometry
• Second Whirled Warp — and another earlier look

Moniliform Maths

by in Arithmetic, Integer Sequences, Mathematics, Powers and tagged , , | Leave a comment

2 = 1/2 + 2/4 + 3/8 + 4/16 + 5/32…

sum(np / 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(np / 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(np / 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(np / 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(np / 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(np / 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(np / 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(np / 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(np / 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)