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Boring Math (11.0) 

((15.5^2)*(2/3))-14^2 ~= -36

Boring Math (11.0) 

(36 - (19*2^pi - 13^2)/(19*2^e - 5^3))/6 ~= 12

Boring Math (11.0) 

2657 = 2000*2^(-e + π) - 5^2

Boring Math (11.0) 

For the above, the values are essentially 11^2 + epsilon & (17/2)^2 + epsilon, which sort of relates back to:

(19^2 + 2)/3 = 11^2
(12^2 + 1/2)*2 = 17^2

Boring Math (11.0) 

factor 14^4-prime(3^4) --> 37997
factor 15^4-prime(6^4) --> 39998 = 2*7*2857

Boring Math (11.0) 

I'm curious about:

28,30,34,43,67,163,953...

Which is based on

floor(sqrt(k*28)), k=163

Exploration is like:

floor(sqrt(k*28)), k=32437
floor(sqrt(k*28)), k=32504

primes near 32470 (7 candidates)

Boring Math (11.0) 

a(n+1) = floor(sqrt(a(n)*163)), a(0)=28

n | a(n)
0 | 28
1 | 67
2 | 104
3 | 130
4 | 145
5 | 153
6 | 157
7 | 159
8 | 160
9 | 161

Where 67, 130, 145, 157 are of interest to me, particularly in the context of

math.stackexchange.com/questio

Boring Math (11.0) 

I'm thinking 145 & 34 are the relevant ones here:

(6^2 - 2)/2 = 17
(12^2 + 1/2)*2 = 17^2

Which may imply 17 is relevant to music theory.

en.m.wikipedia.org/wiki/34_equ

Boring Math (11.0) 

So I'm thinking 72, 77, 80 is relevant here, for:

en.m.wikipedia.org/wiki/Music_
en.m.wikipedia.org/wiki/34_equ

9^2, 80 for 12 notes
5^2, 24/"72" for 17 notes

Boring Math (11.0) 

I'm thinking about oeis.org/A107360

{3, 5, 7, 13, 17, 19, 31, 61, 127 }

2^n - 3 = p --> {3, 4, 6} | {5, 13, 61}
2^n - 1 = p --> {2, 3, 5, 7} | {3, 7, 31, 127}
2^n + 1 = p --> {1, 2, 4} | {3, 5, 17}
2^n + 3 = p --> {-inf, 1, 2, 4} | {3, 5, 7, 19}

3 occurs 3x
5 occurs 3x
7 occurs 2x
All other terms occur 1x

Boring Math (11.0) 

Another aggregation:

2 happens 3x
4 happens 3x
1 happens 2x
3 happens 2x
All others -inf, 5-7 happen 1x

Boring Math (11.0) 

I'm looking at

2^({1, 2, 4, 6, 8, 12, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 48, 54, 56, 60, 64, 66, 72, 78, 80, 84, 88, 90, 96, 100, 104, 108, 112, 120, 126, 128, 132, 140, 144, 150}+2) - 3 (practical numbers)

Example:

factor {5, 13, 61, 253, 1021, 16381, 262141, 1048573, 4194301, 67108861, 1073741821, 4294967293, 17179869181, 274877906941, 4398046511101, 17592186044413, 1125899906842621, 72057594037927933}

Prime for

{5, 13, 61, 1021, 16381, 1048573, 4194301}

Boring Math (11.0) 

It looks like practical numbers may not be relevant, based on

Table[(n-2)*Boole[isprime(2^n - 3)]], n= 2 to 1000

Boring Math (11.0) 

I was looking at

"3, 4, 6 -seq:7 -seq:9 -seq:11 -seq:13 -seq:15" on oeis.org

And realized the 3, 4, 6... may just be prime(n) + 1

Boring Math (11.0) 

There's two features in this picture that interest me:

grantjenks.com/wiki/_detail/pr

One is at coordinate (165, 363), the other is at (318, 477), which may essentially be related to the Heegner number 163 (& 2*163).

Boring Math (11.0) 

You can draw a 45 degree angle and find another feature of interest at (250, 450) -- that's my estimate based on the features at (240, 422) & (254, 437):

This makes me interested in prime(54), and prime(38) / prime(70).

Boring Math (11.0) 

I'm leaning towards 163, 251, 342 now.

Boring Math (11.0) 

Now I'm leaning towards 163, 251, 339:

This image shows a line that essentially is supposed to represent 339 at location (338, 367):

Boring Math (11.0) 

For the above image, you can't see the bottom right of the line segment, but it seems to have 5/20 pixels to the left, unlike the other one which is a 50:50 split.

Boring Math (11.0) 

I'm a little interested in:

prime(20)*e ~= 193
prime(30)*pi ~= 355 = 5*prime(20)

Boring Math (11.0) 

(339+71)/(5/2) = 163+1 = 41*4

Boring Math (11.0) 

So if sigma(n) < 2*n - 1 (powers of 2), I think n cannot be a practical number. I'm curious if there is an upper limit where n must be a practical number.

Creating code isn't that hard, although sigma and practical numbers are obscure-ish.

Boring Math (11.0) 

from itertools import chain, cycle, accumulate, combinations
from typing import List, Tuple

# %% Factors

def factors5(n: int) -> List[int]:
"""Factors of n, (but not n)."""
def prime_powers(n):
# c goes through 2, 3, 5, then the infinite (6n+1, 6n+5) series
for c in accumulate(chain([2, 1, 2], cycle([2,4]))):
if c*c > n: break
if n%c: continue
d,p = (), c
...

Boring Math (11.0) 

...

while not n%c:
n,p,d = n//c, p*c, d + (p,)
yield(d)
if n > 1: yield((n,))

r = [1]
for e in prime_powers(n):
r += [a*b for a in r for b in e]
return r[:-1]

# %% Powerset

def powerset(s: List[int]) -> List[Tuple[int, ...]]:
"""powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3) ."""
return chain.from_iterable(combinations(s, r) for r in range(1, len(s)+1))
...

Boring Math (11.0) 

...

# %% Practical number

def is_practical(x: int) -> bool:
"""Practical number test with factor reverse sort and short-circuiting."""

if x == 1:
return True
if x % 2:
return False # No Odd number more than 1
mult_4_or_6 = (x % 4 == 0) or (x % 6 == 0)
if x > 2 and not mult_4_or_6:
return False # If > 2 then must be a divisor of 4 or 6
...

Boring Math (11.0) 

...

f = sorted(factors5(x), reverse=True)
if sum(f) < x - 1:
return False # Never get x-1
ps = powerset(f)

found = set()
for nps in ps:
if len(found) < x - 1:
y = sum(nps)
if 1 <= y < x:
found.add(y)
else:
break # Short-circuiting the loop.

return len(found) == x - 1
...

Boring Math (11.0) 

...


if __name__ == '__main__':
highest_practical = 1
lowest_practical = 1
highest_nonpractical = 1
lowest_nonpractical = 1

highest_practical_value = 0
lowest_practical_value = 1e10
highest_nonpractical_value = 0
lowest_nonpractical_value = 1e10
...

Boring Math (11.0) 

...

max = 3000
for x in range(1, max + 1):
normalized_sigma_value = sigma(x, 1)/x
if is_practical(x):
if normalized_sigma_value < lowest_practical_value:
lowest_practical = x
lowest_practical_value = normalized_sigma_value
...

Boring Math (11.0) 

...
elif normalized_sigma_value > highest_practical_value:
highest_practical = x
highest_practical_value = normalized_sigma_value
else:
normalized_sigma_value = sigma(x, 1)/x
if normalized_sigma_value < lowest_nonpractical_value:
lowest_nonpractical = x
lowest_nonpractical_value = normalized_sigma_value
...

Boring Math (11.0) 

...
elif normalized_sigma_value > highest_nonpractical_value:
highest_nonpractical = x
highest_nonpractical_value = normalized_sigma_value

print(lowest_practical, " (", float(lowest_practical_value), ") - ", highest_practical, " (", float(highest_practical_value), ")")
print(lowest_nonpractical, " (", float(lowest_nonpractical_value), ") - ", highest_nonpractical, " (", float(highest_nonpractical_value), ")")

Boring Math (11.0) 

Oh, right, the output is:

1 ( 1.0 ) - 2520 ( 3.7142857142857144 )
2999 ( 1.0003334444814937 ) - 1608 ( 2.537313432835821 )

Basically, the lower values (both near 1) are uninteresting.

The upper values seem to increase without bound, but maybe the ratio between them is predictable. highest_nonpractical_value may have a strict bound, but I need to get a more efficient algorithm set to figure that out.

Boring Math (11.0) 

I forgot to paste some modifications from yesterday (mostly after midnight):

Looking at records over time. I think I'm interested in the narrow band starting at the perfect numbers and going up to a factor of 2 of the upper limit: en.wikipedia.org/wiki/Divisor_

Boring Math (11.0) 

Essentially this means I'm looking for a modified definition of practical numbers that excludes powers of 2 and the most composite practical numbers. I'm wondering what the density of this set of numbers is compared to the primes, and also wondering if I can e.g. construct a 1:1 relationship between primes and "modified practical numbers".

Boring Math (11.0) 

Using e**(log(sigma(v, 1))/(log(v)*log(log(v)))):

The values seem to approach ~7/4 and 1/2. think the 1/2 is exact based on en.wikipedia.org/wiki/Divisor_

But I'm not sure about the other one.

Boring Math (11.0) 

I'm wayyy off lol:

oeis.org/A335030

e**(log(sigma(52307529120))/(log(52307529120)*log(log(52307529120)))) ~= 1.39127

Boring Math (11.0) 

I was trying to consider practical numbers where the negative sign could be used e.g.:

1
3-1
3
3+1
9-3-1
9-3
9-3+1
9-1
9

Notably, there is a common math question related to base-3 and a balance scale.

Which likely results in the sequence:

oeis.org/A196149

Although it's still possible there are exceptions.

Follow

Boring Math (11.0) 

This helped me realize a useful property / algorithm optimization of practical numbers:

Divisors of practical numbers seem to increase by at most a factor of 2. I think this is necessary, but I'm not sure if it is sufficient.

Boring Math (11.0) 

And just as I say that I get rekt by 78

Boring Math (11.0) 

78 is practical:

1 | 2 | 3 | 6 | 13 | 26 | 39 | 78 (8 divisors)

Boring Math (11.0) 

Aww yee:

en.m.wikipedia.org/wiki/Comple

Which leads to an interesting sequence:

oeis.org/A203074

I still want to figure out a sequence containing 78 specifically.

Boring Math (11.0) 

This references arxiv.org/abs/1405.2585

Which seems to have a mention of the maximal ratio (and it may be unbounded?)

Boring Math (11.0) 

Table[(2*n)*Boole[Max[Ratios[Divisors[(2*n)]]] > 2 and not isprime(n)]], n=1 to 200

Which leads me to

oeis.org/A317412

Boring Math (11.0) 

I'm curious about:

72-80 & 81 -- 3
81*(9/2) -- 33/2
81*9 -- 63

For

oeis.org/A317412

Boring Math (11.0) 

Not great rn, but:

Table[n^n+1], n=0 to 9

-->

factor { 2, 5, 28, 257, 3126, 46657, 823544, 16777217, 387420490}

Where I'm curious about 5 & 257 as Fermat primes.

This makes me curious about the neighboring 2, 28, 3126

2 = 2*1
28 = 4*7
3126 = 6*521

Boring Math (11.0) 

I started thinking about this because I was comparing primes to practical numbers and noticed the first major difference was the gap between primes 23 & 29 where there are two practical numbers (i.e. 24 & 28).

I also was particulary curious about comparing 8 (practical) to 11 (prime).

Boring Math (11.0) 

But n^n + 1 where n=2*k + 1 (i.e. odd) is interesting because it seems to result in a multiplier of 2*k + 2.

Boring Math (11.0) 

This leads to

Table[((n)^(n)+1)/(n+1)], n=0 to 26

-->

{Indeterminate, 1, 5/3, 7, 257/5, 521, 46657/7, 102943, 16777217/9, 38742049, 10000000001/11, 23775972551, 8916100448257/13, 21633936185161 ...}

Boring Math (11.0) 

One of these numbers looks familiar:

m.youtube.com/watch?v=p-HN_ICa

46657/7 versus 420514/7

The ratio 420514 / 46657 = 9.01288...

420514 - 9*46657 = 601 = prime(11*10)

Boring Math (11.0) 

1367 ~= (521*(3/2 + sqrt(5)/2) + 3)

Boring Math (11.0) 

Of interest:

(2 (17 - 3))/(3 + sqrt(5))
(2 (251 - 3))/(3 + sqrt(5))

Which have numerators that resolve to 28 & 496 (both perfect numbers).

Notably, 6 maps onto 6, but it isn't prime so it's not an "Ibrishimova number".

Boring Math (11.0) 

There's two elements with sub-2 ratios:

Round[sigma(n*2 - 6)/(n*2 - 6), .000001], n=5
Round[sigma(n*2 - 6)/(n*2 - 6), .000001], n=61

7/4
105/58 = (3*5*7)/(2*29)

Boring Math (11.0) 

sigma(n*2 - 6)/(n*2 - 6), n=283

-->

(3*31)/(5*7)

Boring Math (11.0) 

The best way to find the center of the 4 is not a 4-way average (mean or geometric average), but finding the middle two of the 4 and then doing that:

((sigma(13*2 - 6)/(13*2 - 6)) + (sigma(1367*2 - 6)/(1367*2 - 6)))*(1/2)

((sigma(13*2 - 6)/(13*2 - 6)) * (sigma(1367*2 - 6)/(1367*2 - 6)))^(1/2)

2.105715 ~= (23/14)^(3/2) ?

Boring Math (11.0) 

If you add these ratios, except 283, the sum is about 16.01. 283 has a ratio of about 2.65, and the ratio between these two values is about 6.

Not sure if it's meaningful.

Boring Math (11.0) 

I was originally curious about 61 and (61*2 - 6)=116 which is the only non-practical item.

This led me to:

oeis.org/A330870

I was then curious about:

120
118 = 2*59
116 = 4*29
114 = 6*19
112 = 8*14
110 = 10*11
108 = 12*9
// 106 = 2*53

Which notably leads to an observation

{9, 11, 14, 19, 29, 59} + 1 = {10, 12, 15, 20, 30, 60}

The numbers that aren't captured are 1-6.

Boring Math (11.0) 

Table[(1-n)/n], n=1 to 1
Table[(2-n)/n], n=1 to 2
Table[(6-n)/n], n=1 to 3
Table[(12-n)/n], n=1 to 4
(absent)
Table[(60-n)/n], n=1 to 6
...
Table[(5040-n)/n], n=1 to 10

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