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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
...

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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.

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Boring Math (11.0) 

I'm a little interested in:

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

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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.

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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):

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Boring Math (11.0) 

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

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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).

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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).

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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

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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

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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}

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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

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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

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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

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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

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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)

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Boring Math (11.0) 

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

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