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Finished my 6-second game trailer for my Asteroids (1979)-clone in virtual reality:

youtu.be/tGoc2T6ABUc

Boring Math (9.0) 

2 --> 3.5 primes
3 --> 4.5 primes
4 --> 4.5 primes
5 --> 4 primes
7 --> 4 primes
13 --> 8 primes

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Boring Math (9.0) 

I think these 6 equations are reasonably interesting. Combined they may be analogs to the original in

math.stackexchange.com/questio

(( + 12 x^2 + 12 x + 2 )/2 )
(( + 12 x^2 + 12 x - 3 )/3 )
(( + 12 x^2 + 12 x + 4 )/4 )
(( + 12 x^2 + 12 x - 5 )/1 )
(( + 12 x^2 + 12 x + 7 )/1 )
(( + 12 x^2 + 12 x - 13 )/1 )

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Boring Math (9.0) 

Also, I realized the length of the prime runs is halved, because the negative indices are identical to the positive ones (+/-1)

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Boring Math (9.0) 

I guess:

(( + 12 x^2 + 12 x + 4 )/4 )

is also interesting. It starts with 1, but it includes several values that are prime afterwards

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Boring Math (9.0) 

The two equations in question are:

12*x^2 + 12*x - 5
12*x^2 + 12*x - 13

&

12*x^2 + 12*x + 7 (similar, "worse")

(( + 12 x^2 + 12 x + 10 )/2 ) -- interesting

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Boring Math (9.0) 

# Uncomment this to produce dotted primes
= 0
f in factors:
# numfac += f[1]
numfac == 1: radius = 0.4
: radius = 0.1

# Set dot size by number of unique prime factors
npf = len(factors)
radius = pow(1.414, npf) / 12
if is_prime(abs(i+offset)):
plotted_items += circle((x,y), radius, rgbcolor=(0,0,1))
plotted_items += text((i+offset), (x,y))

show(plotted_items)

# hexnet.org/content/hexagonal-n

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Boring Math (9.0) 

plotted_items = circle((0,0), 1, rgbcolor=(0,0,0))
for j in range(n):
i = j + 1
ring = hexring(i) # What ring is i in
perim = ring * 6 # How many numbers are in the perimeter of that ring
ringcount = hexnum(ring) - i # Where is i in that ring
r = ring + 1 - (float(1) / float(perim)) * ringcount
theta = (tau / perim) * ringcount
x = cos(theta) * r
y = -sin(theta) * r
factors = [0]
if i + offset != 0:
factors = factor(i+offset)

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Boring Math (9.0) 

# GRAHAM SPIRAL: The official script
from math import *

tau = 2 * pi
offset = -7
n = 12 # rings
n = (n+1)*(n+1)*(n+1) - n*n*n - 1

# Use this to add an arbitary numerical offset
= input("Offset:")

# What "ring" any given natural number falls into
def hexring(n):
val = 0
while n > 0:
val += 1
n = n - 6 * val
return val

# The nth hex number - 1, based on being the gnomon of the nth cube
def hexnum(n):
return (n+1)*(n+1)*(n+1) - n*n*n - 1

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Boring Math (9.0) 

I was suddenly curious if there are any analogs for

math.stackexchange.com/questio

Except with a hexagonal Ulam spiral.

So I modified some code from the internet to find out.

Based on a preliminary search, I found at least 2: 5 & 13, instead of the original 6: 2, 3, 4, 5, 7, 13.

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Boring Math (9.0) 

I'm also thinking about irregular twin primes:

101, 103, 461, 463, 617, 619, 809, 811, 1151, 1153...

oeis.org/A060012

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Boring Math (9.0) 

I'm thinking of

56/37 ~~= (3/2)

Where the 37 is the first irregular prime

And

56 is the first practical number that does not have a prime +/- 1.

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Boring Math (9.0) 

When I looked at

(7/8)^150 = 2.0008196933445578*^-9

I was a little surprised by the 2 and 8196 ~= 2^13 + 4

I'm really curious if this is mathematical coincidence.

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Boring Math (9.0) 

This means that the world record going from 118 to 154 is 36 more than the previous, which isn't really unheard of, considering the probability if done correctly is (7/8) for each consecutive success, a success by 1 is guaranteed, and we wouldn't be talking about small margins of success (which makes it stand out from the crowd).

You could potentially find that some dice have minor statistical properties beyond this, improving your odds marginally.

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Boring Math (9.0) 

The scrap paper version of this is:

Choose:

2,3,4,5
1,3,4,6

2+1 = 3
2+3 = 5
2+4 = 6
2+6 = 8

3+1 = 4
3+3 = 6
3+4 = 7
3+6 = 9

4+1 = 5
4+3 = 7
4+4 = 8
4+6 = 10

5+1 = 6
5+3 = 8
5+4 = 9
5+6 = 11

Where 2 out of 16 outcomes result in 7.

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Boring Math (9.0) 

I watched

m.youtube.com/watch?v=8Ko3TdPy

And was curious about the world record for "craps" which uses 2 six-sided die.

I found out that if you spin two dice perfectly along one axis you can turn a (36-6)/36 = 5/6 into a (16-2)/16 = 7/8

Which essentially means you can be 1.16x luckier if you have a fortunate habit (intentional or not).

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Boring Math (9.0) 

I'm still a little curious about

c - d + b^3 = 251, c + d + b^5 = 283, c - d + b^7 = 1367, b > 0

b≈2.73215, c≈180.684, d≈-49.9214

From yesterday.

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Boring Math (9.0) 

I tried this

Table[Round[259 + 67^k/4 + 8*(-1)^k, .0001]], k=0 to 4

And was surprised the first two values were close.

This is the equation I was going for (it has some error):

Table[Round[259 + 67^k/4 + 8*(-1)^(k+1), .0001]], k=0 to 4

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Boring Math (9.0) 

+ 16 x^2 − 40 x + (58/2)

5, 13, 53 (, 225...)

(44 + 1) n + (-1)^n/2 + 33/2

17, 61, 107 (, 151...)

[Unknown]

251, 283, 1367 (possibly swapping first two features
values)

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Boring Math (9.0) 

While trying to find an equation that uses 251, 283, 1367, I noticed that 1367 = 2^10 + 7^3:

Table[ 7^k + 2^(16 - 2*k)], k=0 to 8

{65537, 16391, 4145, 1367, 2657, 16871, 117665, 823547, 5764802}

Not guaranteed it's interesting, but it seems like it could be useful.

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Boring Math (9.0) 

I was trying to create three fundamental equations for

oeis.org/A276260

Similar to the three I made for Heegner numbers that always include 37/67.

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