"I'd like to converse with you about your findings, but I have no paper, no lemmas and no theorems. "
It appears there is no lemma or theorem and OP is working at best in 8th grade math.
I'm just a lowly front end developer but I'll take a crack at it.
All the key variables are on two apparent slopes, d4, and 'gold'.
I could see getting estimates on unknown factors by simply taking known products with common variables (d4a, d4u, d4z, d4H, etc), and then finding the smallest of the set, on the basis that would tightly constrain the upper and lower bound of the factors of those variables, but that assumes none of the unknown factors like u, d4, c, z, or H, are less than 1.
I also wrote some functions for converting variables between the gold line, the _gold line, and the d4 line.
def _gold_to_d4(var):
return (var**2)*Hc
def d4_to__gold(var):
return (var/Hc).sqrt()
This followed from the observation that
d4u/Hc/G
Decimal('1.332259228726959456136695243518871469428174838645985237776420270')
(_goldc**2/G)
Decimal('1.332259228726959456136695243518871469428174838645985237776420270')
meaning
d4u/Hc == some _gold**2
Shows d4 and _gold series are interconnected.
Taking any _gold, multiplying by the power of 2, and then multiplying by Hc should give
a d4 variable.
While dividing a d4 variable by Hc, and then rooting it, should give a _gold series variable
if d4 is inverted (less than 1), then (((_goldc)**2)*Hc)/d4 > d4u
To test this I ran the script a few times with various factors.
a
Decimal('84780491')
b
Decimal('7022437637')
>>> (((_goldc)**2)*Hc)/d4 > d4u
True
>>> a
Decimal('247017779')
>>> b
Decimal('8552169383')
(((_goldc)**2)*Hc)/d4 > d4u
True
>>> a
Decimal('39420301')
>>> b
Decimal('8490912585839')
(((_goldc)**2)*Hc)/d4 > d4u
False
As you can see with the above example, both c and d4 are normal, but c is less than d4,
so _cd4 is less than 1, and the outcome is of course false in this case.
Good to know.
And what this means is that when _cd4 is less than 1,
And by doing this we can tell the output of the following:
(((_goldc)**2)*Hc)/d4 > d4u
even if we don't know d4.
Thats all I have to add at the moment.
