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Figurates Up To 10^13


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so as i said i expect the calculation to take 58 days, i thought I'd post my daily log here.

an hour and a half in and i can confirm donk's calculation for 10^10.

i could have started at where donk left off but i want to make sure my numbers line up, especially when i get to around where donk left off.

(expected time to get to 10^12, 6.25 days.)

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Quoting Freeztar:

Apparently, no one wants to admit to the coolness of this.

Or...It's misunderstood. In any case, thumbs up here!

 

Your support is much appreciated Freeztar!

 

But to be fair, we must also remember that some Hypographers

put in a lot of time and effort into this project/experiment,

while others at least supported it with their comments and feedback.

 

Besides Phillip 1882, Yourself and Myself, there are Turtle, Donk, Modest,

Kharakov, Pyrotex, IDMclean, Craig D, Qfwfq, Rade, JMJ Jones 0424, Jay Q,

and if I forgot anybody, well, they know who they are.

 

Many people are afraid of things that are new and threaten the status quo.

Then again, many people are simply not bright enough to understand certain results.

Also, there are those who are simply dishonest and loath to acknowledge anyone's success...

exept their own! :lol:

 

Don.

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I will edit and update this post as Phillip 1882's results come in.

 

The third column shows the predicted values of [math]\varpi(x)[/math] based on

the best current physical measurements of [math]\alpha[/math] which Wikipedia

gives as being between [math]137.035999033^{-1}[/math] and [math]137.035999135^{-1}[/math].

 

Note that by the time we get to [math]x=10^{13}[/math], the predicted values of [math]\varpi(x)[/math]

will vary by about plus or minus 30, whereas at present, they only vary by

about plus or minus 3.

 

The fourth column shows the values that result when we solve for [math]\alpha[/math] in:

 

[math]B(x)*\left(1-\frac{\alpha}{\mu-2*e}\right)=\left(x-\frac{x}{\alpha*\pi*e+e}-\frac{1}{2}*\sqrt{x-\frac{x}{\alpha*\pi*e+e}}\right)*\left(1-\frac{\alpha}{\mu-2*e}\right)[/math]

 

Note that the fluctuations are slowly decreasing and that these values of [math]\alpha[/math]

seem to be converging on some particular value.

 

 

 

 

 

 

 

_______[math]x[/math]____________[math]\varpi(x)[/math]_______[math]B(x)*\left(1-\frac{\alpha}{\mu-2*e}\right)[/math]________[math]\alpha[/math]________

 

100,000,000,000_____64,036,148,166______64,036,147,783_________137.03593392608^-1

200,000,000,000_____128,072,369,864_____128,072,369,683________137.03598372618^-1

300,000,000,000_____192,108,604,710_____192,108,603,778________137.03594625176^-1

400,000,000,000_____256,144,844,029_____256,144,844,185________137.03600571478^-1

500,000,000,000_____320,181,088,566_____320,181,088,626________137.03600111220^-1

600,000,000,000_____384,217,336,898_____384,217,335,932________137.03597169223^-1

700,000,000,000_____448,253,585,852_____448,253,585,409________137.03598831869^-1

800,000,000,000_____512,289,836,587_____512,289,836,605________137.03599946950^-1

900,000,000,000_____576,326,089,252_____576,326,089,206________137.03599821781^-1

1,000,000,000,000___640,362,343,980_____640,362,342,983________137.03598213253^-1

1,100,000,000,000___704,398,597,754_____704,398,597,764________137.03599923616^-1

1,200,000,000,000___768,434,854,386_____768,434,853,414________137.03598530327^-1

1,300,000,000,000___832,471,110,338_____832,471,109,826________137.03599238683^-1

1,400,000,000,000___896,507,366,959_____896,507,366,915________137.03599854497^-1

1,500,000,000,000___???,???,???,???

1,600,000,000,000___???,???,???,???

1,700,000,000,000___???,???,???,???

1,800,000,000,000___???,???,???,???

1,900,000,000,000___???,???,???,???

2,000,000,000,000___???,???,???,???

 

 

 

 

 

 

 

 

 

10,000,000,000,000__?,???,???,???,???___6,403,626,165,690+/-30____???.????????????^-1

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Quoting Freeztar:

 

 

Your support is much appreciated Freeztar!

 

But to be fair, we must also remember that some Hypographers

put in a lot of time and effort into this project/experiment,

while others at least supported it with their comments and feedback.

 

Besides Phillip 1882, Yourself and Myself, there are Turtle, Donk, Modest,

Kharakov, Pyrotex, IDMclean, Craig D, Qfwfq, Rade, JMJ Jones 0424, Jay Q,

and if I forgot anybody, well, they know who they are.

 

Many people are afraid of things that are new and threaten the status quo.

Then again, many people are simply not bright enough to understand certain results.

Also, there are those who are simply dishonest and loath to acknowledge anyone's success...

exept their own! :lol:

 

Don.

 

I've always been a fan of Don and what you are trying to do here. :)

 

Unfortunately, It seems I'm always on the learning end of the rhomboid when it comes to mathematics. Hence I do the only logical thing and explore figurates. ;)

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Even if it takes a few extra weeks to reach [math]10^{13}[/math],

the effort will have been well worth it, because by then,

post #6 in this thread will have 100 entries that we can then

analize in all sorts of ways. (Graphs, correlation coefficients, etc.)

 

It will be a milestone, because by then, it should be relatively clear

as to whether or not the counting function itself is in fact complete

and the difference of plus or minus 30 in the prediction column

may even be sufficient for us to hazzard an educated guess as to

whether or not the actual value of the FSC is above or below

[math]137.035999084^{-1}[/math], which is the median value given by Wikipedia.

 

Mother nature does not easily give up her secrets, but if she did,

then we would not be on this most exiting quest after

the most mysterious number in all of science!

 

Don.

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sure no problem modest.

import time, math
t0 = time.time()
finish = 10**13
max = 10**6
current = max
row = 10
column = 6
inc = 4
total = 0
array = list(range(6,max,3))
check = False
file = open("numbers.txt","w")
while column+inc <= finish:
if column+inc >= current:
	array.sort()
	total += 1
	for i in range(1,len(array)):
		if array[i] != array[i-1]:
			total += 1
               file.write(str(current)+" "+str(total)+"\n")
               print(current,total)
	array = list(range(3*(int(current/3)+1),current+max+1,3))
	current += max
	column = 6
	inc = 4
	row = column*int((current-max-(column+inc))/column +1)+column+inc
	check = True
else:
	if check:
		array += list(range(row,current+1,column))
		column += inc
		inc += 1
		row = column*int((current-max-(column+inc))/column +1)+column+inc
		if inc %3 == 0:
			column += inc
			inc += 1
			row = column*int((current-max-(column+inc))/column +1)+column+inc
	else:
		array += list(range(row,current+1,column))
		column = row
		inc += 1
		row += inc
		if inc%3 == 0:
			column = row
			inc += 1
			row += inc
array.sort()
total +=1
for i in range(1,len(array)):
if array[i] != array[i-1]:
	total += 1
t0 =time.time()-t0
print(total,t0)

if anyone wants to help out, i would greatly apreciate it. (i really don't want to run my pc day and night for 200 days if i can avoid it.)

basically, just change the value of current to something large (perferably 10^13-10^12) and run in python. also you can eliminate the if check else block, along with the check variable. this is for the sole case of being less than 1,000,000. also, there is a small possiblity you may run out of memory. if you do, jsut change the value of max to something smaller. 10^5 should be more than small enough. finnally, change the line:

row = 10

to

row = column*int((current-max-(column+inc))/column +1)+column+inc

and place it below inc.

and change the line

array = list(range(6,current,3)

to

array = list(range(3*(int(current/3)+1),current+max+1,3))

apologies for not making my code more portable. :-)

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uh oh.

i recently passed the 500,000,000 mark, and got a different number.

i did have to pause the program, and restart my computer, and then re enter the program, so it may be a simple bug on my end.

i trust donk's number, and mine is far off. so if i get a different number for 600,000,000 i may have to rethink my algorithm.

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