Strange Numbers

[Tim_Lou]

those listed in the posts 153 to 155 are in the forms of

[math]2^{k-1}ab[/math]

while my theorem only deals with things in the form of

[math]2^{k-1}a[/math]

only one prime....

 

the theorem is indeed simply a special case of all the anomalies...but we are getting there.

well, im working on it... hopfully, as time passes, i'll find the rest of the patterns.

[Tim_Lou]

yes, it is correct. tim's theorem of anomalous strange numbers? hmm, i dont want to take all the credit... perhaps we will name it after we come up with more theorems.....

 

i'm currently working on anomalies in the form of 2^k-1*a*b.... so far i have a expression equating a function of a and b to a function of n,m and k. the problem is... well, its got a and b... more variables= more troubles. i have to express a and b into individual forms....

 

btw, have you guys found any odd strange numbers?? i guess not, perhaps there should be proof on that...

 

and for 54, i'm afraid the theorem wouldn't work on that, since it is not the form of 2^(k-1)*prime...

 

hmmm, 54 is quite an interesting anomaly, are there more anomalies in the forms of 2^(k-1)*a^(n)?

[Tim_Lou]

hmmm.... phat numbers, more readings to do....

 

i suppose that strange numbers are pretty "thin". maybe thats why it was on the list.

 

let's see, if we base on the assumption that no odd perfect number exists, perhaps we can "prove" that no odd strange numbers exist?

 

so much stuffs to consider now... hmm

[Tim_Lou]

nevermind

 

3 pages of scratch work and i thought i found something... well, yeah i found strange numbers in the form of mersenne prime*perfect number.... which is quite useless.....

[Guest chendoh]

I have been consumed by the Fraculator. I have been working on version 2 which includes many new functions. It is letting me brush up on my programming which had been sitting idle for about 12 months. Strange numbers is next on the list. I am going to take the code from you and try and build some interactive analytical tools around it. I will get the working tool to you in some fashion or another.

 

:beer: :eek_big: ;)

 

Now go spot Sasquatch! :QuestionM

 

Bill

 

TBD,

I downloaded Fraculator 2.0 a while back, and forgot about it until recently.

 

It unzipped to a 108kb application file, and when I clicked on it to open, it wouldn't (it failed to initialize properly (0xc0000135)).

XP pro sp2, 1meg ram

 

Used XP's un-zipper to extract......any updated info? Ideas? :shrug:

[TheBigDog]

TBD,

I downloaded Fraculator 2.0 a while back, and forgot about it until recently.

 

It unzipped to a 108kb application file, and when I clicked on it to open, it wouldn't (it failed to initialize properly (0xc0000135)).

XP pro sp2, 1meg ram

 

Used XP's un-zipper to extract......any updated info? Ideas? :hihi:

The latest release of it is here...

 

http://hypography.com/forums/physics-mathematics/5980-fractal-explorations-2.html?highlight=fractalator

 

...along with instructions (post 23).

 

You need to check that .net Framework 2.0 is installed. If it is then the program should just run. Let me know if that works for you.

 

Bill

[Guest chendoh]

.net Framework 2.0 is installed. If it is then the program should just run. Let me know if that works for you. Bill

 

TBD,

 

:hihi: That's what I need to check........When I DL eailier I had 98se, an I didn't want to get involved with net frame.

Now that I have xpsp2..I'll still have to check..

 

Will DL newer V.

steve

[alexander]

ok, i can make this happen, i just wanna know, is it [math]2^{53}*(2^5-2^{10}+2^{54}-1)[/math] or is it more of a [math]2^{53*(2^5-2^{10}+2^{54}-1)}[/math]?

 

i mean thats a difference between 162259276829204419242718052483072 and [math]2^{954763121002492523}[/math]

[alexander]

also i can write a script to generate these numbers, in that case i need to know:

 

if you are using equasion like [math]2^{k-1}*(2^a-2^{2a}+2^k-1)[/math]

how are k and a iterated and from what values, to what values (and i will write a bash script to generate loads of data and leave it running at work over a weekend or something, it will take time and effort, wanna do it right, and wanna do it once :)...

[alexander]

abundant by 56:

[math]2^{5}*(2^3-2^6+2^{6}-1)=224[/math]

[math]2^{6}*(2^3-2^6+2^{7}-1)=4544[/math]

[math]2^{7}*(2^3-2^6+2^{8}-1)=25472[/math]

[math]2^{9}*(2^3-2^6+2^{10}-1)=495104[/math]

[math]2^{15}*(2^3-2^6+2^{16}-1)=2145615872[/math]

[math]2^{18}*(2^3-2^6+2^{19}-1)=137424011264[/math]

[math]2^{21}*(2^3-2^6+2^{22}-1)=8795973484544[/math]

[math]2^{27}*(2^3-2^6+2^{28}-1)=36028789368553472[/math]

[math]2^{42}*(2^3-2^6+2^{43}-1)=38685626227417444939464704[/math]

 

abundant by 992

[math]2^{9}*(2^5-2^{10}+2^{10}-1)=15872[/math]

[math]2^{13}*(2^5-2^{10}+2^{14}-1)=126083072[/math]

[math]2^{16}*(2^5-2^{10}+2^{17}-1)=8524857344[/math]

[math]2^{25}*(2^5-2^{10}+2^{26}-1)=2251766494134272[/math]

[math]2^{28}*(2^5-2^{10}+2^{29}-1)=144114921519448064[/math]

[math]2^{53}*(2^5-2^{10}+2^{54}-1)=162259276829204419242718052483072[/math]

 

do you want me to run the division?.... eeh, i'll do it anyways...

 

This is to 100 decimal places, if you need more.... post .... :)

abundant by 56:

((2^{5}*(2^3-2^6+2^{6}-1))*496)/(28)=3968.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002267

(2^{6}*(2^3-2^6+2^{7}-1))/(28/496)=80493.7142857142857142857142857142857142857142857142857142857142857142857142857142857142857142857142903139

(2^{7}*(2^3-2^6+2^{8}-1))/(28/496)=451218.2857142857142857142857142857142857142857142857142857142857142857142857142857142857142857142857400696

(2^{9}*(2^3-2^6+2^{10}-1))/(28/496)=8770413.7142857142857142857142857142857142857142857142857142857142857142857142857142857142857142857147868807

(2^{15}*(2^3-2^6+2^{16}-1))/(28/496)=38008052589.7142857142857142857142857142857142857142857142857142857142857142857142857142857142857142878861744336

(2^{18}*(2^3-2^6+2^{19}-1))/(28/496)=2434368199533.7142857142857142857142857142857142857142857142857142857142857142857142857142857142857144248210399733

(2^{21}*(2^3-2^6+2^{22}-1))/(28/496)=155814387440493.7142857142857142857142857142857142857142857142857142857142857142857142857142857142857231893935680282

(2^{27}*(2^3-2^6+2^{28}-1))/(28/496)=638224268814375789.7142857142857142857142857142857142857142857142857142857142857142857142857142857143221842439322500451

(2^{42}*(2^3-2^6+2^{43}-1))/(28/496)=685288236028537596070517613.7142857142857142857142857142857142857142857142857142857142857142857142857534450420587735769183152922

 

abundant by 992

(2^{9}*(2^5-2^{10}+2^{10}-1))/(28/496)=287909010.2857142857142857142857142857142857142857142857142857142857142857142857142857142857142857143021662291

(2^{13}*(2^5-2^{10}+2^{14}-1))/(28/496)=2233471561.1428571428571428571428571428571428571428571428571428571428571428571428571428571428571428572704840892

(2^{16}*(2^5-2^{10}+2^{17}-1))/(28/496)=151011758665.1428571428571428571428571428571428571428571428571428571428571428571428571428571428571428657721004951

(2^{25}*(2^5-2^{10}+2^{26}-1))/(28/496)=39888435038949961.1428571428571428571428571428571428571428571428571428571428571428571428571428571428594221962879399977

(2^{28}*(2^5-2^{10}+2^{29}-1))/(28/496)=2552892895487365705.1428571428571428571428571428571428571428571428571428571428571428571428571428571430030224511707066117

(2^{53}*(2^5-2^{10}+2^{54}-1))/(28/496)=2874307189545906855156719786842989.7142857142857142857142857142857142857142857142857142857142857142858785318394026232488660982735338851

 

so these results are not quite accurate though

 

[math]1/(\frac{28}{496})=\frac{496}{28} or (1*496)/28[/math]

you see how there is a pattern that falls apart at the end numbers? there is a degree of inaccuracy due to the way i calculated the numbers, i will rerun them to confirm my suspicions:

 

example:

how it was all ran:

 

(2^(5)*(2^(3)-2^(6)+2^(6)-1))/(28/496) = 3968.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002267

 

how i should have ran it:

((2^(5)*(2^(3)-2^(6)+2^(6)-1))*498)/(28) = 3984.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

 

you see the inaccuracy.....? that is due to the order of operations, since 28/496 creates a large irregular number, it is cut off at 100 digits due to scale, so the rounding at the end of that number is what accounts for the minor inaccuracy, but that inaccuracy is big enough to where makes the numbers look different at the end

 

((2^(53)*(2^(5)-2^(10)+2^(54)-1))*496)/(28) = 2874307189545906855156719786842989.7142857142857142857142857142857142857142857142857142857142857142857142857142857142857142857142857142

[alexander]

do you want me to rerun the calculation, or can you see that all, except for one number on the list, are not divisible by 28/496?

[alexander]

lol, i thought something was strange when you posted the whole 28/496.... but i rolled with it...

 

ok, let me check for abundance :hihi:

[alexander]

actually first i gotta do my physics hw real quick, bbl to run the check :hihi:

[alexander]

so here are some conclusions i have been looking at (sorry for not having pretty math here, worked on it during physics yesterday... 2 hours on adding vectors... what can i say... bewildering boredom!)

 

2^{5}*(2^3-2^6+2^{6}-1) = 224/28 = 8

2^{6}*(2^3-2^6+2^{7}-1) = 4544/28 = 162.28571428571428571428

2^{7}*(2^3-2^6+2^{8}-1) = 25472/28 = 909.71428571428571428571

2^{9}*(2^3-2^6+2^{10}-1) = 495104/28 = 17682.28571428571428571428

2^{15}*(2^3-2^6+2^{16}-1) = 2145615872/28 = 76629138.28571428571428571428

2^{18}*(2^3-2^6+2^{19}-1) = 137424011264/28 = 4908000402.28571428571428571428

2^{21}*(2^3-2^6+2^{22}-1) = 8795973484544/28 = 314141910162.28571428571428571428

2^{27}*(2^3-2^6+2^{28}-1) = 36028789368553472/28 = 1286742477448338.28571428571428571428

2^{42}*(2^3-2^6+2^{43}-1) = 38685626227417444939464704/28 = 1381629508122051604980882.28571428571428571428

 

2^{9}*(2^5-2^{10}+2^{10}-1) = 15872/496 = 32

2^{13}*(2^5-2^{10}+2^{14}-1) = 126083072/496 = 254199.741935483870967741935483870967

2^{16}*(2^5-2^{10}+2^{17}-1) = 8524857344/496 = 17187212.387096774193548387096774193548

2^{25}*(2^5-2^{10}+2^{26}-1) = 2251766494134272/496 = 4539851802690.064516129032258064516129032258

2^{28}*(2^5-2^{10}+2^{29}-1) = 144114921519448064/496 = 290554277256951.741935483870967741935483870967

2^{53}*(2^5-2^{10}+2^{54}-1) = 162259276829204419242718052483072/496 = 327135638768557296860318654199.741935483870967741935483870967

 

conclusion from more testing:

 

when the equasion is such that:

(2^(c-1))*(2^(c/2)-2^©+2^©-1) - the value is divisible by ((2^c-2^(c/2))/2)

reason is, look at what happens when you simplify the equasion:

(2^(c-1))*(2^(c/2)-1)

 

and the result is 2^(c/2)

[alexander]

btw, i can check what the factors for other numbers are, i just gotta write a bc program to do it :evil:

[alexander]

2^(c-1))*(2^(c/2)-2^©+2^©-1) - the value is divisible by ((2^c-2^(c/2))/2)

 

btw, just if you didnt see what i was talking about:

 

take c=10

 

(2^10-2^(5))/2 = 496

 

c-1 c/2 c c

2^{9}*(2^5-2^{10}+2^{10}-1) = 15872/496 = 32

 

but yeah i'll work on a factoring program in bc (probably tomorrow, or later on tonight, shoot it will be a up till one or two, up at 7am knights... been getting lots of those) :evil:

[alexander]

ok, so bc program to factor :)

 

scale=0
print "nNumber to factor: " 
num=read()
numiterations=num/2
if ( num%10==1 || num%10==3 || num%10==5 || num%10==7 || num%10==9 )
{
numiterations=(num+1)/2
}
print "Factors: n"
for (i=1; i<=numiterations; i=i+1)
{
if ((num%i)==0)
{
i
}
}
print "n"
quit

 

lets test it, post like 4-5 numbers and run them on your factoring machine, lets see what you get, and then i'll run it in my bc prog, to make sure it works right (there is a load of problems with setting scale wrong in this prog...