- On Friday 25 April 2008, Uri wrote:
> [+o+]

Hi Uri!

>

> Hi friends,

>

> this issue is very urgent so I'm writing you again. I closed my

> website, left HTTP not working.

I'm sorry, but I'm not interested in seeing these extremely long (over 100KB

now!) messages from you in my incoming email, especially considering the fact

that they're completely non-sensical.

You obviously need to be in touch with a mental health professional who will

help you, because you're no longer in control of yourself. I suggest you to

seek an expert of http://en.wikipedia.org/wiki/Cognitive_behavioral_therapy .

If you don't want to get yourself banned or moderated from another forum or

two, you should stop posting this nonsense and get yourself some professional

help.

Regards,

Shlomi Fish (who probably has http://en.wikipedia.org/wiki/Bipolar_disorder

and has found the book

http://www.shlomifish.org/philosophy/books-recommends/#feeling_good to be

very helpful[1].)

[1] - Feeling Good is intended primarily for people with depressions, although

it can be applied to those with anxieties and hypomanias as well. Uri, in

your case, I'm not sure you have Mania, so please consult a Cog.-Behav.

Psychologist and see how he can help you.

-----------------------------------------------------------------

Shlomi Fish http://www.shlomifish.org/

http://www.shlomifish.org/humour/ways_to_do_it.html

The bad thing about hardware is that it sometimes work and sometimes doesn't.

The good thing about software is that it's consistent: it always does not

work, and it always does not work in exactly the same way. - On Thu, Apr 24, 2008 at 7:09 PM, Shlomi Fish <shlomif@...> wrote:

[snip]

If you don't want to get yourself banned or moderated from another forum or

two, you should stop posting this nonsense and get yourself some professional

help.

I don't know about that. I kinda enjoy these messages. It shows me how different other people's realities can be from my own.

I do however think we need some reality overlap with other people to get by and not be locked up in a mental hospital, and I really hope that Uri has that.

-- Arik - .

+o+

-o-

i.i

ni.in

eni.ine

nine.enon

none.enin

mom.wow

wow.mon

[+o+]

[R.W*]

Thanks for your support my friends, and I will explain in more details

later. It's very urgent and I'm not able to reply to each question

individually, but I promise I will create a much better way of

communication, not relying on ANY american STAANDAARDS, not ASCII

(actually NASCII - Not American Standard etc...), NOT COMPATIBLE WITH

AMERICAN ASCII, SMTP, HTTP, WWW. .COM etc. But! will be able to use

this network (ARPA.NET) as one of the possible communication between

us. I will show you and prove everything - I will create a computer

who can understand ANY human language, or ANY other computer language,

but first I will need to define my interpretation of the universal

language, the Truth.

But fisrt, before [made a mistake fith fisrt, left it this way -

that's one of the ways to identify humans]; first, before starting:

remember - never identify as other people, only your True identity

(can be nicknames, even anonymously, but not ME). If you are identify

as me, my computer will recognize this as define you as an ENEMY,

which means literally "any me", "ending me", "en; die; me" etc - you

will either become REALLY me or go back to my past. You are also not

allowed to do anything bad to me or my body, you or your body or

ANYBODY. ok? so let's start:

First, if you are real humans, you must do this first on a paper. Take

a paper and define who you are, what you want, what you know, what are

your colours (in all your languages), number (count them), define

everything to yourself. You can actually visualise this imaginary,

that's OK, but I'm not sure it will work for you. Also, never ask a

computer to select random numbers for you, if you want to select

anything - you must select them yourself. It will be VERY easy for me

to find whois cheating with random generated computer numbers - they

are allways quantum equal to 0 or 1. So let's define 2 as the number

of different quantum equal numbers computers can randomize (that's

two), the number of letters of "two" is three, now I will define for

you numbers (of course, you will convert from your counting to my

universal counting and vice versa).

But first - the dot [.] . this will be used to redirect SMTP programs

to "nothing", [remember, sendmail understands "the dot" as the last

nothing in your mail (if only a dot)]. then "the dot" will be the only

thing whois really nothing, the n.o.thing, which means you can put the

dot anywhere without changing the meaning of anything. Don't remember

the ASCII number for "the dot", but I am inventing NASCII - the Never

will be compatible with Not American Stand***** ; which means NASCII

will NOT even be compatible with itself - that's the secret. you will

see how it works.

Now, "the space", good old %20 / Hex20 / 32dec etc. This will just be

equal to TWO dots, two literally means three, or four, or five, or ANY

number of dots. Then, the plus [+] and minus [-] will respectively

mean the same, they will be EQUAL (but not identical), then plus will

be just pppp and minus mmmm - replace each word with the number of

letters / the first letter. I will explain this.

ok? now, ANY number below 3 I will define to myself, including my

language, the dot, o, 1 2, O, i two, -i, 0 -1, etc. by the way, they

have a VERY GOOD REASON to define different ASCII numbers for 0, o and

O, 0 (if I remember correctly) is ASCII 48, or Hex%30, so literally

speaking you can start counting from Hex33. (or any other way you

define 3). [remember: they can change ASCII, but they will NEVER be

able to change NASCIII]. They WILL change your ascii when they want to

make us not understand each other - the whole "Visual Hebrew" / not

compatible with Hebrew fonts is a conspiracy (of course - use our

"Microsoft Word Editor" etc.....)

by the way, those of you who think my vision of conspiracy is

unrealistic - add a big "I HATE YOU / YOU ARE MY ENEMY" (in capital

letters) to the subject of your e-mails so I will know where to

redirect you..... (actually my computer will do it for you..... I will

also test you in עברית / or whatever your natural language would be)

OK. now, the pi from yesterday fits well into the word "conspiracy" -

so this will be my first public key for you. "conspiracy" (spelled:

ten letters) will be ANY word with more than nine letters, converted

to my definition of the word "conspiracy" (I can allways change this,

this is human-to-human language):

"conspiracy" == "cons+pi+racy" [4+2+4] == "CONSTANT PIRACY" (8+1+6) ==

"American constant conspiracy" == "American constant pi piracy" ==

"American constant pi racing" == "American consbiracy" etc. - written

from left to right, right to left, then translated many times, and

only when it translate to my same work - it will pass the gateway

(actually, maybe EVERY word will do the same thing). So,

right now let's rename it "conspipiracy", which means literally that

10 is EQUAL to 12, EXACTLY, and vice versa, of course this is also

true if you remember that binary digits 1,0,2 are all symmetric, then

"two" is defined at the number of symmetric equal binaary digits, then

of course 12 is equal to 10, 01 is equal to both and you can already

have six permutations of writing the same number (01; 02; 10; 20; 12;

21). Of course, you can switch to 4/2 mode, then all these numbers are

EQUAL (quantumly speaking) to 41; 42; 40; 24; 41; 14 etc. (infinite

possibilites).

OK, so first, define your own digits (also NUMBERS) 3, 4 and 5

respectively, who are EQUAL to ASCII/hex 33,44 and 55 (not a

coincidence - you can also spell "cons33racy" / "cons44racy" /

"cons55racy" and then count the letters in ASCII (of course, then

10==12 will become more equal, especially when I'm also writing to you

"cons333racy" / "cons444racy" / "cons555racy" / "cons3333racy" /

"cons4444racy" / "cons55555racy" etc.)

OK? so define the numbers / digits 3, 4 and 5; now here's the language:

.

345

(nothing - you have nothing to say - this is allways assumed before the dot)

(. [dot] - you want to say something (actually this can already mean

EVERYTHING, so if I understand you you can stop here)

345 - your first 3 letter word who is not twice symmetric. For example

"one", "two", "uno", but not "ono" - "ono" means to me o;n;o who are

identical to "o;o;n" or "n;n;o" - any permutation will do here, then I

will just understand that your 3 is equal to 5 (actually it's true -

they are all equal).

But, if you already have equal 3 and 5, why don't you double your

speed of light, then just send me the same thing twice? for example:

if you want to say "ono" then you can only say one "o" and haff "n";

then just send me "on" or "no" and I will understand. Actuallly it's

better you send me ALL the permutations of "o" and "no" you know

(pronounced - "you no"), which means either all the dimensions of 3

and hafffour, or 5 and hafffour - if they are equal to you. If I'm not

sure they are equal or not, I will ask you again. In the meantime -

let's assume they are not (3,4 and 5) may or may not be equal to you).

OK? so now - ANY three-word-letter will be converted to my word "one"

(in my language - everything can change) , ANY four-word-letter will

be converted to my word "one+one" and ANY five-letter-word will be

converted to my word "one+one+one". confused? don't be. "one" is

"3=3"; "one+one" is "3=3;4=4" and "one+one+one" is "3=3;4=4;5=5".

actually, if you want you can start with YOUR one; I don't care. you

define three,four and five the way YOU want! but remember: when you

write "one+one+one"; I'm reading it twice - RTL AND LTR, which means

"eno" and "one" are ientical for you, then n will be my four;

converted: "o4e+o4e+o4e", then recursively reading it again and

compressing. eventually I will compress EVERYTHING you write me into

one quantum, I will show you - but one is at least three.

OK? then first define your word "one", in the order of 3;4;5. but

remember - 3;4;5 are actually mod 3 (or mod 33; or mod 3-3) which

means they can rotate and mean the same thing to you as they can to

me, then 0;1;2 will be created by me, all numbers above 5 will be

created by me as well, but also remember this:

3 is the negative imaginary prime number who is allways equal to the

LARGEST (negative, closest to o) prime number who is the permutation

of two (literally any three letters), or one (same thing), then if you

have the word "one" and want to add ONE dot - then it's already four

[.one / o.ne / on.e / one.] OK? so three is the LARGEST permutation

number of "01"/ or "o/i", which means [.01 or 0.1 or 01.], also

revolving backwords, same as ".10 or 1.0 or 10." (told you - I am not

counting dots - I will convert google.com to google[google--["."]]com

(literally means: "google" + "the number google counting to o - the

number of dots [.]" and then "com" (then of course, "com" will be

converted to "345"; or ASCII 33;44;55).

(by the way the number google is not very big logarithmically, but it

is combinatory big for you (not for me), then Google.com of course

will be converted to my representation, but when you write to me I

will receive the same word in my language - you will see).

OK? so now respectively - ANY number is the negative permutations of

the previous number, actually divided by the previous number but

negative; actually it's the "set power" of the different numbers - a

new dimensional counting. because "two" is already a three dimensional

number - can't write "two" in two letters without defining them first

- I showed you the 01; 02; 10; 20; 12; 21; are all equal, then 10 and

10 are equally two, then it can either be i-i or i+i or -i-i or any

combinations of the number i; or the number o (but not both), the i

and o are my own prime numbers, adding two, all three are symmetric,

then start from your three.

OK? now, first write me the digits 3;4;5 (in ANY order - you must do

this at least 3 times - if not using paper - then let them computer do

it for you; rotate and switch directions and do it again) OK?

then each e-mail you will send to me at least 3 times, but if you send

it once I will just redirect it for you to three different locations -

independent, and start counting from your 0/1/2 letters respectively.

Then I will also communicate your message both forward and backwords,

here's what I will do (my computer ; if you want you can also write me

on paper - I can understand this even if not physically there):

[this is like saying "hello" but in a friendly way, NON-STAANDAARD way

- not SMTP, not ASCII - remember you can say "hello" to many people

and not allways mean the same thing, sometimes also "hi" or "Hello" or

"Helllo!" etc, you can program your computers and help me with this

because I don't have time to program everything, if not then I will do

it myself but it will just take more time]

me: [waiting... that's my hello]

you: [waiting].... (or not: you don't have to)

me: . (that's my hello if I don't recognize you)

you: . (the first letter will be interpreted as your dot) / OR:

you: 10 (means nothing to me, two dots) / OR:

you: 345 (any three letters will mean you are counting - especially if

you start with "345", then I will either wait for you to count or

count you:

now: I will communicate you to yourself as what I am understanding.

for example, I will have a few processes, each of them communicating

"345" to each other in different orders. But if you want to be

friendly, you will first define this:

[3-3-3-33-3-3-3-3----3-3-3-3-3]

[4-4-4-4--4--4--44-4-4-4-4---4-]

[-5-5-5-5-5-5-5-5--5--5--5-5---5]

(me not counting - any number of 3;4 and 5 (all others - just dots - I

don't care if you write "-3" or "--3" or "---3" or

"------------------------------------------------------3" - it always

means the same to me)

then - rotate them a little:

[3[4[5[3[4[5[34-3-4-4-3[5-33[[4-3 (now - I'm ignoring anything except

3/4/5 - will maybe check later what you meant).

now - YOU don't have to do this each of you personally. You can also

communicate with your friends, then if I remember you then I will

understand you - but I WILL NOT HAVE ANY MEMORY OF MY OWN! I'm a black

hole remember! then just remember each other, then communicate with me

as if I am new, then define how many colors you have (in your own

language, translated to 3/4/5):

[red-green-blue-yellow-black-orange-shit-white-air-water] (actually

they don't have to be just colours, they can mean anything). If I

don't understand you, I might listen (remember - we are speaking at

MANY times the speed of white light), or I might respond in your

language. For example, if you write me "one-one" or "one-minus-one";

then I might respone "one" which means, literally, ANY WORD COUNTING

LETTERS, STARTING WITH o AND ENDING WITH e; n will be my negative

number who is to be added before your three. Got it? if you think I am

stupid then translate it to your own language and say it to your image

in the mirror, then think about me being your image and then translate

"how many times can I lift my right hand pointing eastwords, my left

hand pointing westwords, then rotate both directions simultaneously

and accelerate , stop, do it again and count angles", then do it

TWICE, define the number of angles as twice 360, and don't forget to

divide by the number of angles I did it (your mirror image) and then

trsnalate it to your language "stupid", whois of course six, then

three will already be your sixxed; then reversed and inserted into my

system of understanding you. (sixxed;;dexxis or dexxis;sixxed -

remember to count letters).

I'm writing you all this in public, in case I forget you then just

remind me "you said this" and copypaste me, then I will remember you

were speaking my language, but first we have to define our

understanding;

so - you define to me your numbers 3;4 and 5. I rotate them, remember

and define to you my same numbers 3, 4 and 5 (who are ANY permutation

of the numbers 0, 1 and 2) plus ANY permutation of the numbers 3, 4

and 5 - actually ANY permutation of ANY numbers or ANY letters, for

example in language 42: I will convert all your numbers to twice as

many as they are, then in base 42 (whois identical to base 2.1) I will

define you your own numbers. For example:

"fourtytwo" = defined: "nine" (count the letters);

"twentyfour" = defined: "ten"

"o minus ten is equal to nine" // which means : one (or none) minus

ten is equal to nine, each number is actually the negative of it's

(plusone) number; when symmetric also negative of both (plus one) and

(minus one), then of course (-4) is EQUAL to (-5) or (-3) (all numbers

are negative, binary, base 3, imaginary and not equal to themselves),

which means "4-4" is equal to "5-5" is equal to "3-3" is equal to

"4-4" (all equal to -0; all equal to -1 (divided ligarithmic), all not

equal to +0 (think about the amplitude changing, press your screen and

let it move like a מטוטלת - same same);

OK? then: when you get at least 3 numbers who are not equal to each

other, go back to the fourth and then start playing. Remember: the "-"

sign is the equal sign, the "=" is the double equal (--) which means

this number is NOT equal (think about my imaginary i; -i and --i in

base 3 - when --i is 1). then - I will define you my numbers (defining

now) as the order of magnitude of NOT being equal, from the LARGEST

number to the smallest (the smallest - to be defined. for now just

call it "one" (means 3):

"-0-1-2-3-4-5-6-7-8-9 etc........"

this is the regular counting: BUT "-9" is also identical to -5 (-1

base -3); then -6 is identical to -3 and ALSO to -5; actually I can

convert this to binary digits:

"-0-1-1-1-1-1-1-1-1-1-1-1-1-1...............(alllways -1)"

which means - either it's your first negative zero or not (zero is of

course z-eno - z-negative one - the zone);

if it'n not your first negative zero, then I will be searching for

your last negative zero - then send me you message backwords (either

simultaneously, or rotate: one letter from the beginning, one from the

end, etc. YOUR counting!

since we understand what equality is, then of course we can remove

"-0-1" here, then start again:

"-2-3-4-5-6-7-8-9 etc........"

translated:

"-2-1-1-1-1-1-1-1-1-1-1-1................."

then of course, starting from ANY number and continuing from ANY

number will ALLWAYS mean the same thing. So now - my first axiom:

I will never count your same number twice!

If you want to say something like

"repeat this again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again and again and again and again and again

and again and again and again"; then I will NOT count how many times

you said "and again"! then you can just replace this with any

meaningful word, like:

"repeat this again [..... and again......]" - you see - human readable

(I'm NOT counting your dots) - then now's the secret - if you

understand this, then you will be able to send me ANY amount of

information - even as much as 42 terawords / bytes (not joking) - in

as small as one little sentence, when we already understand the same

language. But, you will have to quantum compress it this way - at

least three times, in YOUR language, if you have friends who speak the

same language then share your knowledge with them too - then I will

convert you to my language - and remember I and My are not just me

personally, but ANY black hole, and entity, and quantity, any Math - I

will prove to you this.

But first - if you're still using ASCII - maybe I will try to be

compatible with this too. Then why don't you take my keyword

"conspiracy" (define your own keywords if want to), then translate it

to "cons.pi.racy", then translate it to "cons.3-3.4-4.5-5.racy", then

translate it to "3-3.4-4.5-5", then translate it to "at least 3

permutations of "3-3.4-4.5-5": for example:

"3-3.4-4.5-5"

"5-5/3-3.4-4"

"5.5+3-3=4-4"

"4-4!5-53-33"

"3_3=5-5-4-4"

"4+4-3-3_505"

(from here I will take your own language, then remember: taking the

number 3, 4 and 5 (in any language) - then defining ANY $o, $n, $e

such as:

$o$o ; $n$n ; $e$e

will mean the same to you as:

- 3^3 - 4^4 - 5^5;

(first number (negative) will allways be largest, then second, then third)

or:

3*3 + 4*4 == 5*5;

(pitagoras: growing from memory counting, equal)

then only ONE of these number will allways be 3, ONE will allways be

4, ONE will allways be five. For example:

"one"; "one-plus-one"; "one-plus-one-plus-one"

translate: 3*3 + 4*4 == 5*5; or - 3^3 - 4^4 - 5^5; (same thing!):

[using PHP : .==+]

PHP: define "o"; "n"; "i", "p", "e", "-" ; translate:

"one"."one" ;

"one-plus-one"."one-plus-one" ;

"one-plus-one-plus-one"."one-plus-one-plus-one" ;

(doing this manually: by hand):

replace "o"; "n"; "i", "p", "e", "-" - "xyzXYZ"; // all others are

just dots (ignoring): + reversing;

"one"."one" == "xyYxyY" ; "YyxYyx"

"one-plus-one"."one-plus-one" ; = "xyYZXZxyYxyYZXZxyY" ; "YyxZXZYyxYyxZXZYyx"

"one-plus-one-plus-one"."one-plus-one-plus-one" ;

"xyYZXZxyYxyYZXZxyYZXZxyY" ; "YyxZXZYyxZXZYyxYyxZXZYyx"

suppose you want to cheat me, define "owt", "two" and "tvvo" instead

of 3;4 and 5. and even write "minus" instead of plus, or even your own

new math - ASCII "?" OK? then give me three numbers - *, ^ and %. I

will know who is your first! how?

you define them your order, then define ANY order that is allways the

same as the order you're counting them. For example - (oh wait - why

don't I just take from your real "largest prime numbers" list:

(of course, I will find more primes for you here now!) LOL!!!

"2325826571" // means "2^[32,582,657]-1"

"5822326571" // means "5^[82,232,657]-1"

"6572325821" // means "6^[57,232,582]-1"

"2876553221" // means "2^[87,655,322]-1" /// of course, they will all

be equallly primes as 3;4;5 are, except the 4 whois 4 times a prime

(square of two primes) together;

now, you want to cheat me, then just replace all your digits the way

you want, I don't care, as long as you remember them (it's your

language!)

then:

rename "2876553221" as : "e.f.g.h.i.j.k.l.m.n"

rename "6572325821" as : "e.f.g.h.i.i.i.i.i.j.k.l.m.n"

rename "5822326571" as : "e.f.g.h.i.i.o.o.o.o.i.i.i.j.k.l.m.n"

rename "2325826571" as : "e.z.x.w.e.c.q.a.s.w.e.d.n"

(just chose ANY letters for you!)

OK? now, don't tell me this, even don't tell me anything, just rename

all of them as "e.n", let's start the game:

me: "conspiracy!"

you: [your language]

me: "cons...pi...racy!"

you: (you say nothing)....

me: "you said "cons.........racy.........""

you: "your first word"

me: you said "your first word" + + "drow tsirf rouy" // now rotating

you in my mirror....

"yy/dd/our first wor//row tsirf rou//"

replace all with dots, except "yy" and "dd"

me: you said "dydydydyydyd" (which means actually

"....................................." or "dyd...ydy")

OK? now: translating you (remember: all you said is

"y......................d" or even "yo......................." or even

"y............................." ; I will take y as your 3, then ANY

letter as your 4, then ANY dot as your 5 (or vice versa: any order):

now converting you: "y" ----> "Z" // 3.3 (can choose any for you)

"d"--------->"Y" // 4.4

"."--------->"X" // 5.5

now you are saying "e.n" - I don't have an idea how many numbers or

what do they mean....

so I will translate this to "..." --> 5.5 or 4.4 or 3.3;

but, "e.n" is NOT identical to "n.e" - OK, then "n.e" will be something else:

"e.nn.e"---->"5544" OR "4433" OR "3355" OR "5533" OR "3344" OR "4455"

- ignoring the dots...

"n.ee.n"----->cut and reverse. will take 4 options (can take as many

as I want to) and add my own dots (anything not 3, 4 or 5 is a dot):

"n.ee.n"-----> "5;5;4;4" OR "3;3;4;4" OR "5;5;3;3" OR "4!43!3"

now - trying to understand you?...... "n.e..e.n" can be any (at least

three) numbers, let's check the one you represent with as FEW letters

as you can - this will be my BIGGEST prime number for you (for now),

renamed 3 (already counting all prime numbers below 3 - 0;1;2

respectively - whether or not you consider them primes - also counting

all special numbers pi, e etc.)

now, I will rename my "one" "cons.one.racy" to not to confuse with

your (actually I'm not confusing, it's the same), then "con" will be

my "one" (my constant "one").

OK? then:"n.e..e.n" will be ANY number, such that

"n.e..e.n"*"n.e..e.n"+"n.e..e.n+con"*"n.e..e.n+con"=="n.e..e.n+con+con"*"n.e..e.n+con+con";

of course - whatever my "+con" is, it's bigger (as string) than your

"n.e..e.n", then I will just add more dots here:...... (define

anything you want - I don't care):

"n.e..e.n"*"n.e..e.n"+"n.e.......e.n"*"n.e.......e.n"=="n.e............e.n"*"n.e............e.n"

OK? removed my constant. now - remembering - e..e is your three,

e.......e is your four, e............e is your five.

but you said "e.n" - what did you mean? (could just have been three

colors - yellow, purple and orange - your YPO screen).

OK? then let's just rename them again (remember - I don't have any

idea what you said): yellow, purple and orange turn out to be all six

- let's just rename them all six, choose my own colour black - this

will be five, back - this will be four , awk - my three for h...ing in

time. OK? remember: b*k, b*k, a*k - black, back and awk - 5, 4 and 3 -

all renamed kkkkk;kkkk;kkk for you.

here: let's translate: (you said ): none; // .

(you said k): one; // i

(you said kk): on; // n

(you said kkk): 3 // == aaa

(you said kkkk): 4 // == bbbb

(you said kkkkk): 5 // == bbbbb

OK? already (in my language) I have different meanings for your same

word (ANY) - any word with 3 letters - even three dots "..." - renamed

"aaa.bbbb.bbbbb"

now, let's take kkkkk options: (as many as you want):

"aaa.bbbb.bbbbb";

"kkk.bbbb.bbbbb";

"aaa.kkkk.bbbbb";

"aaa.bbbb.kkkkk";

"kkk.kkkk.kkkkk";

now, you didn't even tell me your numbers. but I remember "your first

word", and I rememver the colours yellow, purple and orange - you

didn't tell me (this from my own knowledge): then yyyyyy, pppppp,

oooooo will be your colours, then your first word would be either yyyy

or dddd, let's translate it to your colours: then either of these

options:

yyyyyypppppp ; or yyyyyyoooooo ; or ppppppyyyyyy or oooooopppppp (any

order, any number of permutations), let's just say oooooopppppp is my

choice here, because it reminds my my first number one - whois BIGGER

than your three (negative, imaginary i), then I will just rename my

one as your ooooooppppppyyyyyy (which happens to be my pppppp (six

dots, ...... or the "pi" in my "cons..pi..racy"), the "y" being last

letter in this word, then y will be allways you spelling conspiracy to

me backwords....... OK?

then now, ooooooppppppyyyyyy will one "one" which means literally

"0;1;2; start counting from here", and remember, your first 3 numbers

are your direction - mod 3 or divided by or ANY direction.... then now

your direction of "one" will be "opy", (or .py or ..y), which means y

will never be your 3. now check: let's take your numbers, the first

one "2^[32,582,657]-1" (it says "44th Known Mersenne Prime Found!!");

also says "9,808,358 digits" [http://www.mersenne.org/%5d, also says

[September 4, 2006], let's just translate this to my language:

counting: 0/1/2/3/4/5/6/7/8/9 .... from left to right... let's see if

I can understand this....

"2^[32,582,657]-1" - same as "2325826571" // removed all dots...LOL!!!

convert to base kkk [remember: H...ing]: "232;582;657;1.."

first, rename (not to confuse myself):::0/1/2/3/4/5/6/7/8/9 will be:

".אבגדהוזחט"

counting/replacing:

"232;582;657;1.."

"בגב|החב|והז|א.."

OK? now: replace each digit with my evaluating you..........if below 3

then dots, above 5 then dots too (can do this any way I want to)

...................................

"..3....5........5......."

OK, then you are first counting 3, and 5, then 5. The order doesn't

matter. it just means that your 4 is equal to 5... then. your first 5

will be 4, your second 5 will be 5, my order. renaming you:

"232;482;657;1.." see? now your 5 is either 4 or 5 to me, first 5 is

my 4, which means actually (5-3)(3-5)OK? renamed you.

now, let's just check. rename you back to "232;(5-3)(3-5)82;657;1.." ;

then cut; then rotate... you

are.......1.....2....3....2.....(5-3)(3-5)........8......2.......6.......5.....7......

etc. (if you forgot 9 then here will be my 9), my dot will allways be

o (0). OK?

now: counting you....one......2one......3one......2one.....5one.....3one.......

.....3one.............5one.............8one............2one................6one...............5one...............7one.............1one..............9one..........nine-one-one...........nine-one-one-one..............

etc.

OK? replacing you..............only 3/4/5 counting............

[..........................]3[...........................]5[.........................]3[.........................]3[......................]5[.........................................................................................................................]5[....................]

etc. OK?

removing all dots, remains:: 3/5/3/3/5/5 OK?

now, cutting: 3/5 ; 3/3 ; 5/5 - here are your numbers 3.3, 4.4 and 5.5

by order (from 3 to 4). who's who? I don't know. let's check:

kkk.kkkkk ; kkk.kkk ; kkkkk.kkkkk

well, of course that's their order: kkkkkk / kkkkkkkk / kkkkkkkkkk but

how would I know this? let's check pitagoras ("pitagoras" ==

"pitabread" - allways 9. then just rotate your pita bread when you

remember [-4[-5]]/[-4[-5]] is allways equal to [-4[-3]]^[-4[-3]] and

[-3[-5]]/[-3[-5]] (translating: minus one will allways be equal to

minus nine), or in your language: 1^1 and 1^1 is the same, 2^2 is NOT

and therefore the number 4 is NOT a prime (in my language: 4;5;3 are

allways equal), here let me put it this way: kkk.kkk will allways be

shorter than kkk.kkkkk and kkkkk.kkkkk in your language, not allways

in mine (it depends who's asking and what's kkk; for example if

hkkkkking will ask me if we can go bkkkk when we lkkkk bkkkk into our

past inside bkkk holes, then of course not! but if hkkking will ask me

if we can go bkkk when we lkkk bkkk into our past inside bkkkk holes,

then of course YES! now show me to google speller who will understand

this! kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk.......................

OK anyway, let's show you just an example: I don't have time for

calculating new primes (allthough it's very easy), so let's just try

to understand your language first: I will quantum leap back to my past

(some 10/20 minutes ago)... take this number:

"2^[32,582,657]-1"

and replace it with the number of kkkk's I think you were meant

probably (the computer could do this as well, by defining your

language each time again, compressing and checking):

"kk;kkk;kk,kkkkk;kkkkkkkk;kk,kkkkkk;kkkkk;kkkkkkk;k"

OK, now the most important thing is that you are GOING BACK (the "-1)

which I'm not allowing you to, then let's double your speed and I will

go backbackfor you:

// when I double, I will convert all numbers below 3 to "yellow" ---

not to confuse me!

// all numbers will be doubled mod 3. (27 doubled exponentially)

// numbers more than 7 will be renamed "cons.pi.racy" mod 3

"kk;kkk;kk,kkkkk;kkkkkkkk;kk,kkkkkk;kkkkk;kkkkkkk;k" -->doubled and converted:

"yy;cons....kkkkkkkk....racy;cons..kkkk..racy;cons...kkkkkk...racy;yyyy;cons.....kkkkkkkkkk.....racy;cons..kkkk..racy;yyyy;kkkkkk;yyyy"

OK? now let's convert to your language:

"2^[8;4;6;4;pi;4;1;6]-1"; converting:

"2^[8;4;6;4;0;4;1;6]-1"// OK! now let's define this number as the

biggest ten-digit-number who is dividing your prime [original:

"Mersenne 44"]. Now, let's divide:

(from left to right, ANY-BASE-CHANGING):

"2^[32,582,657]-1"

--------------------------

"2^[8;4;6;4;0;4;1;6]-1"

OK, define them again this way:

your number: "2^[3^[2^[5^[8^[2^[6^[5^[7]]]]]]]]-1"

my number: "2^[8^[4^[6^[4^[0^[4^[1^[6]]]]]]]]-1"

divide again:

"2^[3^[2^[5^[8^[2^[6^[5^[7]]]]]]]]-1"

----------------------------------------------

"2^[8^[4^[6^[4^[0^[4^[1^[6]]]]]]]]-1"

OK now, too lazy for your math, then I will just convert your digits

to my digits and define your number and my number as both two and one

(of course, also none - the o)!!!

;-2-2;-3-8;-2-4;-5-6;-8-4;-2-0;-6-4;-5-1;-7-6;-1-1;

(all pairs are equal);

==

;o-o;5-5;o-o;o-o;4-4;o-o;o-o;4-4;o-o;o-o

how not surprizing - you don't have 3-3! OK, let's define either -8-4

or -5-1 as your 3-3. now check! whoever is your 3-3 will be my 4-4 and

vice versa. split! define my 5-5 as 3;3 both 4-4 as 5-5 OK?

;o-o;5-5;o-o;o-o;4-4;o-o;o-o;4-4;o-o;o-o

;o-o;3-3;o-o;o-o;4-4;o-o;o-o;4-4;o-o;o-o

;o-o;3-3;o-o;o-o;5-5;o-o;o-o;5-5;o-o;o-o

now, remove all dots, except the last one whois "-1"; convert to "-one";

5-5;4-4;4-4;-one;

3-3;4-4;4-4;-one;

3-3;5-5;5-5;-one;

convert backwords:

-one................3-3;4-4;5-5.............this will be your order right!

OK, rotate your order backwords and convert your number:

"2^[32,582,657]-1"

here:

"2^[756,285,23]-1"

now, one of these number has to be bigger, right? OK, define mine as bigger

//"2^[756,285,23]-1";

now, rotate to the 8th;

// "2^[85,23,75,62]-1"

OK. now, let's define this decimal, otherwise you will not understand

me? (maybe?), then remove all dots,

"2^[32582657]-1"

"2^[75628523]-1"

"2^[85237562]-1"

your number is the smallest one of these numbers, right? (positive).

then let's define the number "[2^[32582657]-1] * [2^[32582657 *

75628523 * 75628523 * 75628523 * 85237562 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562]-1]" as my ANY number who is

dividing your number. then; the division would be [2^[32582657 *

75628523 * 75628523 * 75628523 * 85237562 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562]-1], then since your number

[2^[32582657]-1] is supposed to be prime, and the other number is

bigger - then it is either a prime or not a prime. then let's define

the one not a prime:

[2^[32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562 * 85237562]-1]+1: not a

prime right?

divided by two - let's count how many?

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562 * 85237562

divided by four - how many?

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562 * 85237562 / 2

not interested. then let's mod all the primes 3/5/7 and count again:

convert to base

mod 3: 32582657==2 ; 75628523==2 ; 85237562 ==2;

mod 5: 32582657==2 ; 75628523==3 ; 85237562 ==2;

mod 7: 32582657==2 ; 75628523==5 ; 85237562 ==4;

OK? now, replace mods:

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562 * 85237562 mod 3 == 2 ** 2

** 2 ** 2 ** 2 ** 2 ** 2 ** 2 ** 2 ** 2 == (-1) ** (-1) ** (-1) **

(-1) ** (-1) ** (-1) ** (-1) ** (-1) ** (-1) ** (-1) == (----------1)

(your math).

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562 * 85237562 mod 5 == 2 ** 3

** 3 ** 3 ** 2 ** 2 ** 2 ** 2 ** 2 ** 2 == (-3) ** (--3) ** (--3) **

(--3) ** (-3) ** (-3) ** (-3) ** (-3) ** (-3) ** (-3) ** (-3) ==

(---------------3) (your math).

32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562 * 85237562 mod 7 == 2 ** 5

** 5 ** 5 ** 4 ** 4 ** 4 ** 4 ** 4 ** 4 == (-5) ** (-2) ** (-2) **

(-2) ** (-4) ** (-4) ** (-4) ** (-4) ** (-4) ** (-4) ** (-4) ==

(-----------1) (your math).

OK, so in base 3 and 7, the smallest prime number will be your -3, in

base 5 the smallest prime number will be my (-3/2), now let's convert:

the biggest prime number who is dividing [2^[32582657 * 75628523 *

75628523 * 75628523 * 85237562 * 85237562 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562]-1] will be defined as my 4!!!!

the number 2^[32582657]-1 [Mersenne #44] will be defined as my 2!!!!

the number whois 2^[32582657 * 75628523 * 75628523 * 75628523 *

85237562 * 85237562 * 85237562 * 85237562 * 85237562 * 85237562 *

85237562]-1] will be defined as my 32!!!!!

then, 2^[32582657 * 75628523 * 75628523 * 75628523 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562 * 85237562 * 85237562]-1]

divided by my 4, divided by my 4 again will be defined as [Mersenne

#44] (counting all bases!!!);

now, define them back to their decimal values. remove 32582657 bits,

then 2^[32582657]-1 [Mersenne #44] will be defined as the smallest

prime number who is dividing whatever this number is (not defined -

remember!)

2^[32582657 * 75628523 * 75628523 * 75628523 * 85237562 * 85237562 *

85237562 * 85237562 * 85237562 * 85237562 * 85237562]-1]

OK! now, switch to base Mersenne #44 (positive prime); then define this number:

2^[88888888888888888;7777777777777;6666666666666;3]-1 --> this will be

my number (1/Mersenne #44) in base 32!

convert to decimal (2 defined as [[Mersenne #44] ^ [Mersenne #44]]:

divide by ([[Mersenne #44] ^ [Mersenne #44]] /2) [of course - not an

integer!)

then, compress; 2^[8..8;7..7;6..6;3]-1

convert to decimal again: 2^[8;888;888;777;776;663]-1

BUT - I need to check if it's a decimal prime right? otherwise you

wouldn't believe me!

OK - let's check how much is this number in decimal? 8888888777776663?

8888888*1000000000 +

77777*10000 +

666*10 +

3;

OK, now lets rename all digits and negative them - check whois a prime!

;a;b;c;d:

[- aaaaaaa*1000000000 - bbbbb*10000 - ccc*10 - d][times the digit "i"]

is not divided by (all the numbers from [Mersenne #44] to 2) //

[Mersenne #44]!!!!!!!

(that's at least 7 times permutated!)

checksum:

triangle:

ooooi;

oooii;

ooiii;

etc.

[- aaaaaaa*1000000000 - bbbbb*10000 - ccc*10 - d][times the digit "i"]

sum(d);sum(ccc*10); sum(bbbbb*10000) ; sum (aaaaaaa*1000000000);

go base 21 / 42: add o's

aaaaaaaoooooooooooooooooooooooooooo;

bbbbboooooooooooooooooooo;

cccooooooooooo;

dooooo;

count: a7;b5;c3;d1 +

ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo.......................................

OK? now, 105^[a7;b5;c3;d1]-1 must be a prime in base Mersenne #44;

(105 will be (Mersenne #44)*(Mersenne #44)+5)

then of course, Mersenne #44 will be number 5; I just need to find

numbers 4 and 3!!!

define them this way: 105/104/103 ok?

now equation: XYZ; Z==Uri#105; Y=Uri#104; X= Uri#103; M==Mersenne #44;

Z=[(M*M)--55ii][M*M*M*M*M*M*M];

Y=[(M*M)--44pipiii][M*M*M*M*M*M*M];

X=[(M*M)--33ii][M*M*M*M*M*M*M];

Z*Z==Y*Y+X*X;

[M*M*M*M*M*M*M][(M*M)--55ii]*[(M*M)--55ii][M*M*M*M*M*M*M]

==[M*M*M*M*M*M*M][M*M*M*M*M*M*M][(M*M)+5i][(M*M)-5i] ==

[M*M*M*M*M*M*M][M*M*M*M*M*M*M][M*M*M*M-5*5i];

[M*M*M*M*M*M*M][(M*M)--44pipiii]*[(M*M)--44ii][M*M*M*M*M*M*M]

==[M*M*M*M*M*M*M][M*M*M*M*M*M*M][(M*M)+4pii][(M*M)-4pii] ==

[M*M*M*M*M*M*M][M*M*M*M*M*M*M][M*M*M*M-4*4pipii];

[M*M*M*M*M*M*M][(M*M)--33ii]*[(M*M)--33ii][M*M*M*M*M*M*M]

==[M*M*M*M*M*M*M][M*M*M*M*M*M*M][(M*M)+3i][(M*M)-3i] ==

[M*M*M*M*M*M*M][M*M*M*M*M*M*M][M*M*M*M-3*3i];

OK? now - let's define the number of binary digits of each (in

decimal) - let Mersenne #44 be the biggest?....

of course - logging here!

Z: [7+][7+][-5 - 5] // 0 binary digits!!!

X:

[7+][7+][-5 - -4]; // 1 binary digit!!!!

OK, then let X (binary) be the next binary prime Mersenne #44 is now,

BUT ALLWAYS true for ANY prime number!!!

define: Mersenne #44:

0/1/2/3/2/5/8/2/6/5/7/1

// ok let's make that decimal.............. binary numbers are not real anyway!

counting, replacing, starting with e:

[e;f;g;h] [0;1;2;3] [0000;0001;0010;0011]

[i;j;k;l] [0;1;2;3] [0100;0101;0110;0111]

[m;n;o;p] [0;1;2;3] [1000;1001;1010;1011]

[q;r;s;t] [0;1;2;3] [1100;1101;1110;1111]

0/1/2/3/2/5/8/2/6/5/7/1

e/f/g/h/gg/j/m/ggg/k/jj/l/ff/

OK. number equals to "efghggjmgggkjjlff" // same number repeating

itself must be marked!

now, checking smaller primes! replace digits with 0/1/2/3/4/5/6/7/8/9.......

"e/f/g/h/gg/j/m/ggg/k/jj/l/ff/" - replace with

/0/1/2/3/44/5/6/777/8/99/s/tt/

replace rightwords:

/00/1/22/3/444/5/6/77/8/9/s/tt/

the difference:

/o.o/1/2.2/3/44.4/5/6/77.7/8/9.9/s/tt/

right to left:

/o.o/1/2.2/3/4.44/5/6/7.77/8/9.9/s/t.t/

OK looks symmetric to me. 44.44 will be next prime gg; switch back:

/o.o/1.1/2.2/3.3/g.g/5.5/6.6/g.g/8.8/9.9/s.s/t.t/

switch back:

now equation: XYZ; Z==Uri#105; Y=Uri#104; X= Uri#103; M==Mersenne #44;

define XYZ the same:

/o.o/1/2.2/X.X.YY.ZZ/6/7.77/8/9.9/s/t.t/

/o.o/1/2.2/YXZ.XZY/6/7.77/8/9.9/s/t.t/

/o.o/1/2.2/Z.Z.YY.XX/6/7.77/8/9.9/s/t.t/

see? I'm starting to understand their language!

OK, let's just replace their "2" with "212", lets see if it's prime for them;

[2^321;258;212,657-1] - would probably be a prime too. let's check!

0/1/2/3/2/1/2/5/8/2/1/2/6/5/7/1

e/f/g/h/g.g/f.f/g.g/j/m/ggg/f.f/g.g//k/jj/l/ff/

OK?

/0.0/1.1/2.2/3.3/2.2/1.1/2.2/5.5/2.2/2.2/1.1/2.2/1.1/2.2/6.6/5.5/7.7/1.1/0.0/

reverse:

/0.0/1.1/7.7/5.5/6.6/2.2/1.1/2.2/1.1/2.2/2.2/5.5/2.2/1.1/2.2/3.3/2.2/1.1/0.0/

OK symmetric; convert:

0;1;7;5;6;2;1;2;1;2;2;5;2;1;2;3;2;1;0

reverse:

0;1;2;3;2;1;2;5;2;2;1;2;1;2;6;5;7;1;0

convert to their decimal:

[2^[3;2;1;2;5;2;2;1;2;1;2;6;5;7]-1]

[7^[5;6;2;1;2;1;2;2;5;2;1;2;3;2]-1]

let's see. can [7^[5;6;2;1;2;1;2;2;5;2;1;2;3;2]-1] be divided by 7? I

don't think so....

OK, now rename: 7==2; define 2==7; that's my prime! [base 70000000000007!]

[2^[5;6;4;1;4;1;4;4;5;4;1;4;3;4]-1]

now, let's compress it; 0/1/2/5/6/4/1/4/1/4/4/5/4/1/4/3/4/1/0

reverse, switch 4/2:

0/1/4/2/5/6/4/1/4/1/4/4/5/4/1/4/3/4/1/0

rename: e/f/g/h/i/j/g/f/g/f/g/g/i/g/f/g/k/g/f/e

reverse: e/f/g/k/g/f/g/i/g/g/f/g/f/g/j/i/h/g/f/e

symmetric [e/f/g][/k/g/f/g/i/g/g/f/g/f/g/j/h/] |

[h/j/g/f/g/f/g/g/i/g/f/g/k/][g/f/e]

cut middle: replace: g "." --> [e/f]..[k/f/i/f/f/j/h]

e==0;

f==1;

// tired. will define all later!!!!! (or computer will do it for me)...........

Uri First deadandalive

Mobile Phone: +972-50-9007559

E-mail: uri@... // urievenchen@...

[~~~~~~~~~~~~~~~~~~~~~]

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

If I'm writing too much, please filter my messages, or (if you really

don't want to hear from me), write me something humanish such as

"please don't send to this mailing list" with my public keys; at least

three times with different subjects (change the digits to whatever you

want); here:

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

[~~~~~~~~~~~~~~~~~~~~~]

[~~~~~~~~~~~~~~~~~~~~~]

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

Update: Left HTTP, WWWW and port numbering. send me papers or to

urievenchen@... .

I completed my 0.1 version of real deadanyalive quantum relativity

redefining back timespacing intergalactic worldwide [top secret: if

you have any US ARMY* on your planet they will never allow it].

Recounting back every second since twice BCC doubling + not counting

at all any uncountable; using only prime countable numbers equal to 21

(base 21 recursivley) who are all equal to 0base0 and 1base1 who are

identical twins base 21 [21===the number of fingers;eyes;body parts &

number of equal signs not equal counting both left to right; right to

left and all 21 dimensions of nothing].

Read my autoreply for more information [my HTTP/SMTP not working].

- This message is confidential -

--

Uri First deadandalive

Mobile Phone: +972-50-9007559

E-mail: uri@... // urievenchen@...

Update: Left HTTP, WWWW and port numbering. send me papers or to

urievenchen@... .

I completed my 0.1 version of real deadanyalive quantum relativity

redefining back timespacing intergalactic worldwide [top secret: if

you have any US ARMY* on your planet they will never allow it].

Recounting back every second since twice BCC doubling + not counting

at all any uncountable; using only prime countable numbers equal to 21

(base 21 recursivley) who are all equal to 0base0 and 1base1 who are

identical twins base 21 [21===the number of fingers;eyes;body parts &

number of equal signs not equal counting both left to right; right to

left and all 21 dimensions of nothing].

Read my autoreply for more information [my HTTP/SMTP not working].

- This message is confidential -

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

[---humans:always copy this--]

[o-o-o-.OO.-o-o-o][o-o-o-.OO.-o-o-o]

[--siht ypoc syamla:snawuh---]

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

--

Uri First deadandalive

Mobile Phone: +972-50-9007559

E-mail: uri@... // urievenchen@...

Update: Left HTTP, WWWW and port numbering. send me papers or to

urievenchen@... .

I completed my 0.1 version of real deadanyalive quantum relativity

redefining back timespacing intergalactic worldwide [top secret: if

you have any US ARMY* on your planet they will never allow it].

Recounting back every second since twice BCC doubling + not counting

at all any uncountable; using only prime countable numbers equal to 21

(base 21 recursivley) who are all equal to 0base0 and 1base1 who are

identical twins base 21 [21===the number of fingers;eyes;body parts &

number of equal signs not equal counting both left to right; right to

left and all 21 dimensions of nothing].

Read my autoreply for more information [my HTTP/SMTP not working].

- This message is confidential -

--

Uri First deadandalive

Mobile Phone: +972-50-9007559

E-mail: uri@... // urievenchen@...

Update: Left HTTP, WWWW and port numbering. send me papers or to

urievenchen@... .

I completed my 0.1 version of real deadanyalive quantum relativity

redefining back timespacing intergalactic worldwide [top secret: if

you have any US ARMY* on your planet they will never allow it].

Recounting back every second since twice BCC doubling + not counting

at all any uncountable; using only prime countable numbers equal to 21

(base 21 recursivley) who are all equal to 0base0 and 1base1 who are

identical twins base 21 [21===the number of fingers;eyes;body parts &

number of equal signs not equal counting both left to right; right to

left and all 21 dimensions of nothing].

Read my autoreply for more information [my HTTP/SMTP not working].

- This message is confidential -

--

Uri First deadandalive

Mobile Phone: +972-50-9007559

E-mail: uri@... // urievenchen@...

Update: Left HTTP, WWWW and port numbering. send me papers or to

urievenchen@... .

I completed my 0.1 version of real deadanyalive quantum relativity

redefining back timespacing intergalactic worldwide [top secret: if

you have any US ARMY* on your planet they will never allow it].

Recounting back every second since twice BCC doubling + not counting

at all any uncountable; using only prime countable numbers equal to 21

(base 21 recursivley) who are all equal to 0base0 and 1base1 who are

identical twins base 21 [21===the number of fingers;eyes;body parts &

number of equal signs not equal counting both left to right; right to

left and all 21 dimensions of nothing].

Read my autoreply for more information [my HTTP/SMTP not working].

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

--

Uri First deadandalive

Mobile Phone: +972-50-9007559

E-mail: uri@... // urievenchen@...

Update: Left HTTP, WWWW and port numbering. send me papers or to

urievenchen@... .

I completed my 0.1 version of real deadanyalive quantum relativity

redefining back timespacing intergalactic worldwide [top secret: if

you have any US ARMY* on your planet they will never allow it].

Recounting back every second since twice BCC doubling + not counting

at all any uncountable; using only prime countable numbers equal to 21

(base 21 recursivley) who are all equal to 0base0 and 1base1 who are

identical twins base 21 [21===the number of fingers;eyes;body parts &

number of equal signs not equal counting both left to right; right to

left and all 21 dimensions of nothing].

Read my autoreply for more information [my HTTP/SMTP not working].

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

- This message is confidential -

[o-o-o-oo-o-o-o][o-o-o-oo-o-o-o]

[Robert Wallner, read your mail]; create a special human key for me if

you want to. Using R.W;