I mentioned an extremely large number in this comic, and then talked a little about comparing it to another large number in this blag post. But all the numbers discussed had a sense of arbitrariness to them. The xkcd number was just a function that grew rapidly plus a large argument. I’ve been thinking about it, and I want to construct a number that is not only mind-bogglingly large, but* elegantly* large. This is hard to define — the closest I can get is that it shouldn’t feel arbitrary.

Now, as some have pointed out, noncomputable functions like the Busy Beaver function can grow faster than any computable function. By my understanding, this just means they have a higher computational complexity — their values for a little bit may be smaller than a computable function, but you can never prove that, for sufficiently large arguments, a computable function is an upper bound to the sequence.

I decided I wanted to express the largest number I could in about 32 characters, something akin to the MIT big number duel. Here’s what I came up with:

H{F(n)}=F^n(n);H{H{H{H{BB(9)}}}}

In the first half (before the semicolon) I’m defining an operator H, which transforms a function F and an argument n into an n-level recursive call to itself (I’m using F^n() to refer to F(F(F( … recursed n times … ))) as in derivatives. That is, H{} represents the process of n-level recursion. Then I use this recursion several times on the Busy Beaver function until I run out of space. This isn’t just four recursions — each new H{} is a virtually infinite set of recursions, since it has as many levels as the result of the number in the level below it. As the seed, I picked 9 — the largest integer you can fit in a single digit.

I’m a lot happier with this number’s ‘bigness’ than I was with A(g_64, g_64). It *feels* bigger, in some hard-to-describe way (it also feels bigger in the easy-to-describe way: “>”). It uses a more fundamental idea of bigness It’s also, of course, noncomputable. However, you can stick the Ackermann function in place of the Busy Beaver function and get a number that is again much larger than the xkcd number.

So, anyone want to define an elegantly bigger number in about 32 characters, invoking reasonably standard notation and functions? I’m sure it’s possible — it’s not a well-defined contest, and I’m sure there are a lot of other tricks you can use. But at least I’m finally satisfied with my entry.

**Edit: **As a number of commenters have pointed out, my notation here is pretty bad.Â I’m indeed using H{} more like a macro that acts on an expression than a transform that acts on a function, and this leads to difficulties in mixing the levels.Â Fortunately, you guys have waded through the fog and understood what I was getting at 🙂

Here’s a slightly simplified expression that does the same thing and tacks in two extra recursions (so as to use the 32 characters).Â It loses the generality I was going for, but it’s more compact and not ambiguous:

H(n)=BB^n(n);H(H(H(H(H(H(9))))))

Bump to da bumpage!

g_(A(A(g_(g_99!)!,g_(g_99!)!)!,A(g_(g_99!)!,g_(g_99!)!)!))!

Kinda complicated, but kinda epic.

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Sorry. Here’s a bigger one.

A(BB(BB(A(A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$^^A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$,A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$^^A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$)$))$,BB(BB(A(A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$^^A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$,A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$^^A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$)$))$)$^^A(BB(BB(A(A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$^^A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$,A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$^^A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$)$))$,BB(BB(A(A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$^^A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$,A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$^^A((g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$,(g_(A(A(g_(g_99$)$,g_(g_99$)$)$,A(g_(g_99$)$,g_(g_99$)$)$))$)$)$)$))$)$

$ is a superfactorial. If 3! = 1*2*3, then 3$ = 3!^3!^3!.

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Terribly sorry.

g_(A(A(g_(A(BB(BB(A(A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB

(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_100$,g_100$)

$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)

$,A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$

)$,A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)

$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$)$))$,BB(BB(A(A(B

B(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(

g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$

,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$,A(BB(A(A(g_100$,g_100

$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)

$)$^^A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_1

00$)$,A(g_100$,g_100$)$)$)$)$)$))$))$,g_(A(BB(BB(A(A(BB(A(A(g_100$,g_

100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$

)$)$)$^^A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,

g_100$)$,A(g_100$,g_100$)$)$)$)$,A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_

100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g

_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,

g_100$)$)$)$)$)$))$,BB(BB(A(A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)

$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_100$

,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100

$)$)$)$)$,A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100

$,g_100$)$,A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_100$,g_100$)$,A(g_100$

,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$)$))$))$)

$,A(g_(A(BB(BB(A(A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(

A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_100$,g_100$)$,A

(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$,A

(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,

A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$

)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$)$))$,BB(BB(A(A(BB(A

(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_1

00$,g_100$)$)$)$)$^^A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB

(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$,A(BB(A(A(g_100$,g_100$)$

,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$

^^A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$

)$,A(g_100$,g_100$)$)$)$)$)$))$))$,g_(A(BB(BB(A(A(BB(A(A(g_100$,g_100

$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)

$)$^^A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_1

00$)$,A(g_100$,g_100$)$)$)$)$,A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100

$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_10

0$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_1

00$)$)$)$)$)$))$,BB(BB(A(A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$

)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_100$,g_

100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$

)$)$)$,A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$,BB(A(A(g_100$,g

_100$)$,A(g_100$,g_100$)$)$)$)$^^A(BB(A(A(g_100$,g_100$)$,A(g_100$,g_

100$)$)$)$,BB(A(A(g_100$,g_100$)$,A(g_100$,g_100$)$)$)$)$)$))$))$)$)$

)$

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Pingback: the doldrums. » every eager impulse

tan((A(g_64,g_64)-1)/A(g_64,g_64))

Technically, that’s 35 characters. A(x,y) being the Ackerman function again.

Basically, rather than generating a really really huge number, generate a number really really close to 1 (but slightly less), and then take it’s tangent.

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Here’s mine:

âˆ« Î“(B(x),P(x)) â†‘U(x)â†‘ G(x) dx, 1, y=g_64

Where:

B(x) is the bell polynomial B_[x](x), with [x] being the floor function.

P(x) is the number of polyominoes of size â‰¤ x.

â†‘U(x)â†‘ is Knuth’s up-arrow notation to the ultra-factorial degree (note: U(x) is generalized to the reals, as is the up-arrow notation in this definition. I have no idea how to generalize up-arrow notation to the reals… god help us all)

G(x) is the sum of the Goodstein sequence of x. I have no idea if this actually converges…

g_64 is, of course, graham’s number.

If you understand any of this, this function is also useful for calculating your geekiness!

I’d love to know the value of this function for small values of y (like 2). I have a feeling it stays small from 1 to 3…

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if i had my say id give the answer to collin, but i’m pretty sure that that is a variable of some sort. . .

my go at it:

a(g_64^g_64,9^g_64^a(g_64,g_62))

or vice versa, the question then becomes which is bigger? any ideas?

the other point is that i subscribe to the notion that if there is not enough matter in the universe to contain that number (quantum storage or otherwise) then the number is near enough to infinity that it does not matter. i guess these are the number they are talking about “for values approaching infinity”

revised answer:

a(inf^inf,inf^a(inf^a(inf,inf)))

No. . . BIGGER! (yep, definitely a variable!)

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typo in the revised answer, should be:

a(inf^inf,inf^a(inf,a(inf,inf)))

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Damn. I was gonna say 83.

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You guys missed the fact that Graham’s number is “just” the 64th element of a very fast-growing sequence. So fast growing, in fact, that it far, far, faaaaar transcends the Ackermann function. Instead of bothering with the Ackermann function, why not just subscript the g-sequence instead? I.e., Graham’s number is g_64, now we just subscript the sequence recursively: g_g_64, g_g_g_64, g_g_g_g_64, … until we run out of characters. Or, if you like, define an operator on functions that evaluates each iteration and subscripts it to the next level that many times.

People throw around Busy Beaver functions like they have any idea what it implies… do you even *realize* how fast *computable* functions can grow? The puny familiar computable functions that everyone’s familiar with barely scratch the surface of even just computable functions. Simply invoking BB means nothing, the real challenge is in defining *computable* functions that reach for the upper limits of computability. (While keeping in mind that BB transcends ALL of that.)

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How much larger than ackermanns?

What about A (gn_64, gn_64)th number of the Graham’s series :)?

I guess it would rival even some lower Busy Beavers but as I feel I am probably extremely underestimating it….

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And is gn even faster growing than Goodstain’s function G?

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The Golden Selection!

1618033…

Phi, without a decimal.

Fie in Phi.

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Pingback: Desunt Cetera » Blog Archive » Big Numbers

This is probably cheating, but I propose:

the biggest number in 32 chrs+1

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in simplest terms, (∞-1)^(∞-1) and I challenge anyone to find any larger number.

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H{F(n)}=F^F(n)(n);H{H{H{BB(9)}}}

Its a slight variation, using F(n) as the number of iterations per recursion(or is it recursion per iteration?) leaving us room for one less nesting. i’m not sure if it is actually bigger, but to me it feels bigger… plus the algebra hurts more.

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Not much talk has touched on Conway’s chained arrows, but we do know this:

a C.C.A. with 4 chained terms = *boggle*

with 5 chained terms = fuggedaboudit

so 3->3->3->3->3->3 = um… my pineal gland hurts.

So what if we recursed Conway with altered notation?: I’m thinking a single curly arrow with a superscript

nof the number of terms in the chain.Then the pineal-gland-hurting string of 3s above (limited by ascii) would be…

3–^6–>3

but then… we could replace… *fzzzzt-poof*

Oh, there’s the spewed gray matter, ok, must hurry….

Conway 9 to the BB of the Ackermann of the Grahamses.

9–^(BB_(A(G64,G64)))–>9

Oww, not again… *implode*

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OK, realized “recursion” isn’t quite right for that notation. Maybe. I think.

*hating the limits of ascii*

Somewhere tonight (wish I could remember the page) someone mentioned that on the way to making G_64, that G_1 alone represented a value greater than the number of Planck volumes one could divide the known ‘Verse into.

jeez. what’s the point? (rhetorical, this is still Masochistic fun)

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How about “Robin’s number” as big number?

Robin’s number = R_42 with:

R_n = 42 ? R_(n-1)

R_0 = 42

Where ? is an operator defined so that:

a ? 1 = a

a ? 2 = a?a

a ? 3 = a?a?a

a ? n = a(?a)*(n-1) (if you know what I mean)

And this is not an arbitrary number: it uses 42 (a lot), which, as we all know, is the Answer to Life etc.. granted, it’s not as concise as “H(n)=BB^n(n);H(H(H(H(H(H(9))))))”.

@Christ Schlacta, Nokes, PiAndWhippedCream: it must be a finite number, so yours doesn’t really count…

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(The question mark was suppost to look like |->, a combination of arrow up notation and CCA. Damn ASCII! (The char was U+21B1))

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I would submit for the function using 32 character while never repeating a function:

BB(TREE(A(G,G)))=x, x (hyperx) x

That should just fit 32 including spaces. However, in the way I abused syntax, I might as well refer to my thread in the Mathematics section of the XKCD forum about the number ‘a’ (you’ll see why I would not call it a constant if you visited it), and say 1/a.

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I feel like I’m going to have to agree with the idea that doing something like applying a tangent to a number incredibly close to pi/2 rads is the best option here.

However, tan was not my first choice; instead, what if Riemann’s zeta function was invoked? (It’s represented with ?(s), so it takes up less chars than “tan” does.) You could build the closest number to 1, something like this:

scikidus’ Number = F(x)=?(1-(1/ggx!));F^F(G)(ggG)

where gx is the xth Graham’s number, and G is Graham’s Number, which = g64.

Does anyone know which function increases more quickly as it approaches its respective asymptote?

And is my number big enough?

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Well, I’ve decided to put together a list of positive integers in sequential order. Unfortunately, I had to leave out a few in order to reach the level of some of the numbers being discussed here.

The first Ackermann number

2

3

The second Ackermann number

5

10

20

40

80

500

4000

BB(5)

googol

googolplex

A(4, 2)

A(4, 4)

The third Ackermann number

3????3 (g_1)

Graham’s number (g_64)

g_65

g_99

g_googol

A(g_64, g_64)

BB(g_64)

H(n)=BB^n(n);H(H(H(H(H(H(9))))))

f=BB;^^^^^^^^^!9f9f9f9f9f9f9f9f9

The smallest number bigger than any number that can be named by an expression in the language of first order set-theory with fewer than a googol (10^100) symbols.

I didn’t use many of the numbers being discussed here; there were only two that I saw that seemed to be worth mentioning. One was, of course, Randall’s. The other uses prefix notation, which I’d say is an excellent way to go about this.

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23!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1111one

and so on

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Just read this a couple minutes ago. Umm…as far as

H(n)=BB^n(n);H(H(H(H(H(H(9))))))

That can be improved on pretty easy with using all the same notation:

H(n)=BB^n(n);I(n)=H^n(n);I(I(9))

So…two-layer simple recursion on BB. Obviously bigger since I(9) = H(H(H(H(H(H(H(H(H(9))))))))). There’s not really enough space in 32 characters to define a nonsimple recursion on BB without it looking ugly, though.

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The WTFBIGNUM function:

I define fWTFBIGNUM=tan(({pi}-{pi_R})/{pi})

where,

{pi_R}= The RUTHERFORD QUANTITY = that fraction equal to the remainder of PI after (ya kno; comprised of the decimal to the right of) the first string of digits of PI (as evaluated in base 10 numerals) identical to a fully worked out expression of the string of digits representing Graham’s #.

If that made sense, can someone knowledgeable on the subject tell me if I won?

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I was pretty faded last night when I posted the above (as “DaOG”), and I think tangent might not have been what I meant. Anyway, I think it’s more elegantly stated as:

1/pi_R

So does that compete w/ the busy beavers, or no?

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Its a slight variation, using F(n) as the number of iterations per recursion(or is it recursion per iteration?) leaving us room for one less nesting. i’m not sure if it is actually bigger, but to me it feels bigger… plus the algebra hurts more.

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Aleph_Aleph_Aleph_Aleph_Aleph_99

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i agree with the guy who suggested the multiple graham recursions

BB(g_g_g_g_g_g_g_g_g_g_g_g_g_99), where g_99 goes up to the 99th instead of the 64th, i don’t like busy beavers since they are both non computable and remind me of the action your mom gets on weekends, but i threw one in there just for the sake of it because hey, this mathematical pissing contest isn’t already irrelevant enough, is it?

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Forgive me for the horrible math, but this is a devious little way of showing up the original XKCD number*hence forth to be called XKCD1* Adding nothing but some simple use of up notation on Grahams number within XKCD1. I am not sure if this is a proper large number, I’m a musician not a mathematician…

A{(g64 ^^n g64), (g64 ^^n g64)}

If it is a proper large number I have soooo many college math professors to torment 😀

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Kinda complicated, but kinda epic.

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Try this one, its a function that yields a large finite number no matter what you put in.

f(n)=((2(2n+1)^(2))+1)^(|n|+5)^((|n+2|+2)^((|n+2)^(2)))

It isn’t as large as some of the numbers you folks are dealing with but it makes stuff pretty big.

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Pingback: Private Servers

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