# 2.09. Conway's Chain Arrows

"Knuth's up-arrow notation but on crack."

Introduction

We've discussed Graham's number and its history, magnitude, and continuing popularity. But now, it's time to leave Graham's number in the dust with some REALLY massive numbers. I present to you: Conway's chained arrow notation.

How did Conway's chain arrows originate? They were introduced in John Conway and Richard Guy's Book of Numbers, a book all about recreational mathematics and the same book which introduces Conway and Guy's illion proposal. Pages 59 to 62, as we saw in the previous article, discuss some very large numbers, introducing Knuth's arrows, Graham's number, and a novel notation by Conway called chained arrow notation, the last of which is the subject of this article.

Conway has a very unique tone in his book. He does not have the humor and descriptions Jonathan Bowers has, or the exceptionally thorough tone of Sbiis Saibian, or the more scholarly "just-the-facts" tone of Robert Munafo, or however one would describe my own tone. According to Sbiis Saibian's desription of Conway's work, the book has a particular sense of fun and lightness.

As I said in the previous article, the brevity of the section on large numbers stands as an example of how little attention googology gets among mathematicians. Although his description of chain arrows takes up only half a page, today Conway's chain arrows have sort of become the king of the popular large number notations. This is not that it's the best way to generate large numbers, just that it's the most powerful "popular" way. There are far more clever ways to make large numbers, but Conway's chain arrows take quite an appealingly simple approach while still producing mind-blowing numbers.

Here we'll examine Conway chain arrows with examples of numbers they produce, and how much larger they are than numbers like Graham's number.

Definition of Conway Chain Arrows

First off, what are those chain arrows? Here is how Conway defines his chain arrows in his book:

Our own "chained arrow" notation names some even larger numbers. In this, a^^...^^b (with c arrows) is called a->b->c.

a->b-> ... ->x->y-> 1 is another name for a->b-> ... ->x->y

and a ... x->y->(z+1) is defined to be

a ... x if y = 1,

a ... x->(a...x)->z if y = 2,

a ... x->(a...x->(a...x)->z)->z if y = 3,

and so on. The parentheses here may be rubbed out after the numbers inside them have been completely evaluated.

One thing to note is that Conway says "our own chained arrow notation". Does that mean that he and Guy both invented it, or was "our" only used as a word to indicate that the book was made by Conway and Guy in collaboration? This is all uncertain, but Conway is usually given sole credit for devising chained arrow notation. Also, when John Conway confirmed to Sbiis Saibian that he was the inventor of chain arrows, this suggests that they were only Conway's invention.

Also of note here is that Conway gives a by-example definition, and such definitions are frowned upon by googologists. However since Conway's book was intended for laymen to read, this can be excused. Despite the by-example definiton, we can devise a more formal ruleset that will let us evaluate any chain:

Rule 1 (only two numbers): a->b = ab

Rule 2 (last number is 1): #->1 = # where # denotes the rest of the chain (chop off the ending 1)

Rule 3 (second last number is 1): #->1->a = # where # denotes the rest of the chain (chop off the last two numbers)

Rule 4 (otherwise): #->a->b = #->(#->(a-1)->b)->(b-1) (decrease last number by 1, feed chain with second last number decreased by 1 into the second last number)

First off, notice that I did not give the rule a->b->c = a^^...^^b with c ^s. The rules given above in fact make a->b->c equal to a^cb. Also, how did we get the rule a->b = ab? We get it by working backwards. Conway says that 1's at the end of a chain can be removed, implying that a->b would equal a->b->1 which we know equals a^b = ab.

The second thing I'd like to note is that the # in the beginnings of the chains in rules 2, 3, and 4 can be one, two, three, four, or any number of entries.

Last and most importantly, parentheses are vital in the system as they distinguish (for example) 3->3->3->3 vs 3->3->(3->3) vs 3->(3->3)->3, which are three different numbers although their expressions look similar. As Conway says, you can only remove parentheses if the number within them has been evaluated.

Now that we've reviewed the rules of chain arrows, we're ready to look at the numbers they can produce.

Some Basic Examples of Chain Arrows

First let's do an example of Conway chain arrows that can actually be computed:

3->3->2

First we need to check the rules. Rule 1 doesn't apply since the chain is three numbers, and rules 2 and 3 don't apply since the chain's last or second last number isn't 1. Therefore we apply rule 4:

#->a->b = #->(#->(a-1)->b)->(b-1)

Here, a is 3, b is 2, and # is the rest of the chain (aka 3). So let's look at the rule:

#->a->b = #->(#->(a-1)->b)->(b-1)

this means that:

3->3->2 = 3->(3->(3-1)->2)->(2-1)

And then we can solve the subtractions:

3->(3->(3-1)->2)->(2-1) = 3->(3->2->2)->1

Since the chain ends in 1, now we can apply rule 2 and chop off the 1 to get:

3->(3->2->2)->1 = 3->(3->2->2)

Now the chain is only two numbers: 3 and 3->2->2. But hold on. We can't solve the chain until we solve 3->2->2. So now we need to solve that inner chain with rule 4:

3->(3->2->2) = 3->(3->(3->(2-1)->2)->(2-1))

and simplify:

3->(3->(3->(2-1)->2)->(2-1)) = 3->(3->(3->1->2)->1)

Now the chain 3->(3->1->2)->1 within the parentheses can simplify with rule 2 to 3->(3->1->2). So now we have the chain simplified to:

3->(3->(3->1->2))

And in the innermost chain, (3->1->2) has 1 as the second last entry. So we can use rule 3 to simplify 3->1->2 to just 3, since we chopped off the last two numbers. Now we have the chain:

3->(3->3)

Finally we can use rule 1 since the chain (3->3) is only two numbers. We now get:

3->(3->3) = 3->(33) = 3->27

Now we can apply rule 2 one more time to get a chain:

3->27 = 327 = 7,625,597,484,987

So we've solved the chain now. Now let's see how a bigger chain, 3->3->2->2, will solve:

3->3->2->2

= 3->3->(3->3->1->2)->1 (rule 4)

= 3->3->(3->3->1->2) (rule 2)

= 3->3->(3->3) (rule 3)

= 3->3->27

= 3->(3->2->27)->26 (rule 4)

= 3->(3->(3->1->27)->26)->26 (rule 4)

= 3->(3->3->26)->26 (rule 3)

= 3->(3->(3->2->26)->25)->26 (rule 4)

= 3->(3->(3->(3->1->26)->25)->25)->26 (rule 4)

= 3->(3->(3->3->25)->25)->26 (rule 3)

= 3->(3->(3->(3->2->25)->24)->25)->26 (rule 4)

= 3->(3->(3->(3->(3->1->25)->24)->24)->25)->26 (rule 4)

= 3->(3->(3->(3->3->24)->24)->25)->26 (rule 3)

= 3->(3->(3->(3->(3->2->24)->23)->24)->25)->26 (rule 4)

= 3->(3->(3->(3->(3->(3->1->24)->23)->23)->24)->25)->26 (rule 4)

= 3->(3->(3->(3->(3->3->23)->23)->24)->25)->26 (rule 3)

Hopefully you're beginning to recognize the pattern now. The chain 3->3->2->2 is the same value as 3^^^...(27 ^s)...^^^3. Anyway, when you continue simplifying the chain it'll look something like this:

3->(3->(3->(3->............(3->(3->(3->3->1)->2)->3)->4)................->24)->25)->26

Are you dizzy yet? Well, this number isn't anything new! It's just a number expressible with Knuth's up-arrows, and it doesn't even begin to tap into the power of chain arrows!

Examining 4-Number Chains

To better understand the quirks of chain arrows let's try some sample chains. For example let's try the chain 2->2->2->2:

2->2->2->2

= 2->2->(2->2->1->2)->1 (rule 4)

= 2->2->(2->2->1->2) (rule 2)

= 2->2->(2->2) (rule 3)

= 2->2->22 (rule 1)

= 2->2->4

= 2->(2->1->4)->3 (rule 4)

= 2->2->3 (rule 3)

= 2->(2->1->3)->2 (rule 4)

= 2->2->2 (rule 3)

= 2->(2->1->2)->1 (rule 4)

= 2->2->1 (rule 3)

= 2->2 (rule 2)

= 22 (rule 1)

= 4

Hmm. The chain is only 4. How about we try increasing a number in the chain? First let's change 2->2->2->2 into 3->2->2->2. We'll get:

3->2->2->2

= 3->2->(3->2->1->2)->1

= 3->2->(3->2->1->2)

= 3->2->(3->2)

= 3->2->9

now translating this to up-arrows:

= 3^^^^^^^^^2

= 3^^^^^^^^(3^^^^^^^^^1)

= 3^^^^^^^^3

= 3^^^^^^^3^^^^^^^3

= 3^^^^^^^3^^^^^^3^^^^^^3

.......

= 3^^^^^^^3^^^^^^3^^^^^3^^^^3^^^3^^3^3^3

= 3^^^^^^^3^^^^^^3^^^^^3^^^^3^^^3^^7,625,597,484,987

= 3^^^^^^^3^^^^^^3^^^^^3^^^^3^^^3^3^3^3............^3 (7,925,597,484,992 3's total)

?!?!?! Just increasing an entry by 1 changes the chain from 4 to SUCH AN UNFATHOMABLE VALUE??? Weird. Let's try a different chain to make this kind of thing more clear: 2->3->2->2.

2->3->2->2

= 2->3->(2->3->1->2)->1

= 2->3->(2->3->1->2)

= 2->3->(2->3)

= 2->3->23

= 2->3->8

= 2^^^^^^^^3

= 2^^^^^^^2^^^^^^^2

= 2^^^^^^^4 (because 2^^^^...^^^^2 with as many ^s as we please = 4)

= 2^^^^^^2^^^^^^2^^^^^^2

= 2^^^^^^2^^^^^^4

= 2^^^^^^2^^^^^2^^^^^2^^^^^2

= 2^^^^^^2^^^^^2^^^^^4

.....

= 2^^^^^^2^^^^^2^^^^2^^^2^^^2

= 2^^^^^^2^^^^^2^^^^2^^^2^^2^^2^^2

= 2^^^^^^2^^^^^2^^^^2^^^2^^2^^4

= 2^^^^^^2^^^^^2^^^^2^^^2^^65,536

ANOTHER VERY HUGE NUMBER? Holy shit this is pretty strange. But what do we get when we change the 3rd entry to 3?

2->2->3->2

= 2->2->(2->2->2->2)->1

= 2->2->(2->2->2->2)

= 2->2->4 as we saw earlier

= 2^^^^2

= 4

Whoa now. Now it's suddenly only 4 again! Now let's try changing the last entry to 3:

2->2->2->3

= 2->2->(2->2->1->3)->2

= 2->2->(2->2)->2

= 2->2->4->2

= 2->2->(2->2->3->2)->1

= 2->2->(2->2->3->2)

= 2->2->4

= 4

AGAIN with the 4's? What IS it with chain arrows and FOURS??? It seems like any chain that starts with 2->2 will turn into 4! Actually that can be shown to be true. For three-number chains that's obvious: 2->2->X = 2^X2 = 4. To see why for four-number chains consider that:

2->2->A->B (A and B are any number)

= 2->2->(2->2->(A-1)->B)->(B-1)

= 2->2->(2->2->(2->2->(A-2)->B)->(B-1))->(B-1)

...

= 2->2->X->(B-1) where X is some (probably) very large number

and so on, repeating the process until B shrinks down to 1 and you can chop it off, leaving you with 2->2->Y where Y is some large number. A similar process can be done for chains of 5 or more numbers.

It is more common to give examples of Conway chain arrows starting with 3, probably to mimic Graham's number. So here we'll play around with chains starting with 3.

First off, we saw earlier that 3->3->2->2 is just 3->3->27, an up-arrow level number. Now once again, let's try playing around and adjusting each argument:

4->3->2->2

= 4->3->(4->3->1->2)->1

= 4->3->(4->3->1->2)

= 4->3->(4->3)

= 4->3->43

= 4->3->64

= 4^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^3

If something like 10 up-arrows was impossible to even understand in terms of iterating power towers, then how are we to understand something like 64 ^s? Hold on, this is just a warm-up, and we haven't even passed up Graham's number yet!

next:

3->4->2->2

= 3->4->(3->4->1->2)->1

= 3->4->(3->4->1->2)

= 3->4->(3->4)

= 3->4->34

= 3->4->81

= 3^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^4

Yikes!! But this is also just a warm-up.....

next:

3->3->3->2

= 3->3->(3->3->2->2)->1

= 3->3->(3->3->2->2)

= 3->3->(3->3->27)

= 3->3->3^273

= 3^^^^^^^^^... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ^^^^^^^^^3 with 3^^^^^^^^^^^^^^^^^^3 ^s

Crazy, right? Well, this still isn't even as big as Graham's number

3->3->2->3

= 3->3->(3->3->1->3)->2

= 3->3->(3->3)->2

= 3->3->27->2

= 3->3->(3->3->26->2)

= 3->3->(3->3->(3->3->25->2))

etc

How big is that number? First off, remember that 3->3->2->2 = 3->3->27 and 3->3->3->2 = 3->3->(3->3->27). Continuing the pattern we have 3->3->4->2 = 3->3->(3->3->3->2), etc. So 3->3->2->3 = 3->3->27->2 =:

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

27

That's 27 layers total, and "not too much smaller" than Graham's number. Here, "not too much smaller" is used in a very different way from usual. It's only "not too much smaller" than Graham's number in that you only need 38 more layers to pass up Graham's number!

So why can I call it "not too much smaller"? Because, as you can see, it's in about the same realm of numbers as Graham's number.

How do you compare this to Graham's number? Let's use Graham's G(n) function:

G(1) = 3^^^^3 < 3->3->2->2

3->3->2->2 < G(2) = 3^^^...(G(1) ^s)...^^^3 < 3->3->3->2

3->3->3->2 < G(3) = 3^^^...(G(2) ^s)...^^^3 < 3->3->4->2

etc.

With this method it can be shown that 3->3->2->3 is somewhere between G(26) and G(27). Likewise Graham's number (G(64)) is between 3->3->64->2 and 3->3->65->2, the former of which is a common approximation for Graham's number using chained arrows.

But now, let's introduce a number Conway himself used as an example of a number bigger than Graham's number that can be named with Conway's chain arrows.

It is equal to:

3->3->3->3

How would you go about to solving this chain? Let's find out:

3->3->3->3

= 3->3->(3->3->2->3)->2

= 3->3->(3->3->27->2)->2 as we saw earlier

How will we go further in describing this number? Here's how: Let Cn be 3->3->n->2, i.e. just like Graham's function Gn but starting with 27 instead of 3^^^^3. Then C1 = 27, C2 = 3^273, C3 = 3^C23, etc. Note that we can easily tell that C2 falls between G1 and G2, C3 falls between G2 and G3, and so on.

Then:

3->3->3->3

= 3->3->(3->3->27->2)->2

= 3->3->C27->2

= CC27

Hm. We can express this number quite compactly with our C function as CC27. But don't be fooled by it. 3->3->3->3 is really equal to:

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

: : : :

: : : :

: : : :

: : : :

: : : :

: : : :

: : : :

: : : :

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

27

but instead of just 27 layers, you have

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

3^^^... ... ...^^^3

27

layers! AAAAHHHHH!!!!! This is way bigger than Graham's number alright. Speaking of which, where does this number fall in terms of Graham's function?

Since this number is CC27, and C27 alone is between G26 and G27, we can bound it a little better:

First let's upper bound it:

CC27 < GC27, since Cn is always less than Gn. Then GC27 is itself less than GG27. So now we know that CC27 < GG27.

As for the lower bound ... this is a little trickier. Cn is always greater than Gn-1. So we can bound it:

CC27 > GC27-1. C27-1 (that's one less than C27) itself greater than D27, where Dx is like Cx but starting with 26 instead of 27.

Dx is itself greater than Gx-1, so we can bound: GC27-1 > GD27 > GG26. So we have it established:

GG26 < 3->3->3->3 < GG27

3->3->3->3 is a number that has gotten some fame among the googology community. Sbiis Saibian has in fact named it "Conway's tetratri", from "tetra" meaning four and "tri" meaning three, indicating that the chain consists of four threes.

Now let's look at another number, this time 3->3->4->3. How would that solve? Let's find out:

3->3->4->3

= 3->3->(3->3->3->3)->2

= 3->3->CC27->2

= CCC27

Hm, it seems like each time we increment the third argument by 1, we add one more application of the C function. This means that, say, 3->3->10->3, is the unimaginably vast CCCCCCCCC27. Here's what this number would look like in Knuth's up-arrows:

O_o;;;

This is beginning to look a lot like the recursion in up-arrows, but instead of just iterating the power-towers or layers of power-towers or whatever, we're iterating the number of up-arrows and then some!! Graham's number really was just barely starting with the epic recursion!

Hopefully you now get the idea of chains ending with a three. But now let's try chains ending with fours.

First let's try 3->3->1->4:

3->3->1->4 = 27 - since the second last entry is 1 we have to chop off the last two entries.

3->3->2->4:

3->3->2->4 = 3->3->(3->3->1->4)->3 = 3->3->27->3

= CCCCCCCCCCCCCCCCCCCC27 - that's 26 C's!

How about 3->3->3->4? That's equal to 3->3->(3->3->2->4)->3 = 3->3->(3->3->27->3)->3

= CCCCCCCCC ... ... ... ... ... ... CCCCCCCCCCCC27 ... with CCCCCCCCCCCCCCCCCCCC27 C's!!

As you can see our C function is bursting at the seams - therefore we need to devise a D function - we'll define it like so:

D1 = 27

Dn = CCC...CCC27 with D-1 C's

Therefore, we can rewrite 3->3->2->4 as D27, and 3->3->3->4 as DD27. Hopefully you're starting to see a pattern with the last and second last numbers - we'll further look at the pattern and its similarity to Knuth's arrows in a little bit. In the meantime, let's look at another number Conway gives in his book:

4->4->4->4

This is a really awesome number, which is in fact the largest number that is explicitly mentioned in the entire book. It's often cited as the largest number to appear in any published book at all! Sbiis Saibian calls this Conway's tetratet. But how big is it? Let's start solving the value:

4->4->4->4

= 4->4->(4->4->3->4)->3

= 4->4->(4->4->(4->4->2->4)->3)->3

= 4->4->(4->4->(4->4->(4->4->1->4)->3)->3)->3

= 4->4->(4->4->(4->4->256->3)->3)->3

For a better idea of this number's size consider working like so: Let 4->4->x->2 be written as C'x (C'1 = 256, C'2 = 4^2564, C'3 = 4^C'24, etc) - then D'x is 4->4->x->3 (D'1 = 256, D'2 = C'256, D'3 = C'C'256, etc). Then we can imagine this number as D'D'D'256. That doesn't look too bad! But looks are deceiving - this number is really:

YIKES!!! Conway's tetratri looks quite humble now, and Graham's number looks adorable, when conpared to this utter monster!

Anyway, back to the main sequence, let's examine chains ending with 5:

3->3->1->5

= 3->3 = 27

3->3->2->5

= 3->3->(3->3->1->5)->4

= 3->3->27->4

= DDDDDDDDDDDDDDDDDDDDDDDDDD27 (that's 26 D's)

Of course, we now can devise an E function to express t hese numbers more compactly:

E1 = 27

Ex = DE(x-1)

So 3->3->1->5 would be E1, 3->3->2->5 would be E27, and likewise 3->3->3->5 would be EE27, and so on ...

Similarly, with chains ending with 6 we can just have 3->3->1->6 = 27, 3->3->2->6 = EEE...EEE27 with 26 E's, which we can express as F27, 3->3->3->6 = FF27, etc. We can then use G for chains ending with 7, H ending with 8, I for 9, J, K, L, and so on ... - in fact after Z we can just use A27, A28, A29 ... etc, etc, etc.

Are you noticing a pattern here? Each time you increase the fourth number in the chain by one, the increase in the numbers is equivalent to adding an up-arrow. This is not strange or unintuitive at all - in fact it mirrors up-arrows, only that instead of iterating the height of the power-tower you are iterating the number of up-arrows!

If you get all that, hopefully you can use your imagination to think up how you'd evaluate a number like:

3->3->100->100

or anything else like that!

We've said enough about 4-number chains, so now let's transcend anything Conway talked about in his book with 5-number chains ...

5-Number Chains

Chain arrows with 5 numbers lead to some pretty awesome numbers. For example let's try a simple example:

3->3->3->2->2

= 3->3->3->(3->3->3->1->2)->1

= 3->3->3->(3->3->3)

= 3->3->3->3^^^3

= A3^^^3A3^^^327 with the Ak function - SUPER YIKES!!!

This is just like what we've experienced with the beginning of chain arrows - just as we can feed the number of up-arrows into itself, we can feed the number k in Ak into itself! In short, 5-number chains are to 4-number chains as 4-number chains are to Knuth's arrows. But still, we can explore the mechanics of those chains.

First let's try increasing the second-last argument:

3->3->3->3->2

= 3->3->3->(3->3->3->2->2)->1

= 3->3->3->(3->3->3->(3->3->3->1->2)->1)->1

= 3->3->3->(3->3->3->(3->3->3))

= 3->3->3->(3->3->3->(3^^^3))

= AXAX27, where X is A3^^^3A3^^^327

Notice the pattern here? The fourth argument in 5-number chains plays the same role the third does in 4-number chains - if you think back to the rules, the last two entries have the most important role in the chains.

So what do we get with making the last argument 3?

3->3->3->2->3

= 3->3->3->(3->3->3->1->3)->2

= 3->3->3->3^^^3->2

How are we to get a feel of how big this number is? Let's redefine the function Cx as:

C1 = 3^^^3

Cx = ACx-1ACx-127

Then we can imagine each time we increase the number next to C, we feed in the number into k in Ak. This is not strange at all, and it clearly mirrors the feeding-in of up-arrows used to define Graham's number.

We can imagine 3->3->3->2->3 as simply C3^^^3.

And let's try an additional example:

3->3->3->3->3

= 3->3->3->(3->3->3->2->3)->2

= 3->3->3->(3->3->3->3^^^3->2)->2

That would be equal to CC3^^^3.

Here's one more:

3->3->3->2->4

= 3->3->3->(3->3->3->1->4)->3

= 3->3->3->(3->3->3)->3

= 3->3->3->3^^^3->3

when you define Dx as D1 = 3^^^3, D2 = C3^^^3, D3 = CC3^^^3, and so on, this is equal to D3^^^3

I think you can see a pattern here - while 3-number chain arrows are equivalent to Knuth's up-arrows, 4-number chain arrows iterate like Knuth arrows over 3-number chains, 5-number chains do the same to the 4-number chains. Use your imagination to figure out how big these numbers are:

3->3->3->4->5

3->3->3->100->10

3->3->3->(3^^^3)->(3^^^^3)

I think that's about all we need to say about 5-number chains. How would we work with 6-or-more number chains? Let's find out!

6-Number Chains and Beyond

We now have given quite a decent examination of 4- and 5-number chains. With that in mind, it's not hard at all to look at some of the 6-entry chains and get a feel of the size of the numbers they produce. For example, let's look at the chain:

3->3->3->3->2->2

That solves to:

3->3->3->3->(3->3->3->3->1->2)->1

= 3->3->3->3->(3->3->3->3)

I think you can see for yourself how one would continue evaluating this chain, or any chain like that - ditto for chains with 7 or more numbers. Why is that behavior so obvious?

That's because the behavior of the chain arrows is easy to see from here on out - a n-entry chain will iterate the last argument of a (n-1)-entry-chain, and the last two numberrs are always by far the most important in the chain. In other words, while 3-number chain arrows are the same thing as Knuth up-arrows, 4-number chains perform up-arrow-like recursion over the number of up-arrows, 5-number chains do the same to the last number of the 4-number chains, 6-number chains iterate the last number of a 5-entry chain, 7-numbers do the same to 6-numbers, 8-numbers to 7-numbers, and then 9-numbers, 10-numbers, 11, 12, 13, 14 ... and so on with all the iteration.

I think that you get the idea of the recursion from here on now - let's review what we've learned about chain arrows.

Conclusion

We've seen that Conway chain arrows take quite a simple approach at extending the up-arrows - they easily ascend to mind-blowing heights, with iterations on top of iterations on top of iterations and iterations, making the "king of large numbers" Graham's number look tiny!

However, as it turns out, Conway chain arrows don't really take so much advantage of the power of recursion. We'll soon examine notations that will leave chain arrows in the dust (particularly Bowers' arrays), but not before we look at some extensions to chained arrow notation.