Could someone tell me how large this number is?

Question

Context: If you guys didn't know, I'm running a nice little contest to see who can program the largest number. More specific rules if you are interested may be found in my chat room (click here to join). If you are entering, do note that I am accepting entries for quite a while (ignore all the deadlines :-)

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Actually, maybe you are a little too late.

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Now, before judging the size of any other numbers, I want to see how large my number is. It uses Knuth's up-arrow notation and a nice little recursive function I made: (it's name is $\text{THING}_4(a,b,c,d,e)$)

$$f(a,b,c,d,e)=\begin{cases}z&;b\cdot c\cdot d\cdot e<0\\f(z,b-1,c,d,e)&;b>0\\f(z,z,c-1,d,e)&;c>0\\f(z,z,z,d-1,e)&;d>0\\f(z,z,z,z,e-1)&;e\ge0\end{cases}$$

Where $z=a\uparrow^{a+3}(a+1)$. This must also be evaluated top to bottom. For example, you must check if $b>0$ before checking if $c>0$.

An example of how this function goes:

$$f(2,2,2,2,2)=f(2\uparrow^53,1,2,2,2)=f((2\uparrow^53)\uparrow^{(2\uparrow^53)+3}((2\uparrow^53)+1),0,2,2,2)\\x=(2\uparrow^53)\uparrow^{(2\uparrow^53)+3}((2\uparrow^53)+1)\\=f(x,x,1,2,2)\\=f(x\uparrow^{x+3}(x+1),x-1,1,2,2)$$

Clearly, if you keep the expansion up, it is clear this is much larger than Graham's number. Particularly, $f(2,2,1,0,0)>G$.

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My particular number is $f(9,9,9,9,f(9,9,9,9,9))$, and it's name is $SBA_4$, as per the initials of my username.

Does anyone know how to describe this number in terms of FGH or similar notation that sizes up how large a number is? Ultimately, I want to see where my number stands on the list of googolisms, but I cannot tell.

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Updates:

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$2/11/17:$

BEAF looks like a good candidate to compare to my function. Looking into it...

Comparing to BEAF and similar things, I think my number lies somewhere at the chained arrow notation level. I am thinking it lies around maybe $\{a,b,3,2\}$ or higher.

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$2/12/17:$

Actually, I think that my number is larger than Megaior megocal. Particularly,

$$SBA_4>>f(9,9,9,9,0)\stackrel?>\operatorname{forcal}_3(2)$$

which puts my number somewhere around $\{a,b,4+,2\}$.

In general, I think Aarex's Graham generator is a nice comparison (at least in my head, I can compare the two somewhat easily).

I'm also trying to see if I can compare $SBA_4$ to Hypercal. I'm fairly certain that $SBA_4>\text{Hypercal}$, but its a little bit fuzzy in my mind.

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$2/13/17$

I think I've finally found some good bounds:

$$\text{Hypercal}

Hyperior Hyperior Hyperior Hypercal.

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$7/27/2017:$

Not sure why you're still here, but if you just want to see what I've come up just for the heck of it, I've made this program:

def f(a,b=[a,a],c=[a,a])d,*e=b;g,h=c;a.times{a+=(b.class==Fixnum)?a:b.size<2?f(a,d,c):d<0?f(a,e,[g+1,h]):g>0?f(a,[a,d-1,e],[g-1,h]):h<0?f(a,[d-1,e],c):f(a,b,[a,h-1])};a end

When run, $f(n)$ outputs numbers roughly on the scale of $f_{\omega^{\omega^2}+\omega^\omega}(n)$ in the fast growing hierarchy.

In English, the expansions rules basically work as follows:

$$f(n)=f(n,[n,n],[n,n])$$

$b=[b_1,b_2,\dots,b_i]$, $d=b_1$ and $e=[b_2,\dots,b_i]$. Let $n$ be an integer.

$$f(a,n,c)=2^aa$$

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$$f(a,[b],c)=u_a\\u_0=a,u_{n+1}=u_n+f(u_n,b,c)$$

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$$b_1<0\implies f(a,b,[g,h])=v_a\\v_0=a,v_{n+1}=v_n+f(v_n,e,[g+1,h])$$

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$$b_1\ge0\land g>0\implies f(a,b,[g,h])=w_a\\w_0=a,w_{n+1}=w_n+f(w_n,[a,d-1,e],[g-1,h])$$

Note that if $b=[1,2,3]$, then $[a,d-1,e]=[a,0,[2,3]]\ne[a,0,2,3]$.

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$$b_1\ge0\land g\le0\land h\ge0\implies f(a,b,[g,h])=x_a\\x_0=a,x_{n+1}=x_n+f(x_n,b,[a,h-1])$$

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$$b_1\ge0\land g\le0\land h<0\implies f(a,b,[g,h])=x_a\\x_0=a,x_{n+1}=x_n+f(x_n,[d-1,e],[g,h])$$

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To see examples, tr here. To see it worked out, I'll take the time to type out the expansion of $f(2)$ below:

$f(2)=f(2,[2,2],[2,2])\\=2+\underbrace{f(2,[2,1,[2]],[1,2])}+f(2+f(2,[2,1,[2]],[1,2]),[2,1,[2]],[1,2])\\\implies f(2,[2,1,[2]],[1,2])=2+\underbrace{f(2,[2,1,[1,[2]]],[0,2])}+f(2+f(2,[2,1,[1,[2]]],[0,2]),[2,1,[1,[2]]],[0,2])\\\implies f(2,[2,1,[1,[2]]],[0,2])=2+\underbrace{f(2,[2,1,[1,[2]]],[2,1])}+f(2+f(2,[2,1,[1,[2]]],[2,1]),[2,1,[1,[2]]],[2,1])\\\implies f(2,[2,1,[1,[2]]],[2,1])=2+f(2,[2,1,[1,[1,[2]]]],[1,1])+f(2+f(2,[2,1,[1,[1,[2]]]],[1,1]),[2,1,[1,[1,[2]]]],[1,1])\\\vdots$

So as you can see, the amount of recursion gets kinda crazy, and the amount of arrays will slowly explode.

Answer 1

Your function lies at $f_{\omega+4}$ in the fast-growing hierarchy; that is, your $f(n,n,n,n,n)$ is roughly $f_{\omega+4}(n)$.

First, let us set $g(n) = n \uparrow^{n+3} (n+1)$. Note that

$$2 \uparrow^{n-1}(n+1) < f_\omega (n) < 2 \uparrow^{n-1}(2n)$$

so for $n \ge 2$,

$$f_{\omega}(n) = f_n(n) < 2 \uparrow^{n-1}(2n) < g(n) < 2 \uparrow^{n+3}(n+5) < f_\omega(n+4)$$

Now we have $$f(a,0,0,0,0) = f(g(a),g(a),g(a),g(a),-1) = g(g(a))$$

Further, $$f(a,1,0,0,0) = f(g(a),0,0,0,0) = g(g(g(a))) = g^3(a)$$

and by induction we have $$f(a,b,0,0,0) = g^{b+2}(a)$$

In particular we have $f(a,a,0,0,0) = g^{a+2}(a)$, and

$$f_{\omega+1}(a) = f_{\omega}^a(a) < g^a(a) < g^{a+2}(a)$$

The other direction is a little tricker; however we can use the fact that $$f_{\omega}(a+1) > f_{\omega}(a) + 5$$ when $a \ge 2$, so setting $u(a) = f_{\omega}(a)$ and $v(a) = f_{\omega}(a+4)$, we have

$$u^n(a+5) = u^{n-1}(u(a+5)) = u^{n-1}(v(a+1)) > u^{n-1}(v(a)+5) = u^{n-2}(u(v(a)+5)) = u^{n-2}(v(v(a)+1)) > u^{n-2}(v(v(a)+5)) = \ldots > v^n(a)+5$$

Thus $$f_{\omega+1}(a+3) = u^{a+3}(a+3) = u^{a+2}(u(a+3)) > u^{a+2}(a+5) > v^{a+2}(a) > g^{a+2}(a)$$

and we have proven $$f_{\omega+1}(a) < f(a,a,0,0,0) < f_{\omega+1}(a+3)$$

The rest of the letters follow a similar pattern. Define $h(a) = g^a(a)$. Then we have

$$f(a,a,1,0,0) = f(g^a(a), 0, 1,0,0) = f(h(a),0,1,0,0) = f(g(h(a)),g(h(a)),0,0,0) = f(h(g(h(a))),0,0,0,0) = g(g(h(g(h(a)))))$$

$$f(a,a,2,0,0) = f(h(a),0,2,0,0) = f(g(h(a)),g(h(a)),1,0,0) = g(g(h(g(h(g(h(a))))))) = g(gh)^3(a)$$

and by induction again,

$$f(a,a,c,0,0) = g(gh)^{c+1}(a)$$

Now both $gh(a) = g^{a+1}(a)$ and $ggh(a) = g^{a+2}(a)$ lie between $f_{\omega+1}(a)$ and $f_{\omega+1}(a+3)$, so $f(a,a,c,0,0) =g(gh)^{c+1}(a)$ is greater than $f_{\omega+1}^{c+1}(a)$, and by the same argument as the $u,v$ argument above we can show that $f(a,a,c,0,0)$ is less than $f_{\omega+1}^{c+1}(a+4)$.

So $$f_{\omega+2}(a) = f_{\omega+1}^a(a) < f_{\omega+1}^{a+1}(a) < f(a,a,a,0,0) < f_{\omega+1}^{a+1}(a+4) < f_{\omega+1}^{a+1}(f_{\omega+1}(a+2)) = f_{\omega+1}^{a+2}(a+2) = f_{\omega+2}(a+2)$$

Using the same reasoning, we can show that $$f_{\omega+3}(a) < f(a,a,a,a,0) < f_{\omega+3}(a+2)$$ and $$f_{\omega+4}(a) < f(a,a,a,a,a) < f_{\omega+4}(a+2)$$

In terms of Conway chained arrow notation, $f(a,a,a,a,a)$ will be approximately $a \rightarrow a \rightarrow a+1 \rightarrow 5$.

In terms of BEAF, $f(a,a,a,a,a)$ will be approximately $\lbrace a,a+1,4,2 \rbrace$.

Answer 2

In BEAF notation, I have deduced that

$$f(a,0,0,0,0)\approx\{a,2,1,2\}$$

$$f(a,b,0,0,0)\approx\{a,b+1,2,2\}$$

$$f(a,b,1,0,0)\approx\underbrace{\{a,\{a,\{a,\{a,\{}_{\{a,b+1,2,2\}}\dots\}+1,2,2\}+1,2,2\}+1,2,2\}+1,2,2\}$$

$$f(a,b,c,0,0)\approx(c+1)\left\lbrace\tiny\underbrace{\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{}_{\underbrace{\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{}_{\underbrace{\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{a,\{}_{\underbrace{\vdots}_{\{a,b+1,2,2\}}}\dots\}+1,2,2\}+1,2,2\}+1,2,2\}+1,2,2\}}\dots\}+1,2,2\}+1,2,2\}+1,2,2\}+1,2,2\}}\dots\}+1,2,2\}+1,2,2\}+1,2,2\}+1,2,2\}\right.$$

Unless someone can simplify this, BEAF isn't going to be enough, but it looks like a good start.

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Maybe BEAF is a good place, but I'm going to have to understand more about it's nature for more terms in the array and larger numbers at the end.

:-(