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• ## Graham's Number using the Supernova Array

June 29, 2015 by Undeadlift

Graham's Number is a huge number, and it is has been difficult to write its exact value despite the numerous notations already developed for very large numbers. Famous functions such as the fast-growing hierarchy and chained arrow notation only come up with approximations, while the Graham Array Notation and G function give the exact value using complex definitions. The Supernova Array offers an alternative method of writing Graham's Number with a simple set of rules.

• Base Function: $$S(a,b) = a\uparrow^{(b)}a$$, where $$b$$ is the number of $$\uparrow$$'s.
• $$S(a,b,0) = S(a,b)$$
• $$S(a,b,c) = S(a,S(a,b),c-1)$$

The other rules are available here.

We all know that $$3\uparrow \uparrow \uparrow \uparrow 3 = g_1$$, $$g_2 = 3 \uparrow^{(g_1)} 3$$ andâ€¦

• ## Supernova Array

June 23, 2015 by Undeadlift

The Supernova Array is a method of creating very large numbers using recursion of Up-Arrow Notation in a simple linear notation. It is loosely based on the Ultra-Factorial Funcional Array.

• Base Function: $$S(a,b) = a\uparrow^{(b)}a$$, where $$b$$ is the number of $$\uparrow$$'s.
• $$S(a,b,0) = S(a,b)$$
• $$S(a,b,c) = S(a,S(a,b),c-1)$$
• $$S(a,b,c,0) = S(a,b,c)$$
• $$S(a,b,c,d) = S(a,b,S(a,b,c),d-1)$$
• $$S(a,b,c,...x,y,z) = S(a,b,c,...x,S(a,b,c,...x,y),z-1)$$

In general, replace the second to the last input ($$y$$) with the entire function excluding the last input ($$z$$), then subtract 1 from the last input. Do the same for the function within, and the subsequent function, and the function after that, etc. until you reach the base function $$S(a,b)$$.

$$S(1â€¦ Read more > • ## Ultra-Factorial Functional Array June 14, 2015 by Undeadlift The Ultra-Factorial Functional Array (UFA) is a method of creating very large numbers using factorials and recursion in a simple linear notation. • \(|a,b| = a!^{(b)}$$, where $$b$$ is the number of $$!$$'s.

• $$|1,0| = 1$$
• $$|1,1| = 1! = 1$$
• $$|2,1| = 2! = 2$$
• $$|3,1| = 3! = 6$$
• $$|2,2| = 2!! = 2$$
• $$|3,2| = 3!! = 6! = 720$$
• $$|4,2| = 4!! = 24! = 620,448,401,733,239,439,360,000$$
• $$|3,3| = 3!!! = 6!! = 720!$$

In extended notation, another input is added into the array in order to operate a system of recursion.

• $$|a,b,0| = |a,b|$$
• $$|a,b,c| = |a,|a,b|,c-1|$$

In general, if $$c$$ is a positive integer, replace $$b$$ with $$|a,b|$$ and subtract 1 from $$c$$. Repeat as necessary until you reach the base function $$|a,b|$$.

• $$|3,1,0| = |3,1| = 3! = 6$$
• $$|3,1,1| =â€¦ Read more > • ## Factorial Recursion System June 13, 2015 by Undeadlift The Factorial Recursion System (FRS) is a system of notation designed to produce very large numbers using factorials as its base. • \(n\iota 0 = n!$$
• $$n\iota (a+1) = n!^{n\iota a}$$, where there are $$n\iota a$$ number of !s

• $$2\iota 0 = 2! = 2$$
• $$3\iota 0 = 3! = 6$$
• $$2\iota 1 = 2!^{2\iota 0} = 2!^{2!} = 2!! = 2$$
• $$3\iota 1 = 3!^{3\iota 0} = 3!^{3!} = 3!!!!!!$$
• $$3\iota 2 = 3!^{3\iota 1} = 3!^{3!^{3!}} = 3!^{3!!!!!!}$$
• $$3\iota 3 = 3!^{3\iota 2} = 3!^{3!^{3!^{3!}}} = 3!^{3!^{3!!!!!!}}$$ = ultra-3$$\alpha$$

The extended notation is applied similarly to Up-arrow notation:

• $$n\iota \iota a = n\iota (n\iota (n\iota... n$$, where there are $$a$$ number of $$n$$'s
• $$n\iota \iota \iota a = n\iota \iota (n\iota \iota\ (n\iota \iota ... n$$, where there are \â€¦

• ## Very Fast Growing Hierarchy

June 13, 2015 by Undeadlift

A very fast-growing hierarchy (VFGH) is a modification of the fast-growing hierarchy (FGH) designed to compute for even larger numbers than its predecessor.

The following definitions of the VFGH are identical to the FGH:

• $$v_0(n) = n + 1$$
• $$v_\alpha(n) = v_{\alpha[n]}(n)$$ if and only if $$\alpha$$ is a limit ordinal

However, in order to achieve an even faster growth rate, the other definition has been modified with an additional level of iteration:

• $$v_{\alpha+1}(n) = v^{v_{\alpha}(n)} _\alpha(n)$$, where $$v^{v_{\alpha}(n)}$$ denotes function iteration

In general:

• $$v_1(n) = 2n + 1$$
• $$v_2(n) = (2^{2n+1})(n+1) - 1$$
• $$v_3(n) = unknown$$

Following the aforementioned rules and generalities:

• $$v_0(1) = 1 + 1 = 2$$
• $$v_0(2) = 2 + 1 = 3$$
• \(v_1(1) = 2*1 â€¦