Pentation

Pentation is the operation of repeated tetration, just as tetration is the operation of repeated exponentation and is a hyperoperation. Like tetration, it has little real-world applications. It is non-commutative, and therefore has two inverse functions, which might be named the penta-root and the penta-logarithm (analogous to the two inverse functions for exponentiation: nth root function and logarithm). Pentation also bounds the elementary recursive functions.

The word "Pentation" was coined by Reuben Goodstein from the roots penta- (five) and iteration). It is part of his general naming scheme for hyperoperations.

Pentation can be written in Knuth's up-arrow notation as $a\uparrow \uparrow \uparrow b$ or $a\uparrow ^{3}b$.

Extension

It is not known how to extend pentation to complex or non-integer values.

Extension to zero and negative numbers

Using super-logarithm‎s, $a\uparrow ^{3}b$ can be done when b is negative or 0, but doing so is a lot more limited. For all positive integer values of a negative pentation is as follows:

• $a\uparrow ^{3}0=\operatorname {slog} _{a}a=1,\,$ if a > 1.
• $a\uparrow ^{3}-1=\operatorname {slog} _{a}1=0,\,$ if a > 1.
• $a\uparrow ^{3}-2=\operatorname {slog} _{a}0=-1,\,$ if a > 1.

It can also be done when a is negative, but this is only the case when a is equal to -1. For all positive integer values of b, the three possible answers that you can get for $-1\uparrow ^{3}b$ are shown below:

• $-1\uparrow ^{3}b={^{1}{-1}}=-1,\,$ if b is congruent to 1 modulo 3.
• $-1\uparrow ^{3}b={^{-1}{-1}}=0,\,$ if b is congruent to 2 modulo 3.
• $-1\uparrow ^{3}b={^{0}{-1}}=1,\,$ if b is congruent to 0 modulo 3.

Any other case of this type of pentation produces an undefined result.

Selected values

As its base operation (tetration) has not been extended to non-integer heights, pentation $a\uparrow ^{3}b$ is currently only defined for integer values of a and b where a > 0 and b ≥ 0, and a few other integer values which may be uniquely defined. Like all other hyperoperations of order 3 (exponentiation) and higher, pentation has the following trivial cases (identities) which holds for all values of a and b within its domain:

• $1\uparrow ^{3}b=1$
• $a\uparrow ^{3}1=a$

Other than the trivial cases shown above, pentation generates extremely large numbers very quickly such that there are only a few non-trivial cases that produce numbers that can be written in conventional notation, as illustrated below:

• $2\uparrow ^{3}2={^{2}2}=4$
• $2\uparrow ^{3}3={^{^{2}2}2}=^{4}2=65,536$
• $2\uparrow ^{3}4={^{^{^{2}2}2}2}=^{65,536}2=2^{2^{2^{\cdot ^{\cdot ^{\cdot ^{2}}}}}}{\mbox{ (a power tower of height 65,536) }}\approx \exp _{10}^{65,533}(4.29508)$ (shown here in iterated exponential notation as it's far too large to be written in conventional notation. Note $\exp _{10}(n)=10^{n}$)
• $3\uparrow ^{3}2={^{3}3}=7,625,597,484,987$
• $3\uparrow ^{3}3={^{^{3}3}3}={^{7,625,597,484,987}3}=3^{3^{3^{\cdot ^{\cdot ^{\cdot ^{3}}}}}}{\mbox{ (a power tower of height 7,625,597,484,987) }}\approx \exp _{10}^{7,625,597,484,986}(1.09902)$
• $4\uparrow ^{3}2={^{4}4}=4^{4^{4^{4}}}=4^{4^{256}}\approx \exp _{10}^{3}(2.19)$ (a number with over 10¹⁵³ digits)
• $5\uparrow ^{3}2={^{5}5}=5^{5^{5^{5^{5}}}}=5^{5^{5^{3125}}}\approx \exp _{10}^{4}(3.33928)$ (a number with more than 10¹⁰²¹⁸⁴ digits)

References

• Tetration Forum by Jay D. Fox