Hayden's Array Notation (HAN) is a notation based on BEAF made by googology wiki user HaydenTheGoogologist2009. It comes in multiple parts, and currently has at least four versions. An example of a valid array in all versions is 3(1, 2)3.

## Syntax

Hayden explains "it has the format a(#)b where # is an array".

Although it is not clarified, a and b are supposed to be non-negative integers so that expressions in the definition make sense. On the other hand, the precise range of # is complicated, as it is not the set of arrays of non-negative integers. For example, the definition includes an expression "a(1)(1)b" and hence the formal string "1)(1" should be valid for #. However, Hayden later clarified that the formal string "1)(1" is invalid. See #Issues for more details.

Also, the reader should be careful that the definition use # to mean the symbol "#" itself.

## Rules

Here, we explain the definitions of first four functions in Hayden's Array Notation, which Hayden specifies as "Rules".

### Original version (version 1.0)

First, we explain the original version, which is outdated:

1. a()b (no matter what does a or b equal to) = 1
2. 0(b)a (if b is not equal 0 ≠ 1) = 0
3. 1(b)a (if b ≥ 3) = 1
4. a(b)1 (if b ≥ 2) = a

### Updated version (versions after 2.0)

Next, we explain the updated version:

1. a()b (no matter what does a or b equal to) = 1
2. 0(b)a can have these following answers: If b is equal to 0 = 1. If b is equal to 1 = a. If b is equal to 2 = 0. If b is equal to 3 (if a ≠ 0) = 0, otherwise, it's undefined. If b ≥ 4, or it's an array (e.g. a(c)(d)b and a(c, d)b) = Undefined because if a = 0, it will make a power tower of 0 with 0 levels high, which doesn't make any sense. If a > 0, it will be a power tower of 0 with a levels high, and 0^0 is undefined in this context. (Note: 0^0 is defined as 0 or 1 in other contexts in mathematics.)
3. 1(b)a (if b ≥ 3) = 1
4. a(b)1 (if b ≥ 2) = a

Although the second line in the rule overlaps lines in the definition of other parts, it is compatible with them. Therefore the overlapping does not cause a problem.

## Definition

Here, we explain the definitions of all other functions in Hayden's Array Notation. There are a few things to note:

• All operations are solved from left to right.
• ^ indicates arrow notation.
• {} indicates BEAF.

### Original version (versions 1.0 and 2.0)

First, we explain the original version, which is outdated:

a()b = 1

a(0)b = a + 1

a(1)b = a + b

a(2)b = a * b

a(3)b = a^b

a(4)b = a^^b

a(5)b = a^^^b

a(c)b = a^^^ ... ^^^ b (c - 2 arrows) (for c is equal or greater than 3)

a(1)(1)b = {a, b, 1, 2}

a(2)(1)b = {a, b, 2, 2}

a(3)(1)b = {a, b, 3, 2}

a(1)(2)b = {a, b, 1, 3}

a(2)(2)b = {a, b, 2, 3}

a(3)(2)b = {a, b, 3, 3}

a(1)(3)b = {a, b, 1, 4}

a(2)(3)b = {a, b, 2, 4}

a(3)(3)b = {a, b, 3, 4}

a(1)(4)b = {a, b, 1, 5}

a(1)(5)b = {a, b, 1, 6}

a(1)(1)(1)b = {a, b, 1, 1, 2}

a(2)(1)(1)b = {a, b, 2, 1, 2}

a(1)(2)(1)b = {a, b, 1, 2, 2}

a(1)(1)(2)b = {a, b, 1, 1, 3}

a(1)(1)(1)(1)b = {a, b, 1, 1, 1, 2}

a(1)(1)(1)(1)(1)b = {a, b, 1, 1, 1, 1, 2}

a(c)(d)(e) … (x)(y)(z)b = {a, b, c, d, e, … x, y, z + 1}

a(1, 1)b = {a, b (1) 2}

a(1, 1)b / c = {a, b, c (1) 2}

a(1, 1)b / c / d = {a, b, c, d (1) 2}

a(1, 1)b / c / d / … x / y / z = {a, b, c, d, … x, y, z (1) 2}

a(1, 2)b = {a, b (1) 3}

a(1, 2)b / c = {a, b, c (1) 3}

a(1, 2)b / c / d = {a, b, c, d (1) 3}

a(1, 3)b = {a, b (1) 4}

a(1, 3)b / c = {a, b, c (1) 4}

a(1, 3)b / c / d = {a, b, c, d (1) 4}

a(1, 4)b = {a, b (1) 5}

a(1, 5)b = {a, b (1) 6}

a(1, c)b = {a, b (1) c + 1}

a(1, 1 \ 2)b = {a, b (1) 1, 2}

a(1, 2 \ 1)b = {a, b (1) 2, 2}

a(1, 3 \ 1)b = {a, b (1) 3, 2}

a(1, 1 \ 2)b = {a, b (1) 1, 3}

a(1, 2 \ 2)b = {a, b (1) 2, 3}

a(1, 3 \ 2)b = {a, b (1) 3, 3}

a(1, 1 \ 3)b = {a, b (1) 1, 4}

a(1, 2 \ 3)b = {a, b (1) 2, 4}

a(1, 3 \ 3)b = {a, b (1) 3, 4}

a(1, 1 \ 4)b = {a, b (1) 1, 5}

a(1, 1 \ 4)b = {a, b (1) 1, 6}

a(1, 1 \ 1 \ 1)b = {a, b (1) 1, 1, 2}

a(1, 2 \ 1 \ 1)b = {a, b (1) 2, 1, 2}

a(1, 1 \ 2 \ 1)b = {a, b (1) 1, 2, 2}

a(1, 1 \ 1 \ 2)b = {a, b (1) 1, 1, 3}

a(1, 1 \ 1 \ 1 \ 1)b = {a, b (1) 1, 1, 1, 2}

a(1, 1 \ 1 \ 1 \ 1 \ 1)b = {a, b (1) 1, 1, 1, 1, 2}

a(1, c \ d \ e \ … x \ y \ z)b = {a, b (1) c, d, e, … x, y, z + 1}

a(1 | 1, 1)b = {a, b (1)(1) 2}

a(1 | 1 | 1, 1)b = {a, b (1)(1)(1) 2}

a(1 | 1 | 1 | … 1 | 1 | 1, 1)b (c 1’s) = {a, b (1)(1)(1) … (1)(1)(1) 2} (c 1’s)

a(2, 1)b = {a, b (2) 2}

a(3, 1)b = {a, b (3) 2}

a(c, 1)b = {a, b (c) 2}

a(0 ¬ 1, 1)b = {a, b (0, 1) 2}

a(1 ¬ 1, 1)b = {a, b (1, 1) 2}

a(2 ¬ 1, 1)b = {a, b (2, 1) 2}

a(3 ¬ 1, 1)b = {a, b (3, 1) 2}

a(0 ¬ 2, 1)b = {a, b (0, 2) 2}

a(1 ¬ 2, 1)b = {a, b (1, 2) 2}

a(2 ¬ 2, 1)b = {a, b (2, 2) 2}

a(3 ¬ 2, 1)b = {a, b (3, 2) 2}

a(0 ¬ 3, 1)b = {a, b (0, 3) 2}

a(0 ¬ 4, 1)b = {a, b (0, 4) 2}

a(0 ¬ 5, 1)b = {a, b (0, 5) 2}

a(0 ¬ 0 ¬ 1, 1)b = {a, b (0, 0, 1) 2}

a(1 ¬ 0 ¬ 1, 1)b = {a, b (1, 0, 1) 2}

a(0 ¬ 1 ¬ 1, 1)b = {a, b (0, 1, 1) 2}

a(0 ¬ 0 ¬ 2, 1)b = {a, b (0, 0, 2) 2}

a(0 ¬ 0 ¬ 0 ¬ 1, 1)b = {a, b (0, 0, 0, 1) 2}

a(0 ¬ 0 ¬ 0 ¬ 0 ¬ 1, 1)b = {a, b (0, 0, 0, 0, 1) 2}

a(c ¬ d ¬ e ¬ … x ¬ y ¬ z, 1)b (c 0’s) = {a, b (c, d, e, … x, y, z) 2} (c 0’s)

a(1#1, 1)b = a(1##2, 1)b = {a, b ((1)1) 2}

a(2#1, 1)b = {a, b ((2)1) 2}

a(3#1, 1)b = {a, b ((3)1) 2}

a(1#2, 1)b = {a, b ((1)2) 2}

a(1#3, 1)b = {a, b ((1)3) 2}

a(1#1#1, 1)b = a(1##3, 1)b = {a, b (((1)1)1) 2}

a(2#1#1, 1)b = {a, b (((2)1)1) 2}

a(1#2#1, 1)b = {a, b (((1)2)1) 2}

a(1#1#2, 1)b = {a, b (((1)1)2) 2}

a(1#1#1#1, 1)b = a(1##4, 1)b = {a, b ((((1)1)1)1) 2}

a(1#1#1#1#1, 1)b = a(1##5, 1)b = {a, b ((((1)1)1)1) 2}

a(1#1#1# … 1#1#1, 1)b (c 1’s) = a(1##c, 1)b = {a, b ((( … (((1)1)1) … 1)1)1) 2} (c 1’s)

### Updated version (version 3.0)

Next, we explain the updated version, which is also outdated:

a()b = 1

a(0)b = a + 1

a(1)b = a + b

a(2)b = a * b

a(3)b = a^b

a(4)b = a^^b

a(5)b = a^^^b

a(c)b = a^^^ ... ^^^b (c - 2 arrows) (for c is equal or greater than 3)

a(1)(1)b = {a, b, 1, 2}

a(2)(1)b = {a, b, 2, 2}

a(3)(1)b = {a, b, 3, 2}

a(1)(2)b = {a, b, 1, 3}

a(2)(2)b = {a, b, 2, 3}

a(3)(2)b = {a, b, 3, 3}

a(1)(3)b = {a, b, 1, 4}

a(2)(3)b = {a, b, 2, 4}

a(3)(3)b = {a, b, 3, 4}

a(1)(4)b = {a, b, 1, 5}

a(1)(5)b = {a, b, 1, 6}

a(1)(1)(1)b = {a, b, 1, 1, 2}

a(2)(1)(1)b = {a, b, 2, 1, 2}

a(1)(2)(1)b = {a, b, 1, 2, 2}

a(1)(1)(2)b = {a, b, 1, 1, 3}

a(1)(1)(1)(1)b = {a, b, 1, 1, 1, 2}

a(1)(1)(1)(1)(1)b = {a, b, 1, 1, 1, 1, 2}

a(c)(d)(e) … (x)(y)(z)b = {a, b, c, d, e, … x, y, z + 1}

a(1, 1)b = {a, b (1) 2}

a(1, 1)b / c = {a, b, c (1) 2}

a(1, 1)b / c / d = {a, b, c, d (1) 2}

a(1, 1)b / c / d / … x / y / z = {a, b, c, d, … x, y, z (1) 2}

a(1, 2)b = {a, b (1) 3}

a(1, 3)b = {a, b (1) 4}

a(1, 4)b = {a, b (1) 5}

a(1, 5)b = {a, b (1) 6}

a(1, c)b = {a, b (1) c + 1}

a(1, 1 \ 2)b = {a, b (1) 1, 2}

a(1, 2 \ 1)b = {a, b (1) 2, 2}

a(1, 3 \ 1)b = {a, b (1) 3, 2}

a(1, 1 \ 2)b = {a, b (1) 1, 3}

a(1, 2 \ 2)b = {a, b (1) 2, 3}

a(1, 3 \ 2)b = {a, b (1) 3, 3}

a(1, 1 \ 3)b = {a, b (1) 1, 4}

a(1, 2 \ 3)b = {a, b (1) 2, 4}

a(1, 3 \ 3)b = {a, b (1) 3, 4}

a(1, 1 \ 4)b = {a, b (1) 1, 5}

a(1, 1 \ 5)b = {a, b (1) 1, 6}

a(1, 1 \ 1 \ 1)b = {a, b (1) 1, 1, 2}

a(1, 2 \ 1 \ 1)b = {a, b (1) 2, 1, 2}

a(1, 1 \ 2 \ 1)b = {a, b (1) 1, 2, 2}

a(1, 1 \ 1 \ 2)b = {a, b (1) 1, 1, 3}

a(1, 1 \ 1 \ 1 \ 1)b = {a, b (1) 1, 1, 1, 2}

a(1, 1 \ 1 \ 1 \ 1 \ 1)b = {a, b (1) 1, 1, 1, 1, 2}

a(1, c \ d \ e \ … x \ y \ z)b = {a, b (1) c, d, e, … x, y, z + 1}

a(1 | 1, 1)b = {a, b (1)(1) 2}

a(1 | 1 | 1, 1)b = {a, b (1)(1)(1) 2}

a(1 | 1 | 1 | … 1 | 1 | 1, 1)b (c 1’s) = {a, b (1)(1)(1) … (1)(1)(1) 2} (c 1’s)

a(2, 1)b = {a, b (2) 2}

a(3, 1)b = {a, b (3) 2}

a(c, 1)b = {a, b (c) 2}

a(0 ¬ 1, 1)b = {a, b (0, 1) 2}

a(1 ¬ 1, 1)b = {a, b (1, 1) 2}

a(2 ¬ 1, 1)b = {a, b (2, 1) 2}

a(3 ¬ 1, 1)b = {a, b (3, 1) 2}

a(0 ¬ 2, 1)b = {a, b (0, 2) 2}

a(1 ¬ 2, 1)b = {a, b (1, 2) 2}

a(2 ¬ 2, 1)b = {a, b (2, 2) 2}

a(3 ¬ 2, 1)b = {a, b (3, 2) 2}

a(0 ¬ 3, 1)b = {a, b (0, 3) 2}

a(1 ¬ 3, 1)b = {a, b (1, 3) 2}

a(2 ¬ 3, 1)b = {a, b (2, 3) 2}

a(3 ¬ 3, 1)b = {a, b (3, 3) 2}

a(0 ¬ 4, 1)b = {a, b (0, 4) 2}

a(0 ¬ 5, 1)b = {a, b (0, 5) 2}

a(0 ¬ 0 ¬ 1, 1)b = {a, b (0, 0, 1) 2}

a(1 ¬ 0 ¬ 1, 1)b = {a, b (1, 0, 1) 2}

a(0 ¬ 1 ¬ 1, 1)b = {a, b (0, 1, 1) 2}

a(0 ¬ 0 ¬ 2, 1)b = {a, b (0, 0, 2) 2}

a(0 ¬ 0 ¬ 0 ¬ 1, 1)b = {a, b (0, 0, 0, 1) 2}

a(0 ¬ 0 ¬ 0 ¬ 0 ¬ 1, 1)b = {a, b (0, 0, 0, 0, 1) 2}

a(c ¬ d ¬ e ¬ … x ¬ y ¬ z, 1)b (c 0’s) = {a, b (c, d, e, … x, y, z) 2} (c 0’s)

a(1#1, 1)b = a(1##2, 1)b = {a, b ((1)1) 2}

a(2#1, 1)b = {a, b ((2)1) 2}

a(3#1, 1)b = {a, b ((3)1) 2}

a(1#2, 1)b = {a, b ((1)2) 2}

a(1#3, 1)b = {a, b ((1)3) 2}

a(1#1#1, 1)b = a(1##3, 1)b = {a, b (((1)1)1) 2}

a(2#1#1, 1)b = {a, b (((2)1)1) 2}

a(1#2#1, 1)b = {a, b (((1)2)1) 2}

a(1#1#2, 1)b = {a, b (((1)1)2) 2}

a(1#1#1#1, 1)b = a(1##4, 1)b = {a, b ((((1)1)1)1) 2}

a(1#1#1#1#1, 1)b = a(1##5, 1)b = {a, b (((((1)1)1)1)1) 2}

a(1#1#1# … 1#1#1, 1)b (c 1’s) = a(1##c, 1)b = {a, b ((( … (((1)1)1) … 1)1)1) 2} (c 1’s)

### Second updated version (version 4.0)

And finally, we explain the second updated version, which is the current version:

a()b = 1

a(0)b = a + 1

a(1)b = a + b

a(2)b = a * b

a(3)b = a^b

a(4)b = a^^b

a(5)b = a^^^b

a(c)b = a^^^ ... ^^^b (c - 2 arrows) (for c is equal or greater than 3)

a(1)(1)b = {a, b, 1, 2}

a(2)(1)b = {a, b, 2, 2}

a(3)(1)b = {a, b, 3, 2}

a(1)(2)b = {a, b, 1, 3}

a(2)(2)b = {a, b, 2, 3}

a(3)(2)b = {a, b, 3, 3}

a(1)(3)b = {a, b, 1, 4}

a(2)(3)b = {a, b, 2, 4}

a(3)(3)b = {a, b, 3, 4}

a(1)(4)b = {a, b, 1, 5}

a(1)(5)b = {a, b, 1, 6}

a(1)(1)(1)b = {a, b, 1, 1, 2}

a(2)(1)(1)b = {a, b, 2, 1, 2}

a(1)(2)(1)b = {a, b, 1, 2, 2}

a(1)(1)(2)b = {a, b, 1, 1, 3}

a(1)(1)(1)(1)b = {a, b, 1, 1, 1, 2}

a(1)(1)(1)(1)(1)b = {a, b, 1, 1, 1, 1, 2}

a(c)(d)(e) … (x)(y)(z)b = {a, b, c, d, e, … x, y, z + 1}

a(1, 1)b = {a, b (1) 2}

a(1, 1)b / c = {a, b, c (1) 2}

a(1, 1)b / c / d = {a, b, c, d (1) 2}

a(1, 1)b / c / d / … x / y / z = {a, b, c, d, … x, y, z (1) 2}

a(1, 2)b = {a, b (1) 3}

a(1, 3)b = {a, b (1) 4}

a(1, 4)b = {a, b (1) 5}

a(1, 5)b = {a, b (1) 6}

a(1, c)b = {a, b (1) c + 1}

a(1, 1 \ 2)b = {a, b (1) 1, 2}

a(1, 2 \ 1)b = {a, b (1) 2, 2}

a(1, 3 \ 1)b = {a, b (1) 3, 2}

a(1, 1 \ 2)b = {a, b (1) 1, 3}

a(1, 2 \ 2)b = {a, b (1) 2, 3}

a(1, 3 \ 2)b = {a, b (1) 3, 3}

a(1, 1 \ 3)b = {a, b (1) 1, 4}

a(1, 2 \ 3)b = {a, b (1) 2, 4}

a(1, 3 \ 3)b = {a, b (1) 3, 4}

a(1, 1 \ 4)b = {a, b (1) 1, 5}

a(1, 1 \ 5)b = {a, b (1) 1, 6}

a(1, 1 \ 1 \ 1)b = {a, b (1) 1, 1, 2}

a(1, 2 \ 1 \ 1)b = {a, b (1) 2, 1, 2}

a(1, 1 \ 2 \ 1)b = {a, b (1) 1, 2, 2}

a(1, 1 \ 1 \ 2)b = {a, b (1) 1, 1, 3}

a(1, 1 \ 1 \ 1 \ 1)b = {a, b (1) 1, 1, 1, 2}

a(1, 1 \ 1 \ 1 \ 1 \ 1)b = {a, b (1) 1, 1, 1, 1, 2}

a(1, c \ d \ e \ … x \ y \ z)b = {a, b (1) c, d, e, … x, y, z + 1}

a(1 | 1, 1)b = {a, b (1)(1) 2}

a(1 | 1 | 1, 1)b = {a, b (1)(1)(1) 2}

a(1 | 1 | 1 | … 1 | 1 | 1, 1)b (c 1’s) = {a, b (1)(1)(1) … (1)(1)(1) 2} (c 1’s)

a(2, 1)b = {a, b (2) 2}

a(3, 1)b = {a, b (3) 2}

a(c, 1)b = {a, b (c) 2}

a(0 ¬ 1, 1)b = {a, b (0, 1) 2}

a(1 ¬ 1, 1)b = {a, b (1, 1) 2}

a(2 ¬ 1, 1)b = {a, b (2, 1) 2}

a(3 ¬ 1, 1)b = {a, b (3, 1) 2}

a(0 ¬ 2, 1)b = {a, b (0, 2) 2}

a(1 ¬ 2, 1)b = {a, b (1, 2) 2}

a(2 ¬ 2, 1)b = {a, b (2, 2) 2}

a(3 ¬ 2, 1)b = {a, b (3, 2) 2}

a(0 ¬ 3, 1)b = {a, b (0, 3) 2}

a(1 ¬ 3, 1)b = {a, b (1, 3) 2}

a(2 ¬ 3, 1)b = {a, b (2, 3) 2}

a(3 ¬ 3, 1)b = {a, b (3, 3) 2}

a(0 ¬ 0 ¬ 1, 1)b = {a, b (0, 0, 1) 2}

a(1 ¬ 0 ¬ 1, 1)b = {a, b (1, 0, 1) 2}

a(0 ¬ 1 ¬ 1, 1)b = {a, b (0, 1, 1) 2}

a(0 ¬ 0 ¬ 2, 1)b = {a, b (0, 0, 2) 2}

a(0 ¬ 0 ¬ 0 ¬ 1, 1)b = {a, b (0, 0, 0, 1) 2}

a(0 ¬ 0 ¬ 0 ¬ 0 ¬ 1, 1)b = {a, b (0, 0, 0, 0, 1) 2}

a(c ¬ d ¬ e ¬ … x ¬ y ¬ z, 1)b (c 0’s) = {a, b (c, d, e, … x, y, z) 2} (c 0’s)

a(1#1, 1)b = a(1##2, 1)b = {a, b ((1)1) 2}

a(1#*1*#1, 1)b = {a, b (1 (1) 1) 2}

a(0 ¬ 1#*1*#1, 1)b = {a, b (0, 1 (1) 1) 2}

a(1#2, 1)b = {a, b ((1)2) 2}

a(1#3, 1)b = {a, b ((1)3) 2}

a(1#0 ¬ 1, 1)b = {a, b ((1)0, 1) 2}

a(1 | 1#1, 1)b = {a, b ((1)(1)1) 2}

a(1 | 1 | 1#1, 1)b = {a, b ((1)(1)(1)1) 2}

a(2#1, 1)b = {a, b ((2)1) 2}

a(3#1, 1)b = {a, b ((3)1) 2}

a(0 ¬ 1#1, 1)b = {a, b ((0, 1)1) 2}

a(0 ¬ 0 ¬ 1#1, 1)b = {a, b ((0, 0, 1)1) 2}

a(1#1#1, 1)b = a(1##3, 1)b = {a, b (((1)1)1) 2}

a(2#1#1, 1)b = {a, b (((2)1)1) 2}

a(0 ¬ 1#1#1, 1)b = {a, b (((0, 1)1)1) 2}

a(1#1#1#1, 1)b = a(1##4, 1)b = {a, b ((((1)1)1)1) 2}

a(1#1#1#1#1, 1)b = a(1##5, 1)b = {a, b (((((1)1)1)1)1) 2}

a(1#1#1# … 1#1#1, 1)b (c 1’s) = a(1##c, 1)b = {a, b ((( … (((1)1)1) … 1)1)1) 2} (c 1’s)

## Examples

• 2(0)2 = 2 + 1 = 3
• 3(1)3 = 3 + 3 = 6
• 7(2)10 = 7 * 10 = 70
• 2(3)3 = 2^3 = 8
• 5(3)6 = 5^6 = 15,625
• 10(3)100 = 10^100 = googol
• 3(4)4 = 3^^4 = 3^3^3^3 = 3^3^27 = 3^7,625,597,484,987 $$\approx 1.2580143*10^{3,638,334,640,024}$$
• 5(4)3 = 5^^3 = 5^5^5 $$\approx 1.9110126*10^{2,184}$$
• 2(5)2 = 2^^^2 = 2^^2 = 2^2 = 4
• 2(6)2 = 2^^^^2 = 2^^^2 = 2^^2 = 2^2 = 4
• 3(5)2 = 3^^^2 = 3^^3 = 3^3^3 = 3^27 = 7,625,597,484,987
• 2(6)3 = 2^^^3 = 2^^2^^2 = 2^^4 = 2^2^2^2 = 2^2^4 = 2^16 = 65,536
• 3(6)3 = 3^^3^^3 = 3^^7,625,597,484,987 = tritri
• 3(7)3 = 3^^^^3 = 3^^^3^^^3 = 3^^^tritri = grahal
• 10(12)10 = 10 {10} 10 = tridecal
• 10(102)10 = 10 {100} 10 = boogol
• 10(1)(1)100 = {10, 100, 1, 2} = corporal
• 10(10)(9)10 = {10, 10, 10, 10} = general
• 10(10)(10)(9)10 = {10, 10, 10, 10, 10} = pentadecal
• 10(1, 1)100 = {10, 100 (1) 2} = goobol
• 10(1, 9)10 = {10, 10 (1) 10} = emperal
• 10(1, 10 \ 9)10 = {10, 10 (1) 10, 10} = hyperal
• 10(1 | 1, 1}10 = {10, 10 (1)(1) 2} = diteral
• 10(1 | 1, 9)10 = {10, 10 (1)(1) 10} = admiral
• 10(2, 1)10 = {10, 10 (2) 2} = xappol
• 10(10, 1)10 = {10, 10 (10) 2} = dimendecal
• 10(100, 1)10 = {10, 10 (100) 2} = gongulus
• 10(0 ¬ 3, 1)100 = {10, 100 (0, 3) 2} = gangulus
• 10(0 ¬ 0 ¬ 1, 1) = {10, 100 (0, 0, 1) 2} = bongulus
• 10(0 ¬ 0 ¬ 0 ¬ 1, 1) = {10, 100 (0, 0, 0, 1) 2} = trongulus
• 10(1#1, 1)100 = 10(1##2, 1)100 = {10, 100 ((1)1) 2} = goplexulus
• 10(0 ¬ 1#1, 1)100 = {10, 100 ((0, 1)1) 2} = goduplexulus
• 10(1#1#1, 1)100 = 10(1##3, 1)100 = {10, 100 (((1)1)1) 2} = gotriplexulus

## Issues

According to Hayden, the array # in a valid expression a(#)b should not be the formal string "1)(1". However, this contradicts the fact that the definition includes the equation "a(1)(1)b = {a, b, 1, 2}", which should imply that a(1)(1)b is a valid expression.

Also, according to Hayden, the definition of both versions are exactly the same, but Hayden just removed some functions to make it more clean. However, another user pointed out that the updated definition does not seem to have a rule to compute "a(1, 2)b / c", which is a valid expression for the original definition. If it is correct, then the statement that the definition of both versions are exactly the same is wrong.

## History of sources

The number was first described in "Hayden's Array Notation", but was later moved from it to "Hayden's Big Numbers" in the same name.