Hayden's Array Notation (HAN) is a notation based on BEAF made by googology wiki user HaydenTheGoogologist2009. It comes in multiple parts. An example of a valid array is 3(1, 2)3.
Unfortunately, I abandoned this notation on September 15th, 2022 with the statement 'I created this notation when I was too immature.'
a()b (no matter what does a or b equal to) = 1
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.)
1(b)a (if b ≥ 3) = 1
a(b)1 (if b ≥ 2) = a
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.
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)
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 ≈ 1.2580143 * 10^3,638,334,640,024
5(4)3 = 5^^3 = 5^5^5 ≈ 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^^^2 = 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