hyperop.txt

0: [.]      succ(a,  ): add `1` to `a`
1: [+]       add(a, b): ret = a; ret =       succ(ret)    `b`     times
2: [*]       mul(a, b): ret = 0; ret =        add(ret, a) `b`     times
3: [^]       exp(a, b): ret = 1; ret =        mul(ret, a) `b`     times
4: [!] tetration(a, b): ret = a; ret =        exp(ret, a) `b - 1` times (special case b = 0 -> 1)
5: [@] pentation(a, b): ret = a; ret =  tetration(ret, a) `b - 1` times (special case b = 0 -> 1)
6: [#]  hexation(a, b): ret = a; ret =  pentation(ret, a) `b - 1` times (special case b = 0 -> 1)

3 .   -> 4
3 + 4 -> 7                 { ((((3 .  ) .  ) .  ) .  ) }
3 * 4 -> 12                { ((((0 + 3) + 3) + 3) + 3) }
3 ^ 4 -> 81                { ((((1 * 3) * 3) * 3) * 3) }
3 ! 4 -> 7,625,597,484,987 { ((((    3) ^ 3) ^ 3) ^ 3) }
3 @ 5 ->                   { ((((    3) ! 3) ! 3) ! 3) }
3 # 6 ->                   { ((((    3) @ 3) @ 3) @ 3) }

hyperop_.py

from functools import reduce as fold; from itertools import repeat
hyperop = lambda rank: (
    (lambda a, _: a + 1) if not rank else
    lambda a, b: fold(
        hyperop(rank - 1),
        repeat(a, b - (b and rank >= 3)),
        {1:a, 2:0}.get(rank, a if b and rank >= 3 else 1),
    )
)

# incorrect, unfortunately:
    # lambda a, b: fold(hyperop(rank - 1), repeat(a, b), {1:a, 2:0}.get(rank, 1))

succ = hyperop(0)
add  = hyperop(1)
mul  = hyperop(2)
exp  = hyperop(3)
tetr = hyperop(4)
pent = hyperop(5)
hexa = hyperop(6)

[print(f"[{n}] {hyperop(n)(3, 4)=}") for n in range(0, 6)]

hyperop__.py

from functools import reduce as f;from itertools import repeat as n
h=lambda r:(lambda a,b:f(h(r-1),n(a,b-(z:=(b and r>2))),{1:a,2:0}.get(r,a if z else 1))) if r else lambda a,_:a+1