Here's what GPT-4 had to say about the TECO code in the song:

Explain every detail of this TECO code that you can, and speculate on what it does.

[1:i\*^Yu14<q1&377.f"nir'q1/400.u1>^[[8
.-z(1702117120m81869946983m8w660873337m8w1466458484m8
)+z,.f^@fx\*[0:ft^]0^[w^\

[1:i\*^Yu16<q1&77.+32iq1f"l#-1/100.#-1&7777777777.'"#/100.'u1r>6c^[[6
.(675041640067.m6w416300715765.m6w004445675045.m6
455445440046.m6w576200535144.m6w370000000000.m6),.fx\*[0:ft^]0^[w^\

ChatGPT:

TECO (Text Editor and COrrector) is a primitive text editor from the early days of computer programming. TECO code is notoriously difficult to read, as it's a sort of "write-only" language with many commands being single punctuation characters and having different meanings depending on context. Its language is an esoteric one, with little resemblance to most modern programming languages.

Let's try to break down the given TECO code line by line:

[1:i\*^Yu14<q1&377.f"nir'q1/400.u1>^[[8

Here, [1 is saving the current buffer's contents to Q-register 1 (a kind of temporary storage). i\*^Y is inserting the character with ASCII code 25 (^Y stands for decimal 25, \* is for output). u14 is storing the value 14 in Q-register 4.

<q1&377.f"nir'q1/400.u1> is a conditional loop that executes while the value in Q-register 1 (the contents of the buffer) ANDed with 377 (octal, or 255 decimal) is not 0. Inside the loop, it's reading one character from the buffer (.f), if it's not end of file ("n), it's returning the pointer back to the character just read (ir'), otherwise it's storing the value of Q-register 1 divided by 400 in Q-register 1 (q1/400.u1).

>^[[8 at the end is somewhat unclear without context. ^[[ is often a prefix for terminal control sequences (also called escape sequences), but 8 doesn't correspond to a typical terminal control sequence. This might be a TECO-specific sequence.

.-z(1702117120m81869946983m8w660873337m8w1466458484m8)+z,.f^@fx\*[0:ft^]0^[w^\

Here, .-z( to )+z, is calculating a range in the buffer. .-z is the difference between the current pointer position and the end of the buffer, and z is the length of the buffer. 1702117120m81869946983m8w660873337m8w1466458484m8 are arithmetic operations involving multiply (m) and bitwise OR (w), but without the context it's hard to say exactly what this line is doing.

.f^@ reads one character from the buffer and replaces it with ASCII null (^@ stands for decimal 0, and f is for fetching). fx exchanges the Q-register and the buffer, meaning that the contents of the buffer are now in the Q-register and the previous Q-register contents are now in the buffer. \*[0:ft^]0^[w^\ are performing a series of operations involving Q-register 0, fetching and type commands but, without context, it's hard to say exactly what this is doing.

[1:i\*^Yu16<q1&77.+32iq1f"l#-1/100.#-1&7777777777.'"#/100.'u1r>6c^[[6

This is similar to the first line but with a few differences. u16 is storing the value 16 in Q-register 6. <q1&77.+32i is a conditional loop that executes while the value in Q-register 1 ANDed with 77 (octal, or 63 decimal) is not 0. Inside the loop, it's adding 32 to the character in the buffer, essentially converting uppercase ASCII characters to lowercase.

q1f"l#-1/100.#-1&7777777777.'"#/100.'u1r is a series of commands inside the loop. If the fetched character is less than -1, it's performing a series of operations involving division and bitwise AND, and storing the result in Q-register 1. The exact purpose of these operations isn't clear without context.

>6c^[[6 at the end is also unclear. It could be storing the character with ASCII code 6 in Q-register 6, but without context, it's hard to say for sure.

.(675041640067.m6w416300715765.m6w004445675045.m6455444046.m6w576200535144.m6w370000000000.m6),.fx\*[0:ft^]0^[w^\\

Similar to the second line, this line is performing a series of arithmetic operations with the contents of Q-register 6. After that, it's again exchanging the buffer with a Q-register (presumably the Q-register with the result of the arithmetic operations).

In terms of what this code does, it's difficult to determine without context. It's performing a series of operations on the contents of the buffer, and these operations involve arithmetic, bitwise, and possibly character manipulation. But the exact purpose of these operations isn't clear from the code alone. For example, we can see that the code is interacting with the contents of the buffer and Q-registers, but it's not clear what these contents are or what the desired outcome is.

TECO's complex syntax and the lack of modern programming conventions makes it very difficult to determine what a specific piece of TECO code does without additional context or comments.