Largest number in ten bytes of code

Perl, >1.96835797883262e+18

time*time

Might not be the largest answer... today! But wait enough millennia and it will be!

---

Edit:

To address some of the comments, by "enough millenia," I do in fact mean n^100s of years.

To be fair, if the big freeze/heat death of the universe is how the universe will end (estimated to occur ~10¹⁰⁰ years), the "final" value would be ~10²¹⁴, which is certainly much less than some of the other responses (though, "random quantum fluctuations or quantum tunneling can produce another Big Bang in 10¹⁰⁵⁶ years"). If we take a more optimistic approach (e.g. a cyclic or multiverse model), then time will go on infinitely, and so some day in some universe, on some high-bit architecture, the answer would exceed some of the others.

On the other hand, as pointed out, time is indeed limited by the size of integer/long, so in reality something like ~0 would always produce a larger number than time (i.e. the max time supported by the architecture).

This wasn't the most serious answer, but I hope you guys enjoyed it!

Wolfram ≅ 2.003529930 × 10¹⁹⁷²⁸

Yes, it's a language! It drives the back-end of the popular Wolfram Alpha site. It's the only language I found where the Ackermann function is built-in and abbreviated to less than 6 characters.

In eight characters:

$ ack(4,2)

200352993...719156733

Or ≅ 2.003529930 × 10¹⁹⁷²⁸

ack(4,3), ack(5,2) etc. are much larger, but too large. ack(4,2) is probably the largest Ackermann number than can be completely calculated in under an hour.

Larger numbers are rendered in symbolic form, e.g.:

$ ack(4,3)

2↑²6 - 3 // using Knuth's up-arrow notation

The rules say any output format is allowed, so this might be valid. This is greater than 10^1019727, which is larger than any of the other entries here except for the repeated factorial.

However,

$ ack(9,9)

2↑⁷12 - 3

is larger than the repeated factorial. The largest number I can get in ten characters is:

$ ack(99,99)

2↑⁹⁷102 - 3

This is insanely huge, the Universe isn't big enough to represent a significant portion of its digits, even if you took repeated logs of the number.

Python2 shell, 3,010,301 digits

9<<9999999

Calculating the length: Python will append a "L" to these long numbers, so it reports 1 character more than the result has digits.

>>> len(repr( 9<<9999999 ))
3010302

First and last 20 digits:

40724177878623601356... ...96980669011241992192