[isaacfrond]

The number you are looking for is e^(sqrt(163) pi). According to Wikipedia:

In a 1975 April Fool article in Scientific American magazine,[8] "Mathematical Games" columnist Martin Gardner made the hoax claim that the number was in fact an integer, and that the Indian mathematical genius Srinivasa Ramanujan had predicted it – hence its name.

It is not an integer of course.

[toth]

Actually `e^(sqrt(n) pi)` is very close to being an integer for a couple of different `n`s, including 67 and 163. For 163 it's much closer to an integer, but for 67 you get something you can easily check in double precision floats is close to an integer, so I thought it worked better as a joke answer :)

FYI, the reason you get these almost integers is related to the `n`s being Heegner numbers, see https://en.wikipedia.org/wiki/Heegner_number.

[Someone]

> The number you are looking for is e^(sqrt(163) pi) […] It is not an integer of course.

Of course? I’m not aware that we have some theorem other than “we computed it to lots of decimals, and it isn’t an integer” from which that follows.

[less_less]

It's not really "of course", and I don't think we have such a theorem in general. But in this case, I believe the fact that it's not an integer follows from the same theorem that says it's very close to an integer. See eg https://math.stackexchange.com/questions/4544/why-is-e-pi-sq...

Basically e^(sqrt(163)*pi) is the leading term in a Laurent series for an integer, and the other (non-integer) terms are really small but not zero.