3 ms·
It's the coefficient of the linear term in the q-expansion of the j-invariant. The j-invariant of isqrt(163) is an integer exactly, and e^{pisqrt(163)} is the f
by wbl 12y ago
It's the coefficient of the linear term in the q-expansion of the j-invariant. The j-invariant of isqrt(163) is an integer exactly, and e^{pisqrt(163)} is the first term.
- JacobAldridge 12y agoNow I want a t-shirt where the front reads "What's up with 744?" and the back is your response.
- gus_massa 12y agoI use your answer to guide my google searches. The easier explanation I found is in the "details" section of: http://en.wikipedia.org/wiki/Heegner_number#Almost_integers_and_Ramanujan.27s_constant http://en.wikipedia.org/wiki/Heegner_number#Almost_integers_...
- dsuth 12y agoRight. Well, obviously. Off to google...