3 ms·
All their conjectures have low-degree polynomial formulae for both numerators and denominators, with exceptions allowed in the first couple. So I guess the nth
by gjm11 7y ago
All their conjectures have low-degree polynomial formulae for both numerators and denominators, with exceptions allowed in the first couple. So I guess the nth numerator, counting from 0, is -n(2n-3) which goes 0, 1, -2, -9, -20, ..., so n=0 is a special case.
It would seem more natural to rewrite it like this
-4/pi = −1 + 1/(−4 + −2 /(−7 + −9/(−10 + −20/(−13+...))))
which avoids the gratuitously different first numerator -- but I guess they wanted to reproduce results exactly as they happened to emerge from their program.
I think this is, further, equivalent to the following which avoids some gratuitous-looking minus signs:
4/pi = 1 + 1/(4 + −2 /(7 + −9/(10 + −20/(13+...))))
Continuing the fraction using the quadratic polynomial I gave above does indeed seem to make it converge to 4/pi, though not very quickly.
- gjm11 7y agoI wonder whether this one might be equivalent to a known infinite product for pi or pi/4 or something. The ratios between successive convergents are all of the form (a_n+k_n)/a_n where k_n is 1, 1, 3, 3, 15, 15, 105, 105, 315, 315, ... -- i.e., the least common multiples of odd numbers up to n. [EDITED to add:] This should make you think of the Wallis product formula for pi/4, though this doesn't seem to be the Wallis product in disguise. Clearly these k_n divide n!, so maybe the right way to say this is that the conjecture seems to be equivalent to 4/pi = product (a_n+n!)/a_n where (a_n) = (4,130,2464,45448,882528,18410640,...) ... though I don't know what that sequence _is_, haven't shown that it has a nice form, etc. None of (a_n), (a_n+n!), (a_n+n!/2) seems to occur in OEIS or to be a subsequence of anything in OEIS.