19 Apr
2005
19 Apr
'05
2:31 p.m.
inf ==== k \ %pi (2 k + 1) a > tan(-------------) inf / n ==== ==== b \ n = 2 > --------------------------- / s ==== (2 k + 1) k = 0 --------------------------------- s %pi a = +-1, b = +-2 appears to be rational exactly when inf ==== k \ (- a) > ---------- / s ==== (2 k + 1) k = 0 ---------------- is, thus giving four rational sequences s %pi (with somewhat predictable denominators). --rwg