{{{id=1| c = continued_fraction(pi, bits=33); c /// [3, 7, 15, 1, 292, 2] }}} {{{id=2| c.convergents() /// [3, 22/7, 333/106, 355/113, 103993/33102, 208341/66317] }}} {{{id=10| [c.pn(n) for n in range(len(c))] /// [3, 22, 333, 355, 103993, 208341] }}} {{{id=11| [c.qn(n) for n in range(len(c))] /// [1, 7, 106, 113, 33102, 66317] }}} {{{id=9| for n in range(-1, len(c)): print c.pn(n)*c.qn(n-1) - c.qn(n)*c.pn(n-1), /// 1 -1 1 -1 1 -1 1 }}} {{{id=8| for n in range(len(c)): print c.pn(n)*c.qn(n-2) - c.qn(n)*c.pn(n-2), /// 3 -7 15 -1 292 -2 }}} {{{id=3| c = continued_fraction(golden_ratio,bits=15) v = [(i, c.pn(i)/c.qn(i)) for i in range(len(c))] eps=0.5 G = plot(golden_ratio,0,len(v))+line(v,color='grey') + points(v,pointsize=50) show(G,ymin=golden_ratio-eps, ymax=golden_ratio+eps) /// }}} {{{id=4| x = 92902834/292034827; x /// 92902834/292034827 }}} {{{id=5| continued_fraction(x) /// [0, 3, 6, 1, 33, 1, 14, 1, 4, 1, 1, 2, 1, 1, 6, 4, 1, 4, 2] }}}

We do repeated long division using the Euclidean algorithm and output the partial quotients.  As we proved, they are the convergents in the above continued fraction.

{{{id=6| (a, b) = (x.numerator(), x.denominator()) while True: q, r = a.quo_rem(b) print q, a, b = b, r if r == 0: break /// 0 3 6 1 33 1 14 1 4 1 1 2 1 1 6 4 1 4 2 }}} {{{id=7| /// }}}