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%I #5 Nov 02 2014 18:37:51
%S 2,1,5,1,12,1,301,2,78,1,14,1,1,1,10,1,1,2,4,1,4,1,94,1,3,1,1,1,8,1,5,
%T 1,2,2,4,1,10,1,8,1,10,1,13,1,158,1,42,1,18,1,21,1,8,2,2,1,3,1,2,3,23,
%U 1,8,2,39,1,3,1,1,1,7,2,2,1,7,1,5,3,53,1,14,1,6,1,15,1,14,2,5,1,28,1,1,2,4
%N Compute the continued fraction expansion of Pi; multiply each term by i, the square root of -1, compute this new continued fraction and get a number with a real part equal to 0. Then compute the regular continued fraction of the imaginary part of that new number.
%o (PARI) k=contfracpnqn(contfrac(Pi,500)*I);contfrac(imag(k[1,1]/k[2,1]),200)
%K easy,nonn
%O 0,1
%A _Thomas Baruchel_, Mar 31 2004
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