A092142 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.

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, 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, 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

Offset

0,1

Links

Table of n, a(n) for n=0..94.

Prog

(PARI) k=contfracpnqn(contfrac(Pi, 500)*I); contfrac(imag(k[1, 1]/k[2, 1]), 200)

Crossrefs

Sequence in context: A163963 A119763 A363087 * A348497 A376021 A299161

Adjacent sequences: A092139 A092140 A092141 * A092143 A092144 A092145

Keyword

easy,nonn

Author

Thomas Baruchel, Mar 31 2004

Status

approved