OFFSET
0,1
COMMENTS
From Peter Bala, Oct 02 2013: (Start)
A theorem of Kuzmin in the measure theory of continued fractions says that for a random real number alpha, the probability that some given partial quotient of alpha is equal to a positive integer k is given by (1/log(2))*( log(1 + 1/k) - log(1 + 1/(k+1)) ). For example, almost all real numbers have 41.5% of their partial quotients equal to 1, 17% equal to 2, 9.3% equal to 3 and so on. Thus large partial quotients are the exception in continued fraction expansions.
Let now alpha denote the real root of Brillhart's cubic equation x^3 - 8*x - 10 = 0. Then alpha = (5 + (1/9)*sqrt(3*163))^(1/3) + (5 - (1/9)*sqrt(3*163))^(1/3) = 3.31862 82177 50185 65910 .... The continued fraction expansion of alpha is unusual in that there are 8 surprisingly large partial quotients in the first 200 terms, namely [22986, 1501790, 35657, 49405, 53460, 16467250, 48120, 325927]. The explanation for this behavior was given by Stark.
Roberts, p. 227, gives a remarkable product representation for beta := 2 + alpha, namely, beta = q^(1/24) * Product_{n >= 1} (1 + (1/q)^(2*n+1)), where q = exp(Pi*sqrt(163)), and notes that the first factor exp((Pi/24)*sqrt(163)) = 5.31862 82177 50185 63885 ... gives 16 decimal places of beta. Cf. A160514.
Some powers beta^k of beta share with beta the property of having exceptionally large partial quotients early on in their continued fraction expansion. Particularly noteworthy is beta^2, which, like beta, has 8 large partial quotients [126425, 8259853, 1620, 271730, 294038, 90569882, 264667, 1792603] in its first 200 terms. Strangely, apart from the third term 1620, these numbers are almost exactly 5.5 times larger than the corresponding large partial quotients [22986, 1501790, 35657, 49405, 53460, 16467250, 48120, 325927] occurring in the continued fraction expansion of beta.
Also noteworthy is beta^8 = 640320.00000000062437..., very nearly an integer, whose continued fraction expansion begins [640320, 1601600400, 320160, 2135467200, 261949, 10, 1, 20533337, 1, 1, 4, 3, 1, 3, 1369, 2, 2, 14, 3, 9535605, 1, 3, 2, 1, 2, 1, ...].
The cubic irrationals r*beta, where r is rational, also appear to have large partial quotients early on in their continued fraction expansions, which appear to be related to the partial quotients of beta. For example, (2/3)*beta has 8 exceptionally large partial quotients [137919, 9010749, 213951, 296433, 320769, 98803508, 288729, 1955567] in the first 200 terms of its continued fraction expansion and each of these partial quotients is almost exactly 6 times larger than the corresponding large partial quotient in the expansion of beta. (End)
REFERENCES
R. P. Brent, A. J. van der Poorten, and H. J. Te Riele (1996, May), A comparative study of algorithms for computing continued fractions of algebraic numbers. In ANTS-II: International Algorithmic Number Theory Symposium (pp. 35-47), LNCS Vol. 1122, Springer, Berlin, Heidelberg. See Tables 3 and 4.
A. Ya. Khinchin, Continued Fractions, Dover Publications, 1997.
J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 227.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
H. M. Stark, An explanation of some exotic continued fractions found by Brillhart, pp. 21-35 of A. O. L. Atkin and B. J. Birch, editors, Computers in Number Theory. Academic Press, NY, 1971.
LINKS
Harry J. Smith, Table of n, a(n) for n = 0..20000
Adam P. Goucher, Exotic continued fractions.
EXAMPLE
3.318628217750185659109680153... = 3 + 1/(3 + 1/(7 + 1/(4 + 1/(2 + ...)))).
PROG
(PARI) { allocatemem(932245000); default(realprecision, 21000); x=NULL; p=x^3 - 8*x - 10; rs=polroots(p); r=real(rs[1]); c=contfrac(r); for (n=1, 20001, write("b002937.txt", n-1, " ", c[n])); } \\ Harry J. Smith, May 11 2009
(PARI) contfrac(polrootsreal(x^3 - 8*x - 10)[1]) \\ Charles R Greathouse IV, Apr 14 2014
CROSSREFS
KEYWORD
cofr,nonn
AUTHOR
STATUS
approved