A002937 - OEIS

%I M2284 N0903 #36 Oct 21 2023 09:50:39

%S 3,3,7,4,2,30,1,8,3,1,1,1,9,2,2,1,3,22986,2,1,32,8,2,1,8,55,1,5,2,28,

%T 1,5,1,1501790,1,2,1,7,6,1,1,5,2,1,6,2,2,1,2,1,1,3,1,3,1,2,4,3,1,35657

%N An exotic continued fraction (for the real root of x^3 - 8*x - 10).

%C From _Peter Bala_, Oct 02 2013: (Start)

%C 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.

%C 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.

%C 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.

%C 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.

%C 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, ...].

%C 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)

%D 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.

%D A. Ya. Khinchin, Continued Fractions, Dover Publications, 1997.

%D J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 227.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%D 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.

%H Harry J. Smith, <a href="/A002937/b002937.txt">Table of n, a(n) for n = 0..20000</a>

%H Adam P. Goucher, <a href="http://cp4space.wordpress.com/2013/09/04/exotic-continued-fractions/">Exotic continued fractions</a>.

%e 3.318628217750185659109680153... = 3 + 1/(3 + 1/(7 + 1/(4 + 1/(2 + ...)))).

%o (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

%o (PARI) contfrac(polrootsreal(x^3 - 8*x - 10)[1]) \\ _Charles R Greathouse IV_, Apr 14 2014

%Y Cf. A160332 (decimal expansion).

%K cofr,nonn

%O 0,1

%A _N. J. A. Sloane_