A229555 - OEIS

%I #46 Jan 21 2023 12:05:42

%S 1,22,1,31,2,3,1,63,1,10,1,2,1,7,1,160905,2,1,4,58,2,2,1,2,1,7,3,1,3,

%T 1,4,3,1,47,1,214540,1,2,9,1,45,1,3,1,48,1,21,1,9,1,8,1,2,249610,1,1,

%U 1,1,1,3,1,1,1,1,20,1,4,19,1,2,1,1,1,1,3,4,1,1,1

%N An exotic continued fraction for the real root of 6y^3 + 4y^2 - 4y - 7.

%C Among its 148 initial numbers, this sequence has 8 unexpectedly big terms which are considerably bigger than those of the well-known exotic sequence A002937. It is peculiar that the ratio of the biggest of these denominators for the two examples, as well as the ratio of the smallest ones, is almost exactly 7: 115270760/16467250 = 7.000000607... and 160905/22986 = 7.000130514...

%C Deleting the first term in the sequence, we get a continued fraction for the algebraic integer given as the real root of the equation x^3 - 22*x^2 - 22*x - 6 = 0. It has the same set of big denominators, only shifted one position towards the beginning. - _Sergei Duzhin_, Oct 03 2013

%C For all integer cubic polynomials a*y^3 + b*y^2 + c*y + d with 0 < a <= 7, |b| <= 7, |c| <= 7, |d| <= 7, the given one (together with its three modifications obtained by a change of one sign and a palindromic reversal of coefficients) is the only polynomial in this set that, among its first 200 denominators, contains a number greater than 10^8, thus establishing a record. - _Sergei Duzhin_, Oct 04 2013

%H T. D. Noe, <a href="/A229555/b229555.txt">Table of n, a(n) for n = 0..1000</a>

%F y = ((2906 - 126*sqrt(3*163))^(1/3) + (2906 + 126*sqrt(3*163))^(1/3) - 4) / 18. - _Andrey Zabolotskiy_, Jan 21 2023

%p Digits:=500:

%p with(numtheory):

%p x:=fsolve(6*y^3 + 4*y^2 - 4*y - 7);

%p cfrac(x,200,'quotients');

%t r = Roots[6 x^3 + 4 x^2 - 4 x - 7 == 0, x][[1, 2]]; ContinuedFraction[r, 115] (* _T. D. Noe_, Oct 02 2013 *)

%o (PARI)

%o \p 250

%o contfrac(real(polroots(Pol([6,4,-4,-7]))[1])) \\ _Charles R Greathouse IV_, Oct 01 2013

%K nonn,cofr,easy

%O 0,2

%A _Sergei Duzhin_, Oct 01 2013