|
|
A002945
|
|
Continued fraction for cube root of 2.
(Formerly M2220)
|
|
16
|
|
|
1, 3, 1, 5, 1, 1, 4, 1, 1, 8, 1, 14, 1, 10, 2, 1, 4, 12, 2, 3, 2, 1, 3, 4, 1, 1, 2, 14, 3, 12, 1, 15, 3, 1, 4, 534, 1, 1, 5, 1, 1, 121, 1, 2, 2, 4, 10, 3, 2, 2, 41, 1, 1, 1, 3, 7, 2, 2, 9, 4, 1, 3, 7, 6, 1, 1, 2, 2, 9, 3, 1, 1, 69, 4, 4, 5, 12, 1, 1, 5, 15, 1, 4
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
REFERENCES
|
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
|
|
LINKS
|
E. Bombieri and A. J. van der Poorten, Continued fractions of algebraic numbers, In: W. Bosma, A. van der Poorten (eds), Computational Algebra and Number Theory. Mathematics and Its Applications, vol. 325.
|
|
FORMULA
|
Bombieri/van der Poorten give a complicated formula:
a(n) = floor((-1)^(n+1)*3*p(n)^2/(q(n)*(p(n)^3-2*q(n)^3)) - q(n-1)/q(n)),
p(n+1) = a(n)*p(n) + p(n-1),
q(n+1) = a(n)*q(n) + q(n-1),
with a(1) = 1, p(1) = 1, q(1) = 0, p(2) = 1, q(2) = 1. (End)
|
|
EXAMPLE
|
2^(1/3) = 1.25992104989487316... = 1 + 1/(3 + 1/(1 + 1/(5 + 1/(1 + ...)))).
|
|
MAPLE
|
N:= 100: # to get a(1) to a(N)
a[1] := 1: p[1] := 1: q[1] := 0: p[2] := 1: q[2] := 1:
for n from 2 to N do
a[n] := floor((-1)^(n+1)*3*p[n]^2/(q[n]*(p[n]^3-2*q[n]^3)) - q[n-1]/q[n]);
p[n+1] := a[n]*p[n] + p[n-1];
q[n+1] := a[n]*q[n] + q[n-1];
od:
|
|
MATHEMATICA
|
ContinuedFraction[Power[2, (3)^-1], 70] (* Harvey P. Dale, Sep 29 2011 *)
|
|
PROG
|
(PARI) allocatemem(932245000); default(realprecision, 21000); x=contfrac(2^(1/3)); for (n=1, 20000, write("b002945.txt", n-1, " ", x[n])); \\ Harry J. Smith, May 08 2009
|
|
CROSSREFS
|
|
|
KEYWORD
|
cofr,nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
BCMATH link from Keith R Matthews (keithmatt(AT)gmail.com), Jun 04 2006
|
|
STATUS
|
approved
|
|
|
|