OFFSET
0,1
COMMENTS
For x in the open interval (0,1) define the map f(x) = 1 - x*floor(1/x). The n-th term (n >= 0) in the Pierce expansion of x is given by floor(1/f^(n)(x)), where f^(n)(x) denotes the n-th iterate of the map f, with the convention that f^(0)(x) = x.
The present sequence is the case x = 1/2*(5 - sqrt(21)).
Jeffrey Shallit has shown that the Pierce expansion of the quadratic irrational (c - sqrt(c^2 - 4))/2 has the form [c(0) - 1, c(0) + 1, c(1) - 1, c(1) + 1, c(2) - 1, c(2) + 1, ...], where c(0) = c and c(n+1) = c(n)^3 - 3*c(n). This is the case c = 5. For other cases see A006276 (c = 3), A219506 (c = 4) and A006275 (essentially c = 6 apart from the initial term).
The Pierce expansion of {(c - sqrt(c^2 - 4))/2}^(3^n) is [[c(n) - 1, c(n) + 1, c(n+1) - 1, c(n+1) + 1, c(n+2) - 1,c(n+2) + 1, ...].
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..13
T. A. Pierce, On an algorithm and its use in approximating roots of algebraic equations, Amer. Math. Monthly, Vol. 36 No. 10, (1929) p.523-525.
Jeffrey Shallit, Some predictable Pierce expansions, Fib. Quart., 22 (1984), 332-335.
E. W. Weisstein, MathWorld: Pierce Expansion
FORMULA
a(2*n) = (1/2*(5 + sqrt(21)))^(3^n) + (1/2*(5 - sqrt(21)))^(3^n) - 1.
a(2*n+1) = (1/2*(5 + sqrt(21)))^(3^n) + (1/2*(5 - sqrt(21)))^(3^n) + 1.
EXAMPLE
Let x = 1/2*(5 - sqrt(21)). We have the alternating series expansions
x = 1/4 - 1/(4*6) + 1/(4*6*109) - 1/(4*6*109*111) + ...
x^3 = 1/109 - 1/(109*111) + 1/(109*111*1330669) - ...
x^9 = 1/1330669 - 1/(1330669*1330671) + ....
MATHEMATICA
PierceExp[A_, n_] := Join[Array[1 &, Floor[A]], First@Transpose@ NestList[{Floor[1/Expand[1 - #[[1]] #[[2]]]], Expand[1 - #[[1]] #[[2]]]} &, {Floor[1/(A - Floor[A])], A - Floor[A]}, n - 1]]; PierceExp[N[(5 - Sqrt[21])/2 , 7!], 10] (* G. C. Greubel, Nov 14 2016 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 22 2012
STATUS
approved