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 = 2 - sqrt(3).
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 = 4. For other cases see A006276 (c = 3), A219507 (c = 5) 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
F. L. Bauer, Letters to the editor: An Infinite Product for Square-Rooting with Cubic Convergence, The Mathematical Intelligencer, Vol. 20, Issue 1, (1998), 12-14.
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.
Eric Weisstein's World of Mathematics, Pierce Expansion
FORMULA
a(2*n) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n) - 1.
a(2*n + 1) = (2 + sqrt(3))^(3^n) + (2 - sqrt(3))^(3^n) + 1.
From Peter Bala, Jan 18 2022: (Start)
a(2*n+2) = a(2*n)^3 + 3*a(2*n)^2 - 3; a(2*n+1) = a(2*n-1)^3 - 3*a(2*n-1)^2 + 3.
a(2*n) = 6*(Product_{k = 1..n-1} a(2*k))^2 - 3, with a(0) = 1;
a(2*n+1) = 2*(Product_{k = 0..n-1} a(2*k+1))^2 + 3, with a(1) = 5.
sqrt(3) = (1 + 2/3)*(1 + 2/51)*(1 + 2/140451)*(1 + 2/2770663499604051)* .... See Bauer.
1/sqrt(3) = (1 - 2/5)*(1 - 2/53)*(1 - 2/140453)*(1 - 2/2770663499604053)* .... (End)
EXAMPLE
We have the alternating series expansions
2 - sqrt(3) = 1/3 - 1/(3*5) + 1/(3*5*51) - 1/(3*5*51*53) + ...
(2 - sqrt(3))^3 = 1/51 - 1/(51*53) + 1/(51*53*140451) - ...
(2 - sqrt(3))^9 = 1/140451 - 1/(140451*140453) + ....
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[2 - Sqrt[3] , 7!], 10] (* G. C. Greubel, Nov 14 2016 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 22 2012
STATUS
approved