A118242 -id:A118242 - cp's OEIS frontend

A006276 Pierce expansion of (3 - sqrt(5))/2.

2, 4, 17, 19, 5777, 5779, 192900153617, 192900153619, 7177905237579946589743592924684177, 7177905237579946589743592924684179, 369822356418414944143680173221426891716916679027557977938929258031490127514207143830378340325399155217

Offset: 0

N. J. A. Sloane

From Peter Bala, Nov 22 2012: (Start)
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.
Let x = (sqrt(5) - 1)/2, the reciprocal of the golden ratio, and let X = (3 - sqrt(5))/2 so that X = x^2. The Pierce expansion of X^(3^n) is [a(2*n), a(2*n+1), a(2*n+2), ...]. The Pierce expansion of x is A118242 = [1, a(0), a(1), a(2), ...]. The Pierce expansion of x^3 is [a(1), a(2), a(3), ...]. In general, the Pierce expansion of x^(3^n) for n >= 1 is [a(1)*a(3)*...*a(2*n-1), a(2*n), a(2*n+1), a(2*n+2), ...] = [sqrt(a(2*n) - 1), a(2*n), a(2*n+1), a(2*n+2), ...]. Some examples of the associated alternating series are given below.
(End)

			From _Peter Bala_, Nov 22 2012: (Start)
Let x = (sqrt(5) - 1)/2. We have the alternating series expansions
x = 1 - 1/2 + 1/(2*4) - 1/(2*4*17) + 1/(2*4*17*19) - ...
x^2 = 1/2 - 1/(2*4) + 1/(2*4*17) - 1/(2*4*17*19) + ...
x^6 = 1/17 - 1/(17*19) + 1/(17*19*5777) - ...
as well as
x^3 = 1/4 - 1/(4*17) + 1/(4*17*19) - 1/(4*17*19*5777) + ...
4*x^9 = 1/19 - 1/(19*5777) + 1/(19*5777*5779) - ...
4*19*x^27 = 1/5779 - 1/(5779*192900153617) + ....
(End)

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

G. C. Greubel, Table of n, a(n) for n = 0..15
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

Cf. A118242.

Table[c=2*3^Floor[n/2]; 2*Fibonacci[c+1]-Fibonacci[c]-(-1)^n, {n,0,10}] (* Harvey P. Dale, Oct 22 2013 *)
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[(3 - Sqrt[5])/2, 7!], 10] (* G. C. Greubel, Nov 14 2016 *)

r=(1+sqrt(5))/2; for(n=1,10, r=r/(r-floor(r)) print1(floor(r),","))

Let c(0)=3, c(n+1) = c(n)^3-3*c(n) [A001999]; then this sequence is c(0)-1, c(0)+1, c(1)-1, c(1)+1, c(2)-1, c(2)+1, ......
a(n) = 2*F(2*3^floor(n/2)+1)-F(2*3^floor(n/2))-(-1)^n where F(k) denotes the k-th Fibonacci number A000045(k)
Let u(0)=(1+sqrt(5))/2 and u(n+1)=u(n)/frac(u(n)) where frac(x) is the fractional part of x, then a(n)=floor(u(n)). - Benoit Cloitre, Mar 09 2004
a(2*n) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n) - 1.
a(2*n+1) = ((3 + sqrt(5))/2)^(3^n) + ((3 - sqrt(5))/2)^(3^n) + 1. - Peter Bala, Nov 22 2012

More terms from James Sellers, May 19 2000

A140076 Pierce expansion of the cube root of 1/2.

Original entry on oeis.org

1, 4, 5, 7, 8, 18, 384, 7958, 14304, 16623, 18610, 20685, 72923, 883177, 1516692, 2493788, 2504069, 22881179, 110219466, 2241255405, 34982468090, 64356019489, 110512265214, 1142808349967, 3550630472116, 5238523454726, 7129035664265

Offset: 1

Gerard P. Michon, Jun 01 2008

2^(-1/3) = 1-1/4(1-1/5(1-1/7(1-1/8(1-1/18(1-1/384(...))))))

			a(1) is 1 because the floor of 2^(1/3) is 1.
a(2)=4 because 1/(1-2^(-1/3)) is 4.8473221...

G. C. Greubel, Table of n, a(n) for n = 1..500
G. P. Michon, Pierce Expansions.
Eric Weisstein's World of Mathematics, Pierce Expansion.

Cf. A091831, A006283, A006284, A061233, A118242.

$MaxExtraPrecision = 80; x[1] = 2^(-1/3); a[n_] := a[n] = Floor[1/x[n]]; x[n_] := x[n] = 1 - a[n-1]*x[n-1]; Table[a[n], {n, 1, 27}] (* Jean-François Alcover, Dec 12 2011 *)
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^(-1/3), 7!], 25] (* G. C. Greubel, Nov 14 2016 *)

Starting with x(1)=2^(-1/3), a(n) = floor(1/x(n)) and x(n+1) = 1-a(n)x(n).

cp's OEIS Frontend

A006276 Pierce expansion of (3 - sqrt(5))/2.

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