cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A006275 Pierce expansion of sqrt(2) - 1.

Original entry on oeis.org

2, 5, 7, 197, 199, 7761797, 7761799, 467613464999866416197, 467613464999866416199, 102249460387306384473056172738577521087843948916391508591105797
Offset: 0

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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. The present sequence is the case x = sqrt(2) - 1.
The Pierce expansion of (sqrt(2) - 1)^(3^n) is [a(0)*a(2)*...*a(2*n), a(2*n+1), a(2*n+2), ...] = [sqrt(a(2*n+1) - 1), a(2*n+1), a(2*n+2), ...]. The Pierce expansion of (sqrt(2) - 1)^(2*3^n) is [a(2*n+1), a(2*n+2), ...]. Some examples of the associated alternating series are given below.
(End)

Examples

			Let c(0)=6, c(n+1) = c(n)^3-3*c(n); then this sequence is 2, c(0)-1, c(0)+1, c(1)-1, c(1)+1, c(2)-1, c(2)+1, ...
From _Peter Bala_, Nov 22 2012: (Start)
Let x = sqrt(2) - 1. We have the alternating series expansions
x = 1/2 - 1/(2*5) + 1/(2*5*7) - 1/(2*5*7*197) + ...
x^3 = 1/14 - 1/(14*197) + 1/(14*197*199) - ...
x^9 = 1/2786 - 1/(2786*7761797) + 1/(2786*7761797*7761799) - ...,
where 2786 = 2*7*199, and also
x^2 = 1/5 - 1/(5*7) + 1/(5*7*197) - 1/(5*7*197*199) + ...
x^6 = 1/197 - 1/(197*199) + 1/(197*199*7761797) - ...
x^18 = 1/7761797 - 1/(7761797*7761799) + ....
(End)
		

References

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

Crossrefs

Programs

  • 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[Sqrt[2] - 1, 7!], 10] (* G. C. Greubel, Nov 14 2016 *)
  • PARI
    my(r=1+quadgen(8)); for(n=1, 10, print1(floor(r), ", "); r=r/(r-floor(r)));

Formula

Let u(0)=1+sqrt(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
From Peter Bala, Nov 22 2012: (Start)
a(2*n+2) = (3 + 2*sqrt(2))^(3^n) + (3 - 2*sqrt(2))^(3^n) + 1.
a(2*n+1) = (3 + 2*sqrt(2))^(3^n) + (3 - 2*sqrt(2))^(3^n) - 1. (End)
sqrt(2) - 1 = a(0)/a(1) + (a(0)*a(2))/(a(1)*a(3)) + (a(0)*a(2)*a(4))/(a(1)*a(3)*a(5)) + ... = 2/5 + (2*7)/(5*197) + (2*7*199)/(5*197*7761797) + ... . - Peter Bala, Dec 03 2012

Extensions

More terms from James Sellers, May 19 2000