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%I #18 Nov 15 2016 02:41:26
%S 1,3,16,17,72,577,2312,665857,2663432,886731088897,3546924355592,
%T 1572584048032918633353217,6290336192131674533412872,
%U 4946041176255201878775086487573351061418968498177
%N Pierce expansion of 4*(3 - 2*sqrt(2)).
%C 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.
%C Paradis et al. have determined the Pierce expansion of the quadratic irrationality 2*(p - 1)*(p - sqrt(p^2 - 1)), p a positive integer greater than or equal to 3. The present sequence is the case p = 3. For other cases see A219509 (p = 5), A219510 (p = 7) and A219511 (p = 9).
%C Compare this Pierce expansion for 4*(3 - 2*sqrt(2)), with terms determined by quadratic recurrences, with the Pierce expansion of 3 - 2*sqrt(2) given in A006275, where the terms are determined by cubic recurrences.
%H G. C. Greubel, <a href="/A219508/b219508.txt">Table of n, a(n) for n = 0..22</a>
%H J. Paradis, P. Viader, L. Bibiloni <a href="http://www.fq.math.ca/Scanned/36-2/paradis.pdf">Approximation to quadratic irrationals and their Pierce expansions</a>, The Fibonacci Quarterly, Vol.36 No. 2 (1998) 146-153.
%H T. A. Pierce, <a href="http://www.jstor.org/stable/2299963">On an algorithm and its use in approximating roots of algebraic equations</a>, Amer. Math. Monthly, Vol. 36 No. 10, (1929) p.523-525.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PierceExpansion.html">Pierce Expansion</a>
%F a(2*n) = 2*{(1 + sqrt(2))^(2^n) + (1 - sqrt(2))^(2^n) + 2} for n >= 1.
%F a(2*n-1) = 1/2*{(1 + sqrt(2))^(2^n) + (1 - sqrt(2))^(2^n)} for n >= 1.
%F Recurrence equations: a(0) = 1, a(1) = 3 and for n >= 1, a(2*n) = 4*(a(2*n-1) + 1) and a(2*n+1) = 2*{a(2*n-1)}^2 - 1.
%F It follows that a(2*n) = 8*a(2*n-3)^2 for n >=2.
%F 4*(3 - 2*sqrt(2)) = sum {n >= 0} 1/product {k = 0..n} a(k) = 1 - 1/3 + 1/(3*16) - 1/(3*16*17) + 1/(3*16*17*72) - ....
%F a(2*n) = 8*A001601(n-1)^2 for n >= 2
%F a(2*n-1) = A001601(n) for n >= 1.
%t 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[4*(3 - 2*Sqrt[2]) , 7!], 10] (* _G. C. Greubel_, Nov 14 2016 *)
%o (PARI) r=(3 + 2*sqrt(2))/4; for(n=1, 10, print(floor(r), ", "); r=r/(r-floor(r))) \\ _G. C. Greubel_, Nov 15 2016
%Y Cf. A001601, A006275, A219509 (p = 5), A219510 (p = 7), A219511 (p = 9).
%K nonn,easy
%O 0,2
%A _Peter Bala_, Nov 23 2012
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