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Pierce expansion of 1/sqrt(2).
2

%I #25 Jan 05 2025 19:51:37

%S 1,3,8,33,35,39201,39203,60245508192801,60245508192803,

%T 218662352649181293830957829984632156775201,

%U 218662352649181293830957829984632156775203

%N Pierce expansion of 1/sqrt(2).

%C If u(0)=exp(1/m) m integer>1 and u(n+1)=u(n)/frac(u(n)) then floor(u(n))=m*n.

%H G. C. Greubel, <a href="/A091831/b091831.txt">Table of n, a(n) for n = 0..14</a>

%H P. Erdős and Jeffrey Shallit, <a href="http://www.numdam.org/item?id=JTNB_1991__3_1_43_0">New bounds on the length of finite Pierce and Engel series</a>, Sem. Théor. Nombres Bordeaux (2) 3 (1991), no. 1, 43-53.

%H Vlado Keselj, <a href="https://cs.uwaterloo.ca/research/tr/1996/21/cs-96-21.pdf">Length of Finite Pierce Series: Theoretical Analysis and Numerical Computations </a>.

%H Jeffrey Shallit, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/22-4/shallit1.pdf">Some predictable Pierce expansions</a>, Fib. Quart., 22 (1984), 332-335.

%H Pelegrí Viader, Lluís Bibiloni, Jaume Paradís, <a href="http://dx.doi.org/10.2139/ssrn.145561">On a problem of Alfred Renyi</a>, Economics Working Paper No. 340.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PierceExpansion.html">Pierce Expansion</a>

%F Let u(0)=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)).

%F 1/sqrt(2)= 1/a(1) - 1/a(1)/a(2) + 1/a(1)/a(2)/a(3) - 1/a(1)/a(2)/a(3)/a(4)...

%F limit n -> infinity a(n)^(1/n) = e.

%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[2^(-1/2), 7!], 17] (* _G. C. Greubel_, Nov 13 2016 *)

%o (PARI) r=sqrt(2);for(n=1,10,r=r/(r-floor(r));print1(floor(r),","))

%Y Cf. A006275, A006276, A006283.

%Y Cf. A006784 (Pierce expansion definition), A028254

%K nonn

%O 0,2

%A _Benoit Cloitre_, Mar 09 2004