A091831 Pierce expansion of 1/sqrt(2).

1, 3, 8, 33, 35, 39201, 39203, 60245508192801, 60245508192803, 218662352649181293830957829984632156775201, 218662352649181293830957829984632156775203

Offset: 0

Author

Benoit Cloitre, Mar 09 2004

Keywords

• nonn

Comments

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

Links

• G. C. Greubel, Table of n, a(n) for n = 0..14
• P. Erdős and Jeffrey Shallit, New bounds on the length of finite Pierce and Engel series, Sem. Théor. Nombres Bordeaux (2) 3 (1991), no. 1, 43-53.
• Vlado Keselj, Length of Finite Pierce Series: Theoretical Analysis and Numerical Computations .
• Jeffrey Shallit, Some predictable Pierce expansions, Fib. Quart., 22 (1984), 332-335.
• Pelegrí Viader, Lluís Bibiloni, Jaume Paradís, On a problem of Alfred Renyi, Economics Working Paper No. 340.
• Eric Weisstein's World of Mathematics, Pierce Expansion

Crossrefs

Cf. A006275, A006276, A006283.

Cf. A006784 (Pierce expansion definition), A028254

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[2^(-1/2), 7!], 17] (* G. C. Greubel, Nov 13 2016 *)

PARI

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

Formula

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)).

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)...

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