A091846 Pierce expansion of log(2).

1, 3, 12, 21, 51, 57, 73, 85, 96, 1388, 4117, 5268, 9842, 11850, 16192, 19667, 29713, 76283, 460550, 1333597, 1462506, 9400189, 13097390, 30254851, 190193800, 201892756, 431766247, 942050077, 6204785761, 16684400052, 23762490104

Offset: 1

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 = 1..500
• 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, A006284.

Cf. A006784 (Pierce expansion definition), A059180.

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[Log[2], 7!], 25] (* G. C. Greubel, Nov 14 2016 *)

PARI

r=1/log(2);for(n=1,30,r=r/(r-floor(r));print1(floor(r),","))

Formula

Let u(0)=1/log(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)).

log(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.