A091832 Pierce expansion of 1/e^2.
7, 18, 19, 136, 349, 357, 1354, 6996, 7135, 9531, 11558, 15996, 17432, 52118, 151048, 427802, 821834, 877819, 972918, 1046690, 1540789, 3653077, 8200738, 9628573, 164153335, 5607624822, 86457467082, 141885251873, 151882622551
Offset: 1
Keywords
Links
- G. C. Greubel, Table of n, a(n) for n = 1..501 [a(1)=7 inserted by Georg Fischer, Nov 20 2020]
- 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, and Jaume Paradís, On a problem of Alfred Renyi, Economics Working Paper No. 340.
Programs
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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[1/E^2, 7!], 15] (* G. C. Greubel, Nov 14 2016 *) -
PARI
default(realprecision, 100000); r=exp(2); for(n=1, 100, s=(r/(r-floor(r))); print1(floor(r), ", "); r=s) \\ Benoit Cloitre [amended by Georg Fischer, Nov 20 2020]
Formula
Let u(0) = exp(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/e^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->oo} a(n)^(1/n) = e.
Extensions
a(1)=7 inserted by Georg Fischer, Nov 20 2020
Comments