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
Keywords
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
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.
Comments