OFFSET
1,2
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.
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
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.
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 *)
PROG
(PARI) r=1/log(2); for(n=1, 30, r=r/(r-floor(r)); print1(floor(r), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Mar 09 2004
STATUS
approved