%I #36 Apr 30 2023 02:12:05

%S 7,18,19,136,349,357,1354,6996,7135,9531,11558,15996,17432,52118,

%T 151048,427802,821834,877819,972918,1046690,1540789,3653077,8200738,

%U 9628573,164153335,5607624822,86457467082,141885251873,151882622551

%N Pierce expansion of 1/e^2.

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

%H G. C. Greubel, <a href="/A091832/b091832.txt">Table of n, a(n) for n = 1..501</a> [a(1)=7 inserted by Georg Fischer, Nov 20 2020]

%H P. Erdős and Jeffrey Shallit, <a href="http://www.numdam.org/item?id=JTNB_1991__3_1_43_0">New bounds on the length of finite Pierce and Engel series</a>, Sem. Théor. Nombres Bordeaux (2) 3 (1991), no. 1, 43-53.

%H Vlado Keselj, <a href="https://cs.uwaterloo.ca/research/tr/1996/21/cs-96-21.pdf">Length of Finite Pierce Series: Theoretical Analysis and Numerical Computations </a>.

%H Jeffrey Shallit, <a href="http://www.fq.math.ca/Scanned/22-4/shallit1.pdf">Some predictable Pierce expansions</a>, Fib. Quart., 22 (1984), 332-335.

%H Pelegrí Viader, Lluís Bibiloni, and Jaume Paradís, <a href="http://dx.doi.org/10.2139/ssrn.145561">On a problem of Alfred Renyi</a>, Economics Working Paper No. 340.

%F 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)).

%F 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)) ...

%F Limit_{n->oo} a(n)^(1/n) = e.

%t 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 *)

%o (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]

%Y Cf. A006275, A006276, A006283.

%Y Cf. A006784 (Pierce expansion definition), A059194.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Mar 09 2004

%E a(1)=7 inserted by _Georg Fischer_, Nov 20 2020