OFFSET
0,1
COMMENTS
Original definition: Limit of (1/Pi)^...^(1/Pi), n times, as n approaches infinity. Equals exp(-LambertW(log(Pi))).
The value can be obtained by iterating x -> 1/Pi^x with any real starting value, but convergence is linear and slow: about 5 iterations are needed for each additional decimal digit. - M. F. Hasler, Nov 01 2011
According to the Weisstein link, infinite iterated exponentiation such as used here, which is referred to both as an "infinite power tower" and "h(x)" -- with graph and other notations -- "converges iff e^(-e) <= x <= e^(1/e) as shown by Euler (1783) and Eisenstein (1844)" (citing Le Lionnais and Wells references). e^(-e) = A073230. e^(1/e) = A073229. x of interest here = 1/Pi = A049541. (1/A073243)^(1/A073243) = A030437^A030437 = Pi.
If y = h(x) = x^x^x^... converges, then by substitution y = x^y. So x^x^x^... is a solution y to the equation y^(1/y) = x. - Jonathan Sondow, Aug 27 2011
The expressions involving "..." in the above comment are misleading, since the limit is not obtained by applying additional "^x" to the previous expression, i.e., iterating "t -> t^x", but corresponds to iterations of "t -> x^t". - M. F. Hasler, Nov 01 2011
LINKS
Stanislav Sykora, Table of n, a(n) for n = 0..1999
J. Sondow and D. Marques, Algebraic and transcendental solutions of some exponential equations, Annales Mathematicae et Informaticae 37 (2010) 151-164; see the Appendix, arXiv:1108.6096.
Eric Weisstein's World of Mathematics, Power Tower
FORMULA
x = LambertW(log(Pi))/log(Pi), solution to Pi^x=1/x. - M. F. Hasler, Nov 01 2011
EXAMPLE
0.53934349886230120806079568445...
MATHEMATICA
y /. FindRoot[y^(1/y) == 1/Pi, {y, 1}, WorkingPrecision -> 100] (* Jonathan Sondow, Aug 27 2011 *)
First[RealDigits[Exp[-ProductLog[Log[Pi]]], 10, 104]] (* Vladimir Reshetnikov, Nov 01 2011 *)
PROG
(PARI) /* The program below was run with precision set to 1000 digits */ /* n is the number of iterated exponentiations performed. */ /* (n turns out to be 954 with 1E-200 specified here) */ n=0; s=1/Pi; t=1; while(abs(t-s)>1E-200, t=s; s=(1/Pi)^s; n++); print(n, ", ", s)
(PARI) solve(x=0, 1, x-1/Pi^x) \\ M. F. Hasler, Nov 01 2011
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
Rick L. Shepherd, Jul 28 2002
STATUS
approved