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%I #37 Jun 28 2021 08:57:45
%S 5,3,9,3,4,3,4,9,8,8,6,2,3,0,1,2,0,8,0,6,0,7,9,5,6,8,4,4,5,5,5,9,8,4,
%T 2,0,9,8,6,4,5,5,9,7,3,2,9,4,8,4,2,6,1,1,9,4,8,8,1,5,0,1,4,8,7,0,4,6,
%U 2,7,5,4,0,1,7,4,9,0,4,5,5,5,2,8,4,1,5,2,4,2,9,3,6,8,1,7,6,7,7,3,5,4,0,2
%N Decimal expansion of exp(-LambertW(log(Pi))), solution to x = 1/Pi^x.
%C Original definition: Limit of (1/Pi)^...^(1/Pi), n times, as n approaches infinity. Equals exp(-LambertW(log(Pi))).
%C 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
%C 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.
%C 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
%C 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
%H Stanislav Sykora, <a href="/A073243/b073243.txt">Table of n, a(n) for n = 0..1999</a>
%H J. Sondow and D. Marques, <a href="http://ami.ektf.hu/uploads/papers/finalpdf/AMI_37_from151to164.pdf">Algebraic and transcendental solutions of some exponential equations</a>, Annales Mathematicae et Informaticae 37 (2010) 151-164; see the Appendix, arXiv:<a href="https://arxiv.org/abs/1108.6096">1108.6096</a>.
%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PowerTower.html">Power Tower</a>
%F x = LambertW(log(Pi))/log(Pi), solution to Pi^x=1/x. - _M. F. Hasler_, Nov 01 2011
%e 0.53934349886230120806079568445...
%t y /. FindRoot[y^(1/y) == 1/Pi, {y, 1}, WorkingPrecision -> 100] (* _Jonathan Sondow_, Aug 27 2011 *)
%t First[RealDigits[Exp[-ProductLog[Log[Pi]]], 10, 104]] (* _Vladimir Reshetnikov_, Nov 01 2011 *)
%o (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)
%o (PARI) solve(x=0,1,x-1/Pi^x) \\ _M. F. Hasler_, Nov 01 2011
%Y Cf. A000796 (Pi), A049541 (1/Pi), A073240 ((1/Pi)^(1/Pi)), A073241 ((1/Pi)^(1/Pi)^(1/Pi)), A030437 (reciprocal of A073243), A030178 (corresponding limit for 1/e), A030797 (reciprocal of A030178).
%K cons,nonn
%O 0,1
%A _Rick L. Shepherd_, Jul 28 2002
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