

A073243


Decimal expansion of exp(LambertW(log(Pi))), solution to x = 1/Pi^x.


11



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, 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, 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
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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



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 *)


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 1E200 specified here) */ n=0; s=1/Pi; t=1; while(abs(ts)>1E200, t=s; s=(1/Pi)^s; n++); print(n, ", ", s)


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AUTHOR



STATUS

approved



