%I #9 Mar 30 2012 19:00:09

%S 4,8,9,5,3,6,7,9,5,5,5,4,6,1,1,3,4,7,1,9,6,7,1,9,3,3,8,7,2,2,9,8,3,5,

%T 8,4,9,4,7,2,7,3,1,9,5,2,8,0,9,3,7,2,4,4,3,6,3,0,8,4,6,6,4,9,2,9,5,5,

%U 4,1,2,1,0,4,9,5,4,0,9,2,9,3,6,5,3,4,1,1,4,0,8,0,1,2,1,7,9,2,6,1

%N Decimal expansion of y with 0 < x < y and x^y = y^x = 17.

%C Given z > 0, there exist positive real numbers x < y with x^y = y^x = z, if and only if z > e^e. In that case, (x,y) = ((1 + 1/t)^t,(1 + 1/t)^(t+1)) for some t > 0. For example, t = 1 gives 2^4 = 4^2 = 16 > e^e. When x^y = y^x = 17, at least one of x and y is transcendental. See Sondow and Marques 2010, pp. 155-157.

%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.

%e y=4.89536795554611347196719338722983584947273195280937244363084664929554121...

%t x[t_] := (1 + 1/t)^t; y[t_] := (1 + 1/t)^(t + 1); t = t/. FindRoot[x[t]^y[t] == 17, {t, 1}, WorkingPrecision -> 120]; RealDigits[y[t], 10, 100] // First

%Y Cf. A073226 (e^e), A194556 ((9/4)^(27/8) = (27/8)^(9/4)), A194557 (sqrt(3)^sqrt(27) = sqrt(27)^sqrt(3)), A194622 (x with 0 < x < y and x^y = y^x = 17).

%K nonn,cons

%O 1,1

%A _Jonathan Sondow_, Aug 30 2011