A194623 Decimal expansion of y with 0 < x < y and x^y = y^x = 17.

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, 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, 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

Offset: 1

Author

Jonathan Sondow, Aug 30 2011

Keywords

• nonn
• cons

Comments

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.

Examples

			y=4.89536795554611347196719338722983584947273195280937244363084664929554121...
		

Links

• J. Sondow and D. Marques, Algebraic and transcendental solutions of some exponential equations, Annales Mathematicae et Informaticae, 37 (2010), 151-164.

Crossrefs

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).

Programs

Mathematica

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