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%I #11 Mar 03 2018 09:51:12
%S 1,5,7,8,4,4,6,9,1,4,1,9,1,2,7,6,1,8,6,9,1,1,4,7,1,4,5,7,2,5,0,5,8,8,
%T 7,1,8,6,2,5,0,8,5,8,8,1,7,2,6,9,7,2,6,3,7,0,9,1,7,8,2,9,6,2,5,7,9,8,
%U 3,1,3,1,3,0,2,9,8,6,4,6,0,1,8,7,1,0,0,5,1,8,5,6,3,8,8,6,3,7,3,7,1,0,5,5,5
%N Decimal expansion of (W(e-1)/(e-1))^(1/(1-e)), where W(z) denotes the Lambert W function and e = 2.718281828...
%C Solution for x in x^(x^(e-1)) = e.
%C (W((y-1)log(z))/((y-1)log(z)))^(1/(1-y)) = e^(W((y-1)log(z))/(y-1)) so that (W(e-1)/(e-1))^(1/(1-e)) = e^(W(e-1)/(e-1)). - _Ross La Haye_, Aug 27 2008
%C Consider the expression x^x^x^x... where x appears y times. For, say, y = 4 this type of expression is conventionally evaluated as if bracketed x^(x^(x^x)) and is referred to as a "power tower". However, we can also bracket x^x^x^x from the bottom up, e.g., (x^x)^x)^x = x^(x^3). In general, this bracketing will simplify x^x^x^x... to x^(x^(y-1)) when x appears y times in the expression. Solving the equation x^(x^(y-1)) = z for x gives x = (W((y-1)log(z))/((y-1)log(z)))^(1/(1-y)). And setting y = z = e gives the result indicated by this sequence. Special thanks are due to Mike Wentz for introducing me to the "bottom up" bracketing of x^x^x^x... and the motivation for its investigation.
%H G. C. Greubel, <a href="/A141606/b141606.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/LambertW-Function.html">Lambert W-Function</a>
%H Eric Weisstein, <a href="http://mathworld.wolfram.com/PowerTower.html">Power Tower</a>
%e 1.57844691419127618691147145725058871862508588172697263709178296257...
%t RealDigits[(ProductLog[E-1]/(E-1))^(1/(1-E)),10,111][[1]]
%o (PARI) (lambertw(exp(1)-1)/(exp(1)-1))^(1/(1-exp(1))) \\ _G. C. Greubel_, Mar 02 2018
%Y Cf. A001113.
%Y Cf. A143913, A143914, A143915. - _Ross La Haye_, Sep 05 2008
%K cons,nonn
%O 1,2
%A _Ross La Haye_, Aug 21 2008, Aug 26 2008
%E More terms from _Robert G. Wilson v_, Aug 25 2008
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