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A141606
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Decimal expansion of (W(e-1)/(e-1))^(1/(1-e)), where W(z) denotes the Lambert W function and e = 2.718281828...
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4
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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, 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, 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
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OFFSET
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1,2
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COMMENTS
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Solution for x in x^(x^(e-1)) = e.
(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
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.
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LINKS
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EXAMPLE
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1.57844691419127618691147145725058871862508588172697263709178296257...
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MATHEMATICA
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RealDigits[(ProductLog[E-1]/(E-1))^(1/(1-E)), 10, 111][[1]]
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PROG
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(PARI) (lambertw(exp(1)-1)/(exp(1)-1))^(1/(1-exp(1))) \\ G. C. Greubel, Mar 02 2018
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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