[seqfan] Re: A nice continued fraction for φᵠ (and other constants)

[Thomas Baruchel]

On Sat, 15 Apr 2017, Paolo Bonzini wrote:
>> Maybe the following hypergeometric expressions for x^x are useful:
>>   x^x = 1F0( [-x] , [], -x + 1 )
>
> Probably, since this gives phi^phi = 1F0([-phi], [], -1/phi)
>
> http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F0/10/
> has continued fraction expansions of 1F0.

Thank you for these answers; I will have to investigate further in order
to see if both continued fractions are more or less identical, but I am
not absolutely sure of it because mine relies on the function (x/(x-1))^(x-1)
which evaluates to phi^phi when x=phi^2. On the other hand, you provide
a direct x^x function (working for any value of x).

Best regards,

-- 
Thomas Baruchel