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Okay, some thoughts concerning this function.

  1. Are there any numbers equal to the sum of the hyperfactorials of their digits? A brief brute-force search convinces me that there are no such numbers; if there are any, they're very large.
  2. Can we generalize H(n) for the real numbers or the complex plane, the way the gamma function generalizes the factorial?
  3. Is there an equivalent of Stirling's approximation? I figured out , but I'm a beginner at integral calculus and this could be totally wrong.

FB100Ztalkcontribs 19:33, September 6, 2012 (UTC)

Okay, I've got the approximation . It seems pretty accurate, but it's hard to tell with numbers this size. It looks vaguely like the integrand of — could this lead to our continuous hyperfactorial? FB100Ztalkcontribs 21:21, September 6, 2012 (UTC)

This one? Hyperfactorials and the K-function --Cloudy176 (talk) 13:08, September 10, 2012 (UTC)
D'oh, didn't see that. I'm smart. FB100Ztalkcontribs 21:53, September 10, 2012 (UTC)
About the first question: If such numbers exist then it should be less than , because if n is a d-digit number satisfying the conditions, then , and . --Cloudy176 (talk) 13:28, September 10, 2012 (UTC)
1036 is obviously too big for brute force, so I'll have to use a sneakier method (maybe building from the digits up). I'll look into this. FB100Ztalkcontribs 21:53, September 10, 2012 (UTC)
Rather than search through all numbers between 0 and , you can instead search through all combinations of digits of 36 or less, and then check if the sum of the hyperfactorials of those digits has the right combination of digits. The total number of such combinations is , so it's possible to search through.Deedlit11 (talk) 13:22, September 26, 2012 (UTC)
2 digits
Runtime: 0s
3 digits
Runtime: 0s
4 digits
Runtime: 0s
5 digits
Runtime: 0s
6 digits
Runtime: 0s
7 digits
Runtime: 0s
8 digits
Runtime: 1s
9 digits
Runtime: 1s
10 digits
Runtime: 3s
11 digits
Runtime: 5s
12 digits
Runtime: 9s
13 digits
Runtime: 15s
14 digits
Runtime: 26s
15 digits
Runtime: 41s

This might take a while, folks. FB100Ztalkcontribs 20:09, September 27, 2012 (UTC)

Got a few optimizations that shaved 21 digits down to a minute, and working on more. Much to my chagrin, my estimate for total computation time came up to 8-12 hours.
Question: Given a digit sequence d1, d2, d3, d4, ..., dn with each digit greater than or equal to the last, is there a quick way to ensure that s = H(d1) + H(d2) + H(d3) + H(d4) + ... + H(dn) (the sum of their hyperfactorials) has n digits without actually computing the sum? FB100Ztalkcontribs 21:12, September 27, 2012 (UTC)

Hyperderivative[]

The hyperfactorial opens up to some useless analogs of common analytical functions. Define the h-exponential:

\[e_H^x = 1 + x + \frac{x^2}{H(2)} + \frac{x^3}{H(3)} + \ldots\]

and the h-trig functions

\begin{eqnarray} \cos_H(x) &=& \frac{e_H^{ix} + e_H^{-ix}}{2} \\ \sin_H(x) &=& \frac{e_H^{ix} - e_H^{-ix}}{2i} \\ \tan_H(x) &=& \frac{\sin_H(x)}{\cos_H(x)} \\ \end{eqnarray}

FB100Ztalkcontribs 20:00, June 6, 2013 (UTC)

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