--- In

primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

>

> I tried 1/n^c:

>

> v=[1237.1, 328.7, 105.4, 28.01, 6.22 , 1.510, 0.439, 0.0939];

> print(vector(8,k,-log(v[k]/10^6)/(k+4)/log(10)));

>

> [0.5815, 0.5805, 0.5682, 0.5691, 0.5785, 0.5821, 0.5780, 0.5856]

>

> and then A/n^c, using the first datum to remove A:

>

> print(vector(7,k,-log(v[k+1]/v[1])/k/log(10)));

>

> [0.5756, 0.5348, 0.5484, 0.5747, 0.5827, 0.5750, 0.5885]

>

> In both cases c =~ Euler looks rather convincing,

> given the statistics. Well spotted, Sir!

You did very much what I did.

I nearly fell off the chair when I averaged everything out and saw 0.57... :-)

> How strongly are you committed to A = 1, for the average,

> given the variability of the overall factor with a?

Not very.

Would you buy an appeal to Occam's razor, mon vieux?

Mike