[seqfan] Re: DigitMean of a(n)

[Giovanni Resta]

On 05/10/2015 03:34 PM, David Wilson wrote:
> There's a fairly fast way to compute these.

Yes, that is the method I used (in Mathematica) to compute
the terms up to 10^1000.

By the way, I have submitted two drafts, one for this
sequence https://oeis.org/draft/A257829
and one for the similar sequence involving the
geometric mean instead of the average
https://oeis.org/draft/A257829

this other sequence is less easy to compute because
while the possible sums for numbers
of d digits are 9*d, the number of possible non-zero
products seems to be, empirically,  equal to (T_{d+1})^2, where
T_n is the n-th triangular number

(I do not know why!! Does anybody know why ?)

Giovanni