[seqfan] Re: DigitMean of a(n)

Giovanni Resta g.resta at iit.cnr.it
Sun May 10 16:50:49 CEST 2015

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

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 ?)


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