[seqfan] Re: DigitMean of a(n)
Giovanni Resta
g.resta at iit.cnr.it
Sat May 9 21:18:04 CEST 2015
On 05/09/2015 07:35 PM, Eric Angelini wrote:
>
> Let's say that the "DigitMean" of a(n)
> is the sum of the digits of a(n) divided
> by the quantity of digits in a(n).
> ...
> We could look for integers N such
> that DigitMean(N) _starts_ with the
> digits of N:
As you said the numbers increase fast.
If I'm not wrong, the first terms are
1, 2, 3, 4, 5, 6, 7, 8, 9, 45, 566, 1500, 2250, 3750, 18000, 383333,
4428571, 11250000, 788888888, 1000000000, 2000000000, 3000000000,
4000000000, 5000000000, 6000000000, 7000000000, 8000000000,
9000000000, 44545454545, 358333333333, 4461538461538, 42857142857142,
43571428571428, 44285714285714, 1687500000000000, 7500000000000000,
44705882352941176, 4473684210526315789, 45000000000000000000,
4454545454545454545454, 44782608695652173913043,
308333333333333333333333, ....
where terms like 1000000000 are included if we do not consider
the zero before the decimal dot in the DigitMean of a number, so
in this case the mean is 0.10000000000... => 1000000000.
This also means that the sequence is infinite because it contains all
the numbers of the form {1,2,...,9} * 10^(10^k-1).
I computed all the terms with 1000 or less digits:
Up to 10^100 there are only 148 terms in the sequence,
up to 10^1000 there are only 876 terms.
Giovanni
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