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• ... for(n=0,10000,x=0;t=0;while(1,x=x+1;t=t+sigma(x,n);if(x 2&&t%x==0,prin ... I used subsampling via a swift change to nn=0,10000,n=36*nn+8 inter alia. ...
Message 1 of 3 , Jul 2, 2002
--- jbrennen <jack@...> wrote:
for(n=0,10000,x=0;t=0;while(1,x=x+1;t=t+sigma(x,n);if(x>2&&t%x==0,prin
> t1(x,", ");break)))

I used subsampling via a swift change to
nn=0,10000,n=36*nn+8
inter alia.

> Interesting things I noted about the sequence:
>
> - The values 4 and 8 only occur in the first three terms
> - If n == 0 (mod 6) or n == 1 (mod 6), and n>1, then S[n] = 9
> - If n == 3 (mod 6) or n == 4 (mod 6), then S[n] = 7
> - If n == 5 (mod 6), then S[n] <= 36
> - If n == 14 (mod 18), then S[n] <= 38
> - If n == 2 (mod 36), then S[n] <= 74
>
> This leaves the "interesting" subsets where
> n == 8, 20, or 26 (mod 36) -- these appear more chaotic at
> first glance, although it seems reasonable to believe they
> too are periodic and that a "covering set" exists.

Note - when I say "repeats", I mean ther seems to be a period, but
occasionally a smaller value also satisfies the relation, i.e. "<="
as above.

Looking at 8(mod 36)
31 repeats every 5 terms from 8
186 repeats every 5 terms from 44
62 repeats every 5 terms from 80
50 repeats every 5 terms from 116
xx is a gap every 5 terms from 152

So looking at 152(mod 180)
xx is a gap every 2 terms from 152
123 repeats every 2 terms from 332

So looking at 152(mod 360)
200 repeats every 2 terms from 152
17 repeats every 2 terms from 512

Done.

20(mod 36) gives a similar pattern, obviously for the
62, 50, xx, 31, 186
92(mod 180) gives
123, xx,
272(mod 180) gives
17, 200

Actualy it's clear that 123, 17, 200 provide a covering set, so it
will never exceed 200.

> Note that I have not found any value of S[n] exceeding 200,
> searching through about n==5000. The value 200 occurs for
> the first time at S[632].
>
>
> So the question is... is the sequence periodic (after the
> first three terms of course)? If so, what is its period?
> Are there any members exceeding 200? Note that if a covering
> set exists, the period of the sequence will certainly be much
> larger than the period of the covering set, due to the
> presence of sequence members not in the covering set, but
> smaller than the covering member of the covering set.

It must be periodic as each value that can occur at all will cause a
periodic pattern of residues when summed.

> If it is periodic, it has period >= 1607760, since this
> is the period of the subset where n == 2 (mod 36).

Only 200 numbers to check, shouldn't be too hard to find all their
periods, and then munge them together in a big LCM.

However that required more than 5-character edits to your script, and
so were deemed 'hard' :-)

Phil

=====
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• ... I explored this sequence further, and have essentially fully characterized it... It is periodic, as Phil already figured out. The period is 14802818350320
Message 2 of 3 , Jul 3, 2002
--- In primenumbers@y..., "jbrennen" <jack@b...> wrote:
>
> Let S[n] be the smallest number x>2 such that:
>
> x divides: Sum {i=1..x} (sigma_n(i))
>
> Here sigma_n(i) denotes the sum of the nth powers of the positive
> integer divisors of i.
>
> So for any particular n, we are looking for the smallest x>2 such
> that sigma_n(1), sigma_n(2), ... sigma_n(x) have an integer average.

I explored this sequence further, and have essentially fully
characterized it...

It is periodic, as Phil already figured out. The period is
14802818350320 == 2^4*3^2*5*7^2*11*13*23*29*53*83.
That is, for n>=3, S[n+14802818350320] = S[n].

There are 32 numbers which appear in the sequence. The numbers
4 and 8 only appear within the initial (non-periodic) subsequence
of length 3. So 30 numbers appear in the periodic portion of
the sequence. The numbers are: 7, 9, 11, 17, 23, 29, 31,
36, 38, 46, 50, 51, 59, 61, 62, 69, 71, 74, 89, 107, 113, 123,
150, 158, 167, 169, 186, 188, 197, 200.

The last number to show up as the sequence is enumerated is the
number 197, which doesn't appear until S[19592].
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