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• In my brainstorming over sigma-related problems, I came up with a neat sequence. Let S[n] be the smallest number x 2 such that: x divides: Sum {i=1..x}
Message 1 of 3 , Jul 2, 2002
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In my brainstorming over sigma-related problems, I came up with
a neat sequence.

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.

The sequence can be generated in PARI/GP:

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)))

The sequence starts out:

4, 8, 8, 7, 7, 36, 9, 9, 31, 7, 7, 11, 9, 9, 38, 7, 7, 17, 9, 9 ...

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 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.

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

Anyway, I thought this might be of interest....

Jack
• ... 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
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--- 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 1 of 3 , Jul 3, 2002
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--- 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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