- 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 - --- 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:

Note - when I say "repeats", I mean ther seems to be a period, but

>

> - 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.

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,

It must be periodic as each value that can occur at all will cause a

> 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.

periodic pattern of residues when summed.

> If it is periodic, it has period >= 1607760, since this

Only 200 numbers to check, shouldn't be too hard to find all their

> is the period of the subset where n == 2 (mod 36).

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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"One cannot delete the Web browser from KDE without

losing the ability to manage files on the user's own

hard disk." - Prof. Stuart E Madnick, MIT.

So called "expert" witness for Microsoft. 2002/05/02

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http://fifaworldcup.yahoo.com - --- In primenumbers@y..., "jbrennen" <jack@b...> wrote:
>

I explored this sequence further, and have essentially fully

> 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.

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].