- --- 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

=====

--

"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

__________________________________________________

Do You Yahoo!?

Yahoo! - Official partner of 2002 FIFA World Cup

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].