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Old 2018-08-30, 21:42   #5
garambois
 
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Oct 2011

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Quote:
Originally Posted by LaurV View Post
Below 20 (and inclusive) the only sequences which do not terminate are 6^(7, 9, 11, 15, 19).

They are all at the point where I left them, except for few terms of 6^15 added by the elves where the last cofactor was under 110 digits. I work on 10077696 (6^9), and I reserved 95280 years ago, when 279936 (6^7) merged with it. I brought it to 148 digits (currently with a C140 cofactor, I still have it reserved, but not in priority list due to the 2^4*31 driver).
OK, I wrote your name on the page.

Quote:
Originally Posted by LaurV View Post

Interesting that even powers (including higher, to 30 or 50, can't remember) all terminate, some in very large primes. Also, odd powers between 21 and 31 were left after C>100, and were advanced a bit by the DB elves.
Yes, If n is even and if and only if n takes the form n=m^2 or n=2*m^2, then sigma(n)-n will be odd.
If n is odd and if and only if n takes the form n=m^2, then sigma(n)-n will be even.

See the proof, but sorry, in french.

So, because n=6^i always is even, and n takes the form n=m^2 only when i is even, then, sigma(n)-n will be odd and the aliquot sequence will go down. If i is odd, then we will probabily have an Open-End aliquot sequence.

If for example, we take n=5^i which always is odd, when i is even then n takes the form n=m^2, so sigma(n)-n will be even, so the aliquot sequence will probabily be Open-End.

If for example, we take n=2^i which always takes the form m^2 (if i is even) or the form 2 * m^2 (if i is odd), then, sigma(n)-n will always be odd, so the aliquot sequence will probabily always go down.

I would like to find one i with 2^i an open-end sequence, but I haven't found such an aliquot sequence yet.
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