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Old 2020-07-07, 08:19   #862
sweety439
 
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Nov 2016

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A Sierpinski form (k*b^n+1)/gcd(k+1,b-1) has algebra factors if and only if k*b^n is perfect odd power A070265 or of the form 4*m^4 A101046

A Riesel form (k*b^n-1)/gcd(k-1,b-1) has algebra factors if and only if k*b^n is perfect power A001597

Situations in which srsieve will make that statement:

1. For any k on any Riesel base that is a perfect square, when sieving, all n-values that are divisible by 2 can be manually removed.
2. For any k on any Sierpinski base that is of the form 4*m^4, all n-values that are divisible by 4 can be manually removed.
3. For any k on any (Riesel or Sierpinski) base that is a perfect cube, all n-values that are divisible by 3 can be manually removed.
4. (etc.) for k's that are perfect 5th powers, 7th powers, 11th powers, or any prime power where p=power, any n's divisible by p can be manually removed.

Taken to an extreme, for k=2048, which is 2^11, you could manually eliminate all n-values that are divisible by 11.

A special case is k=1:

* For Riesel base, all n-values that are not prime can be manually removed.
* For Sierpinski base, all n-values that are not power of 2 can be manually removed.

To put the above in a different way: You can only eliminate the k if manually removing all of these n-values leaves you with no n's remaining, which would have been the case had you attempted to sieve 900*67^n-1. Had you sieved it, you would have ended up with a sieve file with very few EVEN n-values remaining and zero ODD n-values remaining. Once you manually removed the n's divisible by 2, you would have had nothing left. That means that the k-value can be removed from conjecture testing because it has partial algebraic factors that combine with a numeric factor to make a full "covering set" of factors.

Analysis on both k=125 and 729 shows that they should remain because there is no factor or factors that eliminate the n's that the algebraic factors do not.

When sieving, as per the above, on k=125, you can manually remove all n-values divisible by 3. On k=729, that is one of the few that you can eliminate n-values that are divisible by 2 -or- that are divisible by 3. In effect, you're only left with n-values that are n==(1 or 5 mod 6). If you manually remove those n's, you'll stop getting that message from srsieve.

k=729 would normally be extremely low weight except for the fact that it is divisible by 3, which eliminates any possibility of a factor of 3 for all n-values. That increases the weight to something like a k that is a perfect square.

A k-value that would be extremely low-weight is one that is a perfect square and cube but is not divisible by 2 or 3. I think the lowest one of that nature would be 5^6=15625, which would only be an issue on very few bases.

Last fiddled with by sweety439 on 2020-07-07 at 08:26
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