 mersenneforum.org > Math k*b^n+/-1, Bases 271 and 11971
 Register FAQ Search Today's Posts Mark Forums Read  2013-11-05, 14:28   #12
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

763510 Posts Quote:
 Originally Posted by robert44444uk ....Without having seen Bob's reply..... Sorry folks, I think I am going totally down the wrong line here with this. It is not going to go down the lines I was hoping for, which was to determine a base for which M(x)=1
That question is (or seems to be) purely numerical. AFAIK, this sub-forum is for the discussion of the mathematics behind all of the computations, rather than the actual computations.

If there is some theoretical reason to believe that a base exists with
your desired properties, then the theory would certainly merit discussion here.
But a discussion of a purely numerical search is off-topic.

Perhaps I assume too much? Do people not understand the distinction
between the math and the computing?   2013-11-05, 14:30   #13
robert44444uk

Jun 2003
Suva, Fiji

111111110102 Posts Quote:
 Originally Posted by R.D. Silverman All of the members? This is an infinite set. How does one show that yM(x)2^n+1 or yM(x)2^n-1 never has a factor q whose order to the base 2 is less than (a pre-specified) z for ALL n and ALL y?
No, y is fixed. Payam numbers deal with either - or +. So it is for all n for either - or +.

I did look into "AntiBriers" which are the complement of Brier numbers. Here there are y which meet the requirement for both - and + at the same time, but this is still for fixed y.

Last fiddled with by robert44444uk on 2013-11-05 at 14:32   2013-11-05, 14:32   #14
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

3×5×509 Posts [QUOTE=robert44444uk;358470]
Quote:
 Originally Posted by R.D. Silverman All of the members? This is an infinite set. How does one show that yM(x)2^n+1 or yM(x)2^n-1 never has a factor q whose order to the base 2 is less than (a pre-specified) z for ALL n and ALL y? QUOTE] No, y is fixed. Payam numbers deal with either - or +. So it is for all n for either - or +.
Even with y fixed it is still an infinite set, so the query about an existence
proof still remains.   2013-11-05, 15:43   #15
robert44444uk

Jun 2003
Suva, Fiji

2·1,021 Posts [QUOTE=R.D. Silverman;358471]
Quote:
 Originally Posted by robert44444uk Even with y fixed it is still an infinite set, so the query about an existence proof still remains.
Its the same line of argument around covering sets that works for Sierpinski numbers but the other way around. Sierpinski works off the recurrence of certain factors as n increases and uses CRM to define a k where there is cover at all possible n values in the covering period.

If we have a look at the opposite..the Payam situation; 7 can only be a factor of a member of the power series k*2^n+1 if k is 3,5 or 6 mod 7. If k=11 for example, i.e k=4mod7, then 7 is never a factor, albeit n goes to infinity.

All that we are doing in defining a payam number is picking a k using CRM so that all primes with a multiplicative order base 2 of less than a given value are never factors. If they are never factors, then the smallest factor that any member of the series can have is greater than the given value. This changes the chance of the member being prime, and overall, the series should produce on average more primes than a given k of a similar size. And so they do.   2013-11-05, 15:54   #16
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

3·5·509 Posts [QUOTE=robert44444uk;358477]
Quote:
 Originally Posted by R.D. Silverman Its the same line of argument around covering sets that works for Sierpinski numbers but the other way around. Sierpinski works off the recurrence of certain factors as n increases and uses CRM to define a k where there is cover at all possible n values in the covering period. If we have a look at the opposite..the Payam situation; 7 can only be a factor of a member of the power series k*2^n+1 if k is 3,5 or 6 mod 7. If k=11 for example, i.e k=4mod7, then 7 is never a factor, albeit n goes to infinity.
The mystery (to me) is how one can generate a covering set where each
element has a bounded order. Reference please.   2013-11-05, 16:36   #17
robert44444uk

Jun 2003
Suva, Fiji

2·1,021 Posts [QUOTE=R.D. Silverman;358479]
Quote:
 Originally Posted by robert44444uk The mystery (to me) is how one can generate a covering set where each element has a bounded order. Reference please.
Payam numbers do not aim to generate a covering set. Just to exclude certain primes as factors, contrasting the way that Sierpinski ensures their inclusion.

Payam factors are excluded for all n, Sierpinski factors are included and are factors of different members of a covering set.

Sierpinski will guarantee zero primes, Payam certainly does not guarantee infinite primes or even a guaranteed higher density. For example, I think it is unlikely but there could be a covering set of non Payam factors (all greater than the E level) that generate a Sierpinski covering set and guarantee that a Payam power series has zero primes.   2013-11-06, 11:03 #18 robert44444uk   Jun 2003 Suva, Fiji 7FA16 Posts In fact what I found is a rather useless property, the two primes in question are the first instance primes p that are not multiplicative order p-1 for all prime bases less than 43 and 73 respectively. In terms of the Payam number formula, that means that the two primes are not part of the M(x) formula, for those prime bases mentioned above In terms of what I was trying to achieve, i.e. M(x)=1, I might have a look at that when I get a little more time.   2013-11-07, 10:43   #19
robert44444uk

Jun 2003
Suva, Fiji

37728 Posts Quote:
 Originally Posted by R.D. Silverman If there is some theoretical reason to believe that a base exists with your desired properties, then the theory would certainly merit discussion here. But a discussion of a purely numerical search is off-topic.
Thinking a bit more about this, except base 4 and base 9, bases that are squares exhibit the property of having no prime members in M(x), other than 2 and 3. And bases 4 and 9 have one of those two primes in M(x)

2 and 3 are always members given the members of M(x) exhibit the property of being multiplicative order of p-1 the base, or 1 the base. 2 and 3 always show these properties.

Putting aside the question of non square composite bases, the question remains whether there is a prime base which has no other members in M(x) other than 2 and 3.   2013-11-07, 12:23   #20
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

3·5·509 Posts Quote:
 Originally Posted by robert44444uk Thinking a bit more about this, except base 4 and base 9, bases that are squares exhibit the property of having no prime members in M(x), other than 2 and 3. And bases 4 and 9 have one of those two primes in M(x) 2 and 3 are always members given the members of M(x) exhibit the property of being multiplicative order of p-1 the base, or 1 the base. 2 and 3 always show these properties. Putting aside the question of non square composite bases, the question remains whether there is a prime base which has no other members in M(x) other than 2 and 3.
This is still just numerology. Is there any mathematics that you wish to discuss?

In particular, the assertion in the second paragraph "2 and 3 always show these properties" needs PROOF. Even a heuristic/probabilistic argument
suggesting why the assertion should be true would be something.

The question in paragraph 3 is also somewhat interesting. Or at least it
would be if it were backed by some REASONING.   2013-11-07, 12:26   #21
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

3·5·509 Posts Quote:
 Originally Posted by R.D. Silverman This is still just numerology. Is there any mathematics that you wish to discuss? In particular, the assertion in the second paragraph "2 and 3 always show these properties" needs PROOF. Even a heuristic/probabilistic argument suggesting why the assertion should be true would be something. The question in paragraph 3 is also somewhat interesting. Or at least it would be if it were backed by some REASONING.
Perhaps the moderators should set up a sub-forum for discussion of
purely numerical OBSERVATIONS, free from any mathematical reasoning???

This thread would more properly belong there.   2013-11-07, 15:01   #22
robert44444uk

Jun 2003
Suva, Fiji

2×1,021 Posts Quote:
 Originally Posted by R.D. Silverman Perhaps the moderators should set up a sub-forum for discussion of purely numerical OBSERVATIONS, free from any mathematical reasoning??? This thread would more properly belong there.
I think you are probably right. This is all about observations, there is not a lot of maths here.

I can't write proofs to the standards expected by mathematicians. I lack the training to do this.

The reasoning behind me saying what I did about square bases was as follows:

If a constituent prime number member p of M(x) in the Payam notation for the power series (first series) y*M(x)*b^n+/-1, [y fixed, M(x) fixed, n from 1 to infinity], is defined in terms of having multiplicative order p-1 base b

then if you square b and use this as a base, thereby defining a second power series, then the constituent members of M(x) in the first series will have a multiplicative order of (p-1)/2 base b^2 in this second power series and thereby no longer qualify to be part of M(x) for that series as p-1 =/= (p-1)/2. The fact that it is (p-1)/2 can no doubt be proven - this is my observation.

If a prime is not in M(x) in the first power series, and has order (p-1)/2 base b, then it appears to have order (p-1)/2 base b^2. (observation)

If a prime is not in M(x) in the first power series, and has order q base b, with q=/=(p-1) then it appears to have order q/2 base b^2 or q base b^2. Both observations at this stage.

This covers all possibilities of values of multiplicative order - all either decrease the value of the multiplicative order or leave it the same, and therefore as p-1 is the highest possible value for the order for a prime p, then p-1 is no longer a valid multiplicative order for base b^2.

Think 2 and 3 are special cases as mentioned before.

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