> On 2011-02-27, pamela fluente <pamelaflue...@libero.it> wrote:
> > On 27 Feb, 10:10, Tim Little <t...@little-possums.net> wrote: > >> In a certain sense, sequences where no solution ever exists are > >> "special", having density 0.
> > This is actually a very interesting statement, what's the reasoning > > behind ?
> For a given initial finite sequence, there is a well-defined density > of next sequence values that yield no solution. Then you can define a > density product of finitely many "next" terms, and prove that as the > number of terms increases, it is bounded above by a sequence > converging to zero. The only exception is sequences starting with > c(1) = 1, for which no solution can ever exist.
> > Actually, i think the most familiar and simplest sequence we might > > think about could be the following one. > > Let p(i) be the sequence of primes. Take an initial value r (any r >= > > 1000), and then define:
> > can anyone prove that i will have *at least* 1 solution, for *at > > least* one integer h >= r ? ;-))
> Is m here a variable, or fixed in advance? Is it the same thing as z > from the previous problem description?
> -- > Tim
m is just an "enumerator".
c(m) = p(h + m-1) indicates the m-th term of the sequence of the c's (in this case, the (h+m+1)-th prime).
In the previous posts, viewing the problem a growing set of z equations, z was also indicating the unknown which expresses the number of necessary equations to get 1 solution, as well as the z-th term of the sequence of the c's.
On 2011-03-01, pamela fluente <pamelaflue...@libero.it> wrote:
> m is just an "enumerator".
> c(m) = p(h + m-1) indicates the m-th term of the sequence of the c's > (in this case, the (h+m+1)-th prime).
Ah okay. In that case z=1 will provide a solution for any h, with x(1) = 1 and y(1) = p(h+1) - p(h). The condition 1 < y < p(h) always holds by Bertrand's Postulate, and y(1) =/= p(h) - 2 because y(1) is even and p(h) is odd.
> On 2011-03-01, pamela fluente <pamelaflue...@libero.it> wrote:
> > m is just an "enumerator".
> > c(m) = p(h + m-1) indicates the m-th term of the sequence of the c's > > (in this case, the (h+m+1)-th prime).
> Ah okay. In that case z=1 will provide a solution for any h, with > x(1) = 1 and y(1) = p(h+1) - p(h). The condition 1 < y < p(h) > always holds by Bertrand's Postulate, and y(1) =/= p(h) - 2 because > y(1) is even and p(h) is odd.
> This does seem a fairly trivial solution, though.
> -- > Tim
Thanks Tim. Beautiful argument! Actually as you rightly point out, the case of greater interest is for z larger than a threshold.
Would it possible to extend a similar argument to the case z > k (k any positive integer) or is it an impossible task ?
On 2011-03-02, pamela fluente <pamelaflue...@libero.it> wrote:
> Would it possible to extend a similar argument to the case z > k (k > any positive integer) or is it an impossible task ?
Since all the coefficients are prime, c(z+1) is obviously not a multiple of any of them.
The remaining constraint was y =/= c(m)-2. That may be unsatisfied as it is possible that c(z+1) = -2 mod c(m) for some m, but it becomes less common for larger primes and there are plenty of prime candidates c(z+1) for which it never holds.
I can't prove that there are infinitely many, but I did find examples greater than 10^9.
> On 2011-03-02, pamela fluente <pamelaflue...@libero.it> wrote:
> > Would it possible to extend a similar argument to the case z > k (k > > any positive integer) or is it an impossible task ?
> Since all the coefficients are prime, c(z+1) is obviously not a > multiple of any of them.
> The remaining constraint was y =/= c(m)-2. That may be unsatisfied as > it is possible that c(z+1) = -2 mod c(m) for some m, but it becomes > less common for larger primes and there are plenty of prime candidates > c(z+1) for which it never holds.
> I can't prove that there are infinitely many, but I did find examples > greater than 10^9.
> -- > Tim
Thanks a lot, very nice results indeed Tim. (Let me know in case, in future, some argument comes up in your mind.)
I think that the Bertrand postulate you mentioned is pretty useful.
On 3 Mar, 01:02, Tim Little <t...@little-possums.net> wrote:
> On 2011-03-02, pamela fluente <pamelaflue...@libero.it> wrote:
> I can't prove that there are infinitely many, but I did find examples > greater than 10^9.
It would be incredibly good if given a solution at z, we could at least say what is the count s, where z + s is another solution. It's pretty weird this problem is so elusive that it's even so hard just to count the numbers to the next occurrence. After all, the remainders form a cyclically ordered sequence and it seems strange we can't "keep track" of them.
On 2011-03-05, pamela fluente <pamelaflue...@libero.it> wrote:
> It would be incredibly good if given a solution at z, we could at > least say what is the count s, where z + s is another solution.
For your family of sequences, usually s=1.
> It's pretty weird this problem is so elusive that it's even so hard > just to count the numbers to the next occurrence. After all, the > remainders form a cyclically ordered sequence and it seems strange > we can't "keep track" of them.
It's not elusive at all, it's just vaguely stated. I expect I could write a program that tells which z's give solutions almost as fast as the sequence values can stream in through the Ethernet port of my computer.
The only "elusiveness" is that you're talking about an infinite family of sequences, and speculating about common properties of their collective solutions.
> On 2011-03-05, pamela fluente <pamelaflue...@libero.it> wrote:
> > It would be incredibly good if given a solution at z, we could at > > least say what is the count s, where z + s is another solution.
> For your family of sequences, usually s=1.
> > It's pretty weird this problem is so elusive that it's even so hard > > just to count the numbers to the next occurrence. After all, the > > remainders form a cyclically ordered sequence and it seems strange > > we can't "keep track" of them.
> It's not elusive at all, it's just vaguely stated. I expect I could > write a program that tells which z's give solutions almost as fast as > the sequence values can stream in through the Ethernet port of my > computer.
For simplicity, let's take the version r=h=1, so the c's are just the plain and simple sequence of primes 2 3 5 ... Let z > 2 be any integer which satisfies the system:
c(m) * x(m) = c(z+1) - y(m) 1 <= y(m) <= c(m)-1, y(m) <> c(m)-2 for all m= 1,2,3..., z
Can we really say immediately (i mean without going through each one of the values following z) for each z which is the next (z + s) also satisfying the system ?
Or can we jump directly from one solution to the next one ?
[ What seems strangely elusive to me is determining s = F(z) ]
On 2011-03-05, pamela fluente <pamelaflue...@libero.it> wrote:
> Can we really say immediately (i mean without going through each one > of the values following z) for each z which is the next (z + s) also > satisfying the system ?
No. But again, why would you expect to be able to predict which values of a sequence give solutions without being permitted to look at them?
> [ What seems strangely elusive to me is determining s = F(z) ]
I don't see anything strangely elusive about that at all. The problem is just that artificial restriction you're setting of not being permitted to look at the values of c.
> On 2011-03-05, pamela fluente <pamelaflue...@libero.it> wrote:
> > Can we really say immediately (i mean without going through each one > > of the values following z) for each z which is the next (z + s) also > > satisfying the system ?
> No. But again, why would you expect to be able to predict which > values of a sequence give solutions without being permitted to look at > them?
> > [ What seems strangely elusive to me is determining s = F(z) ]
> I don't see anything strangely elusive about that at all. The problem > is just that artificial restriction you're setting of not being > permitted to look at the values of c.
> -- > Tim
Well ok, even allowing looking at ALL the values up to c(z+1), z > 2, I believe that it will not give any advantage. I think you still will find very difficult to compute (not necessarily with a closed form formula, but at least with a converging algorithm) the s (distance from next solution).
The reason why this seems a little strange to me is that anyway the remainders, from a given solution follow cyclically a precise (and simple) sequence. So the elusive part i see is in the fact that, while from c(z+1) one we can know perfectly each of these sequences (because we know perfectly all the remainders at that point), still it seems difficult or impossible to compute directly at what distance the same condition would occur.
On 2011-03-06, pamela fluente <pamelaflue...@libero.it> wrote:
> Well ok, even allowing looking at ALL the values up to c(z+1)
It's the values *after* c(z+1) that matter just as much as the ones before. Why do you expect to be able to determine whether c(z+2), c(z+3), and so on allow solutions without being permitted to know what values they have?
On 7 Mar, 05:57, Tim Little <t...@little-possums.net> wrote:
> On 2011-03-06, pamela fluente <pamelaflue...@libero.it> wrote:
> > Well ok, even allowing looking at ALL the values up to c(z+1)
> It's the values *after* c(z+1) that matter just as much as the ones > before. Why do you expect to be able to determine whether c(z+2), > c(z+3), and so on allow solutions without being permitted to know what > values they have?
Given any prime p(n), we can know (by division) all the remainders of the divisions by the preceding primes. Denote these remainders are: r(1) ... r(n-1) respectively.
Now the next prime p(n+1) could also be found as p(n) + d where the distance d is the smallest positive integer j such that:
( r(s) + j ) mod p(s) <> 0 for all s=1,...,n
or equivalently: ( remainder of (p(n)/u) + j ) mod u <> 0 for all u=2,3,..
[ so we don't really need to know the values of the primes p(1) ... p(n-1) ].
So we have a formula (algorithm) for any next prime (which actually does not require knowledge of previous primes), and further we know such a prime exists (algorithm is guaranteed to converge). But, as soon as we add just 1 simple additional condition on remainders, like for instance: ( r(s) + j ) mod p(s) <> p(s)-2
then it seems to me we cannot assure that the next prime, satysfying also the above, exists at all. And this seemed a little strange. isn't it ?
On 2011-03-07, pamela fluente <pamelaflue...@libero.it> wrote:
> So we have a formula (algorithm) for any next prime (which actually > does not require knowledge of previous primes), and further we know > such a prime exists
Yes, as originally shown by Euclid using a rather ingenious argument. It was not always obvious that there is no limit to the primes, and Euclid's argument is very "fragile": change the properties of the set even slightly, and it ceases to be useful at all (even if the set is in fact infinite).
> But, as soon as we add just 1 simple additional condition on > remainders, like for instance: > ( r(s) + j ) mod p(s) <> p(s)-2
> then it seems to me we cannot assure that the next prime, satysfying > also the above, exists at all. And this seemed a little strange. isn't > it ?
It seems very likely that infinitely many such primes exist. I merely stated that I could not prove it, a statement to be taken in the context that I devote only a limited amount of time to newsgroup posts.
Note that there are very many unsolved conjectures concerning infinitude of various classes of primes. I see nothing strange in the idea that another such class might or might not be infinite, and that I might not know whether it is or not. Do you?
> On 2011-03-07, pamela fluente <pamelaflue...@libero.it> wrote:
> > So we have a formula (algorithm) for any next prime (which actually > > does not require knowledge of previous primes), and further we know > > such a prime exists
> Yes, as originally shown by Euclid using a rather ingenious argument. > It was not always obvious that there is no limit to the primes, and > Euclid's argument is very "fragile": change the properties of the set > even slightly, and it ceases to be useful at all (even if the set is > in fact infinite).
> > But, as soon as we add just 1 simple additional condition on > > remainders, like for instance: > > ( r(s) + j ) mod p(s) <> p(s)-2
> > then it seems to me we cannot assure that the next prime, satysfying > > also the above, exists at all. And this seemed a little strange. isn't > > it ?
> It seems very likely that infinitely many such primes exist. I merely > stated that I could not prove it, a statement to be taken in the > context that I devote only a limited amount of time to newsgroup posts.
> Note that there are very many unsolved conjectures concerning > infinitude of various classes of primes. I see nothing strange in the > idea that another such class might or might not be infinite, and that > I might not know whether it is or not. Do you?
> -- > Tim
:-)) well, "intuitively" i could bet anything that such a prime will always exist, as 1) a consequence of the infinity of primes And 2) due to the fact that such infiniteness together with remainders' periodicity (replicated" patterns") will certainty cause this condition to re-occur again and again. But actually putting out a rigorous argument for that is all another matter, and probably elusive for most minds!
In particular, I think it's condition 2) which will cause any possible condition of this type (or even any multiple combination of similar conditions) to be verified on infinite primes (clearly "rarer and rarer").
The basic problem is probably to find a good "method" to "count" within this context of simultaneous cyclic patterns... the whole problem could be "reduced" to mere "counting" the integers occurring within the occurrence of the various conditions...
