## Rv: Conjecture for triplets of primes

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• ... De: dgdegiorgi Para: Sebastian Martin Ruiz Enviado: mié,2 febrero, 2011 19:13 Asunto: Re: Conjecture for
Message 1 of 2 , Feb 2, 2011
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De: dgdegiorgi <degiorgi@...>
Para: Sebastian Martin Ruiz <s_m_ruiz@...>
Asunto: Re: Conjecture for triplets of primes

The Ruiz conjecture is independent with the Goldbach conjecture, as it does also
involve the position of the primes and not only their value.

Tn,1 contains three consecutive primes, Tn,2 contains first third and fifth of
five consecutive primes, and so on.

Unfortunately the conjecture seems to be false and some candidates for
counterexamples are 227, 2099, 5641.

227 is p_49, and it is easy to verify that T49,k (k=1..48) is not symmetric,
thus 229 does not appear as the middle element of a symmetric Tn,k.

All T49-k,k are also not symmetric and thus 227 daes not appear as the largest
element of a symmetric Tn,k.

The Tn,k with 227 as smallest element for k=1..4 are

227, 229, 233, -2

227, 233, 241, -2

227, 239, 257, -6

227, 241, 269, -14

with the fourt value the distance of 2pn from the sum of the other primes. for k
= 1000..10000 in step of 1000 we get

227, 8377, 17881, -1354

227, 17881, 38351, -2816

227, 27943, 59879, -4220

227, 38351, 82307, -5832

227, 49121, 105337, -7322

227, 59879, 128683, -9152

227, 71191, 152293, -10138

227, 82307, 176557, -12170

227, 93763, 200867, -13568

227, 105337, 225343, -14896

As the distance alway grows it should not be too difficult to prove that
equality can not be attained.

Daniel

--- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
>
> Examples:
>
> n= 6 k= 2 {7,13,19}     7+19=13*2
> n= 7 k= 2 {11,17,23}   11+23=17*2
> n= 10 k= 3 {17,29,41}       " "
> n= 10 k= 5 {11,29,47}
> n= 11 k= 3 {19,31,43}
> n= 12 k= 6 {13,37,61}
> n= 13 k= 3 {29,41,53}
> n= 13 k= 4 {23,41,59}
> n= 16 k= 1 {47,53,59}
> n= 16 k= 7 {23,53,83}
> n= 17 k= 7 {29,59,89}
> n= 18 k= 4 {43,61,79}
> n= 19 k= 8 {31,67,103}
> n= 20 k= 3 {59,71,83}
> n= 20 k= 4 {53,71,89}
> n= 20 k= 10 {29,71,113}
>
>
> Note: I did not find any example for the prime number 227.
>
> Sincerely
>
> Sebastián Martín Ruiz
>
> ________________________________
>
> De: Carlos Rivera <cbrfgm@...>
> Para: Sebastian Martin Ruiz <s_m_ruiz@...>
> Enviado: dom,30 enero, 2011 01:57
> Asunto: Re: Conjecture for triplets of primes
>
> 2011/1/29 Sebastian Martin Ruiz <s_m_ruiz@...>
>
> Hello all:
> >
> >Let's consider the triplet of prime numbers: Tn,k = {p (n+k), pn, p (n-k)}  k
><n
> >positive integers.
> >
> >We will say that Tn,k is symmetric if p(n+k) +p (n-k) =2pn.
> >( pi= is the i-th prime number).
> >
> >Conjecture:
> >
> >For all q>2 prime number exists n, k such that q is in the simmetric Tn,k.
> >
> >
> >Sincerely:
> >
> >Sebastián Martin Ruiz
> >
>
>
>
> --
> C.Rivera
> www.primepuzzles.net
> cbrfgm@...
>
>
> Examples?
>
>
>
>
> [Non-text portions of this message have been removed]
>

[Non-text portions of this message have been removed]
• ... [snip] Other counterexamples are 7,19, 23,43,47, ..... unless I somehow misunderstood your conjecture. But with such easy counterexamples I don t
Message 2 of 2 , Feb 2, 2011
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--- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
>
>
> De: dgdegiorgi <degiorgi@...>
> Para: Sebastian Martin Ruiz <s_m_ruiz@...>
> Enviado: mié,2 febrero, 2011 19:13
> Asunto: Re: Conjecture for triplets of primes
>
> The Ruiz conjecture is independent with the Goldbach conjecture, as it does also
> involve the position of the primes and not only their value.
>
> Tn,1 contains three consecutive primes, Tn,2 contains first third and fifth of
> five consecutive primes, and so on.
>
> Unfortunately the conjecture seems to be false and some candidates for
> counterexamples are 227, 2099, 5641.
>

[snip]

Other counterexamples are 7,19, 23,43,47, ..... unless I somehow misunderstood your conjecture. But with such easy counterexamples I don't understand how you could make it a conjecture in the first place?
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