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• Apologies if this is a big yawn to you all but I would like to know if ther is a quick way to determine if r  = 6*m*n + m + n where r, m, n are whole numbers.
Message 1 of 3 , Nov 1, 2009
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Apologies if this is a big yawn to you all but I would like to know if ther is a quick way to determine if
r  = 6*m*n + m + n where r, m, n are whole numbers.
(One of the three dominant patterns in primes (6a + 1))

• It s symmetrical and can be seen as a bilinear function f(m,n)=6*m*n+m+n with a relation r=f(m,n) where for n,m 0 f(n+1,m)=f(m,n)+7 f(n,m)=f(m,n) ... Thanks,
Message 2 of 3 , Nov 2, 2009
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It's symmetrical and can be seen as a bilinear function
f(m,n)=6*m*n+m+n
with a relation
r=f(m,n)
where for n,m>0
f(n+1,m)=f(m,n)+7
f(n,m)=f(m,n)
...
Thanks, it's a lot of value(s).
JD

--- In univalg@yahoogroups.com, William Ftabob <fatbobforman@...> wrote:
>
> Apologies if this is a big yawn to you all but I would like to know if ther is a quick way to determine if
> r  = 6*m*n + m + n where r, m, n are whole numbers.
> (One of the three dominant patterns in primes (6a + 1))
>
• Message 3 of 3 , Nov 2, 2009
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--- In univalg@yahoogroups.com, "Jens" <jd@...> wrote:
>
> It's symmetrical and can be seen as a bilinear function
> f(m,n)=6*m*n+m+n
> with a relation
> r=f(m,n)
> where for n,m>0
> f(n+1,m)=f(m,n)+7
> f(n,m)=f(m,n)
> ...
> Thanks, it's a lot of value(s).
> JD
>
> --- In univalg@yahoogroups.com, William Ftabob <fatbobforman@> wrote:
> >
> > Apologies if this is a big yawn to you all but I would like to know if ther is a quick way to determine if
> > r  = 6*m*n + m + n where r, m, n are whole numbers.
> > (One of the three dominant patterns in primes (6a + 1))
> >
>
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