Re: [PrimeNumbers] Re: Primes in the concatenation with the digits of Pi
>From: Andy Swallow <umistphd2003@...>It is as easy to de-value as it is to criticize.
>Subject: Re: [PrimeNumbers] Re: Primes in the concatenation with the digits
>Date: Tue, 3 Feb 2004 10:21:21 +0000
> > I find this to be a huge result way bigger than say Wiles proof of FLT
> > because that proof will never
> > be understood by more than a few people. Besides, it was based on
> > conjecture on modular
> > forms which under certain conditions implied the truth of FLT and Wiles
> > proved that little result.
>Well I see your point, the proof of FLT is a little complex for most
>people. But that point of view could also be said to de-value the
>achievments of some of the more "detailed" results on primes in A.P.'s
>Does this mean you now agree that the infinitude of primes can be
> > I am not meaning to fix n and vary a and k. I want to vary all three
> > Is there only one k and a for which the statement is true?
>Well no, it was just unclear what you wanted to vary when. Fixing *any*
demonstrated by Dirlichlet's
theorem? How about the 6n+1 and 6n+5 argument which you do not address?
>k and a with (k,a)=1, and varying only n, you get infinitely manyWell, while 6n+4 does not satisfy the Dirichlet criteria, I can combine it
> > for k =0 and a =1 we have for n=0,1,2,...,0n+1
> > 1,1,1,1,1,1,1,1,1,...
>Well do you think this sequence contains infinitely many primes? Of
>course not. Every integer divides zero. Or perhaps no integer divides
with all the
(k,a) = 1 sets if I so desire as I did with 0n+1 just to complete the
integers formed by the
second term. This was redundancy. But redundancy is allowed as in the two
sets 2n+1 and 3n+1
where there are repeated numbers. It could well have been left out since 1
is not prime anyway
and the infinite dirichlet produced set of second terms of the rows formed
1n+1 -> 2
2n+1 -> 3
3n+1 -> 4
was sufficient to prove my point. Well, had I done that some of your thunder
would have been
>zero. Is zero an integer? Anyway, you can't say that (0,1)=1.True. However, this does not preclude me to combine the progression 0n+a
>Dirichlet's proof depended on the characters modulo k. If k=0, there are
>no characters, and therefore the proof fails.
Dirichlet allowed ones. 0n+7 would produce infinitly many primes albiet not
My statement still stands:
You can prove that there is an infinite number of primes using Dirichlet's
you can prove the infinitude in a infinite number of ways. well, if you have
time and space.
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> My statement still stands:Dirichlet's theorem proves that there are an infinite number of primes
> You can prove that there is an infinite number of primes using
> Dirichlet's theorem.
in certain infinite subsets of the integers. So the fact that the
total number of primes is infinite follows immediately...
Apologies if I misunderstood your original message and its'