proof sequence

[kohmoto]

    Hi, Seqfans.

    The following formula is well known for proving "Existence of infinite 
primes".
         Product_{1<=i<=n} p_i + 1
    I think that it is elegant and the easiest.
    But I suppose that Seqfans might like the most complicated one.
    So, I submit this sequence.

    Yasutoshi

    %I A000001
    %S A000001 3, 175, 2336191, 26093310174834487
    %N A000001  a(n) = (Product_{0<=e_i<=1}  (Product_{1<=i<=n} p_i^e_i + 
Product_{1<=i<=n} p_i^(1-e_i)))^(1/2) * (Sum_{1<=i<=n} 
(1/p_i*Product_{1<=k<=n} p_k) )
                      Where p_i means i-th prime.
    %C A000001 This is a "Proof of existence of infinite primes" sequence.
                      Proof. Let N = (Product_{0<=e_i<=1} 
(Product_{1<=i<=n} p_i^e_i + Product_{1<=i<=n} p_i^(1-e_i)))^(1/2) * 
(Sum_{1<=i<=n}  (1/p_i*Product_{1<=k<=n} p_k) )  . Suppose there are only a 
finite number of primes p_i, 1<=i<=n. If N is prime, then for all i, not 
(N=p_i). Because, for all i, p_i<N. If N is composite, then it must have a 
prime divisor p which is different from primes p_i. Because, for all i, not 
(N=0, Mod p_i).

    %e A000001 a(3)= 
((1+p_1*p_2*p_3)*(p_3+p_1*p_2)*(p_2+p_1*p_3)*(p_2*p_3+p_1)*(p_1+p_2*p_3)*(p_1*p_3+p_2)*(p_1*p_2+p_3)*(p_1*p_2*p_3+1))^(1/2) 
* (p_2*p_3+p_1*p_3+p_1*p_2)
                 = 
(1+p_1*p_2*p_3)*(p_3+p_1*p_2)*(p_2+p_1*p_3)*(p_2*p_3+p_1) * 
(p_2*p_3+p_1*p_3+p_1*p_2)
                 =        31*11*13*17*31

    %Y A000001 A111392
    %K A000001 none
    %O A000001 1,1
    %A A000001 Yasutsohi Kohmoto   zbi74583 at boat.zero.ad.jp
 
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