# COMMENT on A111392.

Sun Nov 20 04:14:26 CET 2005

```    Neil
Did you see Robert G. Wilson v's mail?
He edited my sequence and Gerald McGarvey extended it.

Yasutoshi

----- Original Message -----
From: "Robert G. Wilson v" <rgwv at rgwv.com>
To: "Neil J. A. Sloane" <njas at research.att.com>
Cc: "Yasutoshi Kohmoto" <zbi74583 at boat.zero.ad.jp>; "Gerald McGarvey"
<Gerald.McGarvey at comcast.net>
Sent: Monday, November 14, 2005 5:07 AM
Subject: COMMENT on A111392.

> Neil,
>
> by
> way of Gerald McGarvey. I also changed the title from "a(n) =
> Product_{1<=i<n}
> (Product_{1<=k<=i} p_k + Product_{i<k<=n} p_k).". Therefore please find
> the following.
>
> Thanx, Bob.
>
>
> %I A111392
> %S A111392
> 2,5,187,162319,10697595389,63619487169453143,74365399061678006800073593,
> %T A111392 11864736003419293844093922527852416537,
> %U A111392 601642845102734414280661105098046392912578705726003
> %N A111392 a(n) = Product_{i=1..n-1} (Product_{k=1..i} p_k +
> Product_{k=i+1..n} p_k).
> %C A111392 This is a "Proof of existence of infinite primes" sequence.
> Proof. Let N = Product_{1<=i<n} (Product_{1<=k<=i} p_k + Product_{i<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_1=0, Mod p_i).
> %C A111392 a(1) may equal 1 or 2 by definition. The author chose 2.
> %o A111392
> t=10;for(n=2,t,print(prod(i=1,n-1,prod(k=1,i,prime(k))+prod(k=i+1,n,prime(k)))));
> Gerald McGarvey
> %t A111392 f[n_] := Product[(Product[Prime[k], {k, i}] + Product[Prime[k],
> {k, i + 1, n}]), {i, n - 1}]; Array[f, 9] (* RGWv *)
> %Y A111392 Cf. A024451.
> %K A111392 nonn
> %O A111392 1,1
> %A A111392 Yasutsohi Kohmoto zbi74583(AT)boat.zero.ad,jp
> %E A111392 Corrected and extended by Gerald McGarvey
> (Gerald.McGarvey(AT)comcast.net) & RGWv (rgwv(at)rgwv.com), Nov 12 2005
>
>

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