## Re: [PrimeNumbers] Euler Totient Result Corrected

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• ... A good strategy can find better extremes than brute force. The prime sums p(n+1)+p(n) and p(n)+p(n-1) are both even (assuming n 2). Considering the
Message 1 of 4 , Jul 8, 2008
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Sebastian Martin wrote:
> I have tested for n=1 to 200000 and I have new possible bounds:
> 1/E < Phi(p(n+1)+pn) /Phi(pn+p( n-1)) < E
> E=2.71828...

> Forget it, I go away to the beach, I need a few vacations.
> P.D. There are any bound?

A good strategy can find better extremes than brute force.
The prime sums p(n+1)+p(n) and p(n)+p(n-1) are both even (assuming n>2).
Considering the behaviour of Phi, we want both sums to have known
factorization, and for an extreme we want one of them to have lots of
small odd prime factors, and the other to have none.

Let p = 3970433188*2411#+1, one of millions of prp's found by Paul Underwood
and Markus Frind in 2003 during an AP8 search.
p-2, p, p+2466 are consecutive primes.

p+2466 + p = 2*3970433188*2411#+2468 = 2^2*617*117541*125887*q,
where q is a 1023-digit prime.
p + p-2 = 2^3*992608297*2411#.

Phi(p+2466 + p)/Phi(p + p-2) = 6.94684...
Proof:

? p=3970433188*prod(i=1,primepi(2411),prime(i))+1;
? p1=nextprime(p-2); p2=nextprime(p1+1); p3=nextprime(p2+1);
? \\ p1, p2, p3 are now consecutive prp's
? print("p+"p1-p", p+"p2-p", p+"p3-p": ",\
eulerphi(p3+p2)/eulerphi(p2+p1)*1.0)
p+-2, p+0, p+2466: 6.94684253993807043

Computing p3 may take some time. The impatient can replace by p3=p+2466.
The factorizations of the prime sums are easy to handle for PARI/GP's
eulerphi.

PrimeForm found all prp's and proved p and p-2.
Marcel Martin's Primo proved p+2466 and q.

--
Jens Kruse Andersen
• ... Below is a similar case with the low value 0.14394..., close to 1/6.94684. ? p=5280767204*prod(i=1,primepi(2411),prime(i))-1; ? p1=nextprime(p-2466);
Message 2 of 4 , Jul 8, 2008
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I wrote:
> Let p = 3970433188*2411#+1
...
> Phi(p+2466 + p)/Phi(p + p-2) = 6.94684...

Below is a similar case with the low value 0.14394..., close to 1/6.94684.

? p=5280767204*prod(i=1,primepi(2411),prime(i))-1;
? p1=nextprime(p-2466); p2=nextprime(p1+1); p3=nextprime(p2+1);
? \\ p1, p2, p3 are now consecutive prp's
? print("p+"p1-p", p+"p2-p", p+"p3-p": ",\
eulerphi(p3+p2)/eulerphi(p2+p1)*1.0)
p+-2466, p+0, p+2: 0.143947882881142336

Here p = 5280767204*2411#-1.
p-2466, p, p+2 are consecutive primes with p+2 found by
Paul Underwood and Markus Frind in 2003.

p+2 + p = 2^3*19^2*3657041*2411#.
p + p-2466 = 2^3*167*q, where q is a 1034-digit prime.
PrimeForm proved p and p+2. Primo proved p-2466 and q.

--
Jens Kruse Andersen
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