- --- Jon Perry <perry@...> wrote:
> It was actually a fish for some voluntary labour. Pari chokes on

Use 'calc' instead. Or bc. Or the other 'calc'. Or use gp and use

> primes>1000000, and I have no other means of performing such high powers

> tests that would be required, let alone access to a modern machine.

'p=nextprime(p+1)' rather than 'forprime(p='.

> I did check it over a range of values, and my bounds were deduced from these

One of (p-1) and (p+1) is divisible by 2,

> tests. As phi(n)/n has no limit, I would assume this doesn't either, but I

> was surprised by the narrow region of results.

the other is divisible by 4 or 2^i i>2

One of (p-1) and (p+1) is divisible by 3.

The 2s combine such that

Phi((p-1)*(p+1)) = Phi(2.2^i.(p-1)/2.(p+1)/2^i)

= 2^i.Phi((p-1)(p+1)/2^(i+1))

The factor of three gives you a 2/3 factor.

Therefore the highest value you will find will be

1/2*2/3 = 1/3 from p=3,5,17

and 1/3-eps from numbers with a few prime factors larger than 2 or 3 in p+1

and p-1.

e.g.

499637 0.3333266618578673045764089881

(23:01) gp > factor(499637-1)

%3 =

[2 2]

[124909 1]

(23:02) gp > factor(499637+1)

%4 =

[2 1]

[3 1]

[83273 1]

Note that by HL it will reach 1/3-eps infinitely often

Lower bound - anyone care for a stab? There should be some bound somewhere,

I'm sure.

Phil

=====

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http://mailplus.yahoo.com - '£10 to Oxfam for each statement you persuade me to retract. I trust you'll

reciprocate?'

What... Oxfam will pay me £10 to persuade you to retract them?

I believe they are both correct, hence I will not allow Oxfam to waste their

money on me, and this in turn leads me to believe that f(p) could have an

average value.

Jon Perry

perry@...

http://www.users.globalnet.co.uk/~perry/maths/

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