## Re: [PrimeNumbers] Re: p^p +(p+1)^(p+1)

Expand Messages
• ... [snip] Everything becomes rather clearer if we don t restrict p to be prime. I studied the divisibiity properties of those forms extensively 3 years ago
Message 1 of 5 , May 11, 2005
In an email dated 10/5/2005 5:44:29 pm GMT Daylight time, "jbrennen" <jb@...> writes:

>--- In primenumbers@yahoogroups.com, Mark Underwood wrote:
>> PS on the positive side, no wonder I couldn't prove that 'result'! :)
>
>But you could prove that p^p+(p+1)^(p+1) is never divisible by
>2, 3, or 7... In general, numbers of this form seem to be
>overall less likely to be divisible by any particular prime.
>
>Here's a table showing the actual proportion of such numbers
>which are divisible by each prime q, along with the ratio of
>that proportion to 1/q:
>
[snip]

Everything becomes rather clearer if we don't restrict "p" to be prime.

I studied the divisibiity properties of those forms extensively 3 years ago but seem to have lost my notes :(

I think it's relatively easy to prove by Fermat's little theorem that the values of n divisible by (odd) primes q repeat with a cycle of length (q-1)*q.

Using pfgw -d I have just found (must go to work now...) that n^n+(n+1)^(n+1) is divisible by primes q as follows:-

q fraction
3 1/(2*3)
5 2/(4*5)
7 4/(6*7)
11 8/(10*11)

As your tables exemplify, there is no such pattern if "p" is constrained to be prime.

-Mike Oakes
Your message has been successfully submitted and would be delivered to recipients shortly.