[seqfan] Re: oddness of iterations of sigma()

[Max Alekseyev]

Vladimir,

Your identity is correct.
In fact, for odd p >= 3,
(p^2 + (p-1)/2)^2 < 1+p+p^2+p^3+p^4 <= (p^2 + (p+1)/2)^2,
implying that
ceil(sqrt(1+p+p^2+p^3+p^4)) = p^2 + (p+1)/2
and
ceil(sqrt(1+p+p^2+p^3+p^4))^2 = (p^2 + (p+1)/2)^2 = p+p^2+p^3+p^4+((p-1)/2)^2.

Regards,
Max

On Thu, Nov 14, 2013 at 6:26 AM, Vladimir Shevelev <shevelev at bgu.ac.il> wrote:
> I observated a possible identity: if p is odd (not necessarily is prime), then
>  p+p^2+p^3+p^4+((p-1)/2)^2=
> (ceil(sqrt(1+p+p^2+p^3+p^4)))^2.
> I say "possible", since yet I have not proved it, but
> by handy I verified it for many p, including , e.g., p=203.
> If , indeed, it is identity, than, at least, for even s the pairs (r=4, arbitrary primes p>3) are impossible.
>
> Regards,
> Vladimir
>
>
> ----- Original Message -----
> From: Donovan Johnson <donovan.johnson at yahoo.com>
> Date: Wednesday, November 13, 2013 17:20
> Subject: [seqfan]  oddness of iterations of sigma()
> To: Seqfan <seqfan at list.seqfan.eu>
>
>> Vladimir Letsko asked:
>> 3. Note that sigma(3^4) = 11^2. Does there exists another pair (p,r)
>> such that p is prime, r > 1 and sigma(p^r) = q^s where q is
>> prime and
>> s > 1?
>>
>>
>> Regarding question #3:
>> sigma(3^4) = 11^2 is the only solution for p^r < 10^16.
>>
>> Donovan
>>
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>
>  Shevelev Vladimir‎
>
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