[seqfan] Re: Searched further in A176245 Primes of the form A174881(k)-1 or A174881(k)+1

[Jonathan Post]

Dear Farideh Firoozbakht,

Thank you for your comments.  I realized the factorization A174881(k)
- 1 after I'd submitted, having suspected beforehand but not done the
elementary algebra.  I also strongly suspected that k must be of the
form 2^n, but still checked every value with Alpertron through k = 90.
And then asked for help.

I hope that your comments are added to the seq, when it is corrected.

I also hope to see more of you on Prime Curios!

On Sun, Apr 18, 2010 at 6:29 AM, Farideh Firoozbakht
<f.firoozbakht at sci.ui.ac.ir> wrote:
> Dear Dr. Post,
>
> There is no prime of the form A174881(k)-1. Because k(k+1)-1 divides
> A174881(k)-1=(k(k+1))^k-1.
> So I think definition of the sequence A176245 should be changed.
>
> Also if A174881(k)+1 is prime then k must be of the form 2^n.
>
> Between the first 14 such integers k, only for k = 2^0, 2^1 & 2^2
> A174881(k)+1 is prime.
> So next term of the sequence (if it exists) is greater than
> ((2^14)*(2^14+1))^(2^14). Namely it has more than 138099 digits.
>
> Note that (40*(41))^8+1 divides ((40*(41+1))^40+1 and ((90*(90+1))^2+1
> divides ((90*(90+1))^90+1, so both numbers ((40*(41+1))^40+1 and
> ((90*(90+1))^90+1 are composite.
>
>
> Farideh
>
>
> Quoting Jonathan Post <jvospost3 at gmail.com>:
>
>> No more than the 3 prime values shown through ((90*(90+1))^90) + 1.
>> Factorization gets slow for these, exceeding 200 digits.
>>
>>
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