[seqfan] Re: Primes and prime remainders

[Daniel Joyce]

Hi Eric.

A few more terms --

157/23 = 19
197/29 = 23
277/31 = 29
401/37 = 31

On Sun, Oct 5, 2014 at 11:18 AM, Daniel Joyce <hlauk.h.bogart at gmail.com>
wrote:

> Hi Eric,
>
> Why not represent all sequential primes in the remainder.
> Also where the next divisor is a sequential prime.
>
> 5/3 = 2
> 13/5 = 3
> 19/7 = 5
> 29/11 = 7
> 37/13 = 11
> 47/17 = 13
> 131/19 = 17
>
> Dan
>
>
>
> On Sun, Oct 5, 2014 at 5:03 AM, Eric Angelini <Eric.Angelini at kntv.be>
> wrote:
>
>> Hello SeqFans,
>> Primes p(n) such that the remainder
>> of p(n)/p(n-1) is prime.
>> The seq P starts with p(1)=3 and is always
>> extended with the smallest possible
>> prime.
>>
>> P=3,5,7,17,19,41,43,89,...
>>
>> Example:
>> 5/3--> remainder 2
>> 7/5--> remainder 2
>> 17/7--> remainder 3
>> 19/17--> remainder 2
>> 41/19--> remainder 3
>> 43/41--> remainder 2
>> 89/43--> remainder 3
>> etc.
>> Hope I didn't mistake somewhere.
>> Best.
>> É.
>>
>>
>> Catapulté de mon aPhone
>>
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>>
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>>
>
>