The first incident of p+n equals a prime after the previous n

[djoyce099]

When n =2,+2+2+2+...Prime gaps taken in order of n.

Starting with p=3
p+2 ----3\5
P+4 ---7\11
P+6 ---23\29
p+8 ---89\97
P+10 ---139\149
P+12 --- 99\211
P+14 ---293\307
P+16 ---1831\1847
p+18 ---1913\1931
p+20 ---3089\3109
p+22 ---3229\3251
p+24 ---4177\4201
p+26 ---5531\5557
..
In other words, gaps taken in the order of n
when n =2+2+2...for each iteration.
After p =14, when will p be a multiple of the next p?
p1/p =>2.

Dan

[Barry Schwarz]

On Wed, 22 Sep 2021 12:02:57 -0700 (PDT), djoyce099
<hlauk.h...@gmail.com> wrote:

>When n =2,+2+2+2+...Prime gaps taken in order of n.
>
>Starting with p=3
>p+2 ----3\5
>P+4 ---7\11
>P+6 ---23\29
>p+8 ---89\97
>P+10 ---139\149
>P+12 --- 99\211

You have a typo at p+12. It should read 199\211.

>P+14 ---293\307

I think you have your n's and p's confused. The first occurrence of
p+14 occurs at 113\127, not 293\307. If the title of your post had
read "after the previous p," then that would have been excluded since
113 is not after the previous p of 199 at p+12. But your title read
"after the previous n" and 14 is obviously after the previous n of 12.

>P+16 ---1831\1847
>p+18 ---1913\1931
>p+20 ---3089\3109
>p+22 ---3229\3251
>p+24 ---4177\4201
>p+26 ---5531\5557
>..
>In other words, gaps taken in the order of n
>when n =2+2+2...for each iteration.
>After p =14, when will p be a multiple of the next p?

What does this mean? p is not 14; n is. Did you mean p+14? Even
with that, p is a prime. No p will ever be a multiple of any other p.

>p1/p =>2.

I don't know what this is supposed to mean either. Is the => symbol
synonymous with the more common >= symbol? What are p1 and p? Does /
represent normal division?

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[djoyce099]

On Wednesday, September 22, 2021 at 8:34:35 PM UTC-4, Barry Schwarz wrote:
> On Wed, 22 Sep 2021 12:02:57 -0700 (PDT), djoyce099
> <hlauk.h...@gmail.com> wrote:
>
> >When n =2,+2+2+2+...Prime gaps taken in order of n.
> >
> >Starting with p=3
> >p+2 ----3\5
> >P+4 ---7\11
> >P+6 ---23\29
> >p+8 ---89\97
> >P+10 ---139\149
> >P+12 --- 99\211
> You have a typo at p+12. It should read 199\211.
>
> >P+14 ---293\307
>
> I think you have your n's and p's confused. The first occurrence of
> p+14 occurs at 113\127, not 293\307. If the title of your post had
> read "after the previous p," then that would have been excluded since
> 113 is not after the previous p of 199 at p+12. But your title read
> "after the previous n" and 14 is obviously after the previous n of 12.

> >P+14 ---293\307

>
> >p+18 ---1913\1931
> >p+20 ---3089\3109
> >p+22 ---3229\3251
> >p+24 ---4177\4201
> >p+26 ---5531\5557
> >..
> >In other words, gaps taken in the order of n
> >when n =2+2+2...for each iteration.
> >After p =14, when will p be a multiple of the next p?
> What does this mean? p is not 14; n is. Did you mean p+14? Even
> with that, p is a prime. No p will ever be a multiple of any other p.

No it is not.
P+14 ---293\307 clearly shows the two primes ((p)=293)\((p1)=307) and 14 is just n=14 gap.
(\) is used as a separator of the 2 primes
The next gap is n=16 ---- 1831\1847

> >p1/p =>2.
>
> I don't know what this is supposed to mean either. Is the => symbol
> synonymous with the more common >= symbol? What are p1 and p? Does /
> represent normal division?

(p1) The next prime >(p) p1/p =>2 where p1 is 2 or more times larger then p.

[Barry Schwarz]

On Thu, 23 Sep 2021 09:54:19 -0700 (PDT), djoyce099

What is not what? I asked what does "p=14" (that you wrote 4 lines
up) mean since p is supposed to be a prime.

>P+14 ---293\307 clearly shows the two primes ((p)=293)\((p1)=307) and 14 is just n=14 gap.
>(\) is used as a separator of the 2 primes

It is customary to place comments after the material the comment
applies to. I never asked about "\". I asked about the "/" in you
wrote in "p1/p" two lines down.

>The next gap is n=16 ---- 1831\1847
>> >p1/p =>2.
>>
>> I don't know what this is supposed to mean either. Is the => symbol
>> synonymous with the more common >= symbol? What are p1 and p? Does /
>> represent normal division?
>(p1) The next prime >(p) p1/p =>2 where p1 is 2 or more times larger then p.

Is this what you are asking?
For n_0=2, we have p_0 is prime and p_0+n_0 is also prime (as you
show).
Let n_1 = n_0+2 and we have a new p_1 and p_1+n_1 that are both
prime with the additional constraint that p_1 > p_0.
Let n_2 = n_1+2 ... (and so on)
We want to find the first p_k after 293 and p_(k+1) that comply
with these conditions plus the condition that p_(k+1) is more than
twice p_k.

For p+36 we get 12853\12889
and for p+38 we 30593\30631

p+44 (36389\36433) and p+46 (81463\81509) is the next occurrence.

There are no others within the first million primes and n <= 800

[djoyce099]

The next would be p+37? The p+37 where that p is twice or more larger than
the previous p in p+36

> p+44 (36389\36433) and p+46 (81463\81509) is the next occurrence.

The next would be p+45?

[Barry Schwarz]

On Fri, 24 Sep 2021 08:34:42 -0700 (PDT), djoyce099

<hlauk.h...@gmail.com> wrote:

>On Thursday, September 23, 2021 at 3:15:56 PM UTC-4, Barry Schwarz wrote:
>> On Thu, 23 Sep 2021 09:54:19 -0700 (PDT), djoyce099
>> <hlauk.h...@gmail.com> wrote:
>>
>> >On Wednesday, September 22, 2021 at 8:34:35 PM UTC-4, Barry Schwarz wrote:
>> >> On Wed, 22 Sep 2021 12:02:57 -0700 (PDT), djoyce099
>> >> <hlauk.h...@gmail.com> wrote:
>> >>
>> >> >When n =2,+2+2+2+...Prime gaps taken in order of n.
>> >> >
>> >> >Starting with p=3
>> >> >p+2 ----3\5
>> >> >P+4 ---7\11
>> >> >P+6 ---23\29
>> >> >p+8 ---89\97
>> >> >P+10 ---139\149

<snip>

>> Is this what you are asking?
>> For n_0=2, we have p_0 is prime and p_0+n_0 is also prime (as you
>> show).
>> Let n_1 = n_0+2 and we have a new p_1 and p_1+n_1 that are both
>> prime with the additional constraint that p_1 > p_0.
>> Let n_2 = n_1+2 ... (and so on)
>> We want to find the first p_k after 293 and p_(k+1) that comply
>> with these conditions plus the condition that p_(k+1) is more than
>> twice p_k.
>>
>> For p+36 we get 12853\12889
>> and for p+38 we 30593\30631
>
>The next would be p+37? The p+37 where that p is twice or more larger than
>the previous p in p+36

If p is a prime greater than 3, p+37 must be even and therefore cannot
be a prime. None of your examples included an odd n. Why did you
decide to add that wrinkle at this late date. If you are going to
allow non-primes, you are going to have to provide a detailed
explanation of how one qualifies (or fails to qualify) to be
considered.

>> p+44 (36389\36433) and p+46 (81463\81509) is the next occurrence.
>
>The next would be p+45?

Not under the existing rules. Did you forget to tell us about double
secret probation?