Talk:Successive prime differences

Different results for 6,4,2

I am getting slightly different results for the last group than the Python example and am finding it hard to see where I may be wrong. I am finding 337 groups. Am I just mising something? --Thundergnat (talk) 01:20, 27 April 2019 (UTC)

Doh! It's me, I glossed over the 'successive' part of the problem. Update to follow. --Thundergnat (talk) 01:25, 27 April 2019 (UTC)

It's my birthday soon so I googled my age and found:

• It's a prime.
• It's a twin prime.

I searched Rosetta Code and found that there was no twin prime task! (I had expected that someone would have already started it). I resolved to wait until closer to my birthday then put up a twin primes task and left it at that.

A few days later I started to think of what a generalisation around the idea of twin primes would be and hit on a difference; then multiple differences; then really liked how my solution to generating a sliding group of <count> items from a list actually did come from the Python fundamentals:

<lang python>zip(*(lst[n:] for n in range(count)))</lang>

I finished the code and played with the differences then firmed up what the task details would become. I wrote the task and added extra explanations and emphasis to try and help the reader grasp the details, then went to bed.

Today I've just done a search of the primes generated from differences of `2, 4` on OEIS to find that it is known to some degree, but expressed differently and not as generally as here - I guess recreational maths peeps think alike :-)

Enjoy. --Paddy3118 (talk) 06:54, 27 April 2019 (UTC)

It is very well studied, but you must state it slightly differently. Let P2 be the infinite sequence of successive primes (p2_a,P2_b) such that P2_b-P2_a=2. and P4 be the similar infinite sequence (P4_a,P4_b) such that P4_b-P4_a = 4. The your generalization to P2P4 as 3 successive primes with Pa,Pb,Pc with Pb-Pa=2 and Pc-Pb=4 is a search through P2 and P4 to find P2_b=P4_a. An interesting study would be to compute over a large range the length of P2 and P4 and thus predict the length of P2P4. For a given range should the length of P2P4 be the same as P4P2?--Nigel Galloway (talk) 13:28, 27 April 2019 (UTC)
I was going to add a twin prime task   (and cousin prime task, a difference of four),   but was somewhat preempted with addition of the   sexy prime   task   (a difference of six),   so I dithered a bit.   There are other named difference primes such as   devil   (also called   beast),   centennial,   and   millennial   primes.   However, having a Rosetta Code task just for twin primes would make the code a   lot   cleaner and simpler,   not to mention faster.   This would've made the task solutions more easier to compare   (and I think more useful for people who wanted a clean and robust code for just concerning the generation of twin primes).   Plus it would be easier to find when people are looking for a simple twin prime generator.     -- Gerard Schildberger (talk) 21:25, 27 April 2019 (UTC)
I googled my age, and found it to be   highly totient.     And I'll never be   highly totient.   again.   Sigh.     -- Gerard Schildberger (talk) 21:38, 27 April 2019 (UTC)
I enjoyed the google to work that out :-)
--Paddy3118 (talk) 06:06, 28 April 2019 (UTC)

Is the first solution for 6,4,2 not 7,13,17,19?--CalmoSoft (talk) 10:40, 7 September 2021 (UTC)

No, because the next prime after 7 is 11. --Tigerofdarkness (talk) 11:30, 7 September 2021 (UTC)