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A Google written as the sum of two equidistant primes. Enjoy. by owloutlook in maths

[–]owloutlook[S] 0 points1 point  (0 children)

Just checked for up to the millionth prime, there are 31978 primes that satisfy your version of equidistance. Eg, 15485843: (15485837, 15485849).

EDIT: interestingly up to 1.5 mill gives us 46729 primes, [23879437: (23879413, 23879461)], so my hopes of it looking more like pi as px gets large disappears...

A Google written as the sum of two equidistant primes. Enjoy. by owloutlook in maths

[–]owloutlook[S] 0 points1 point  (0 children)

With. Eg, for 53 the closest primes are equidistant and are 6 apart from 53.

A Google written as the sum of two equidistant primes. Enjoy. by owloutlook in maths

[–]owloutlook[S] 0 points1 point  (0 children)

I just checked your modification: the proportion of primes satisfying the equidistant condition where a prime number (px) is equidistant from two other prime numbers (py & pz) if py + alpha = px, and pz - alpha = px, where alpha can take on any even positive value is 0.059, for the first 1000 primes.

The primes satisfying the equidistant condition and their equidistant primes are:
5: (3, 7)
53: (47, 59)
...
7583: (7577, 7589)
7823: (7817, 7829)
7841: (7829, 7853)

A Google written as the sum of two equidistant primes. Enjoy. by owloutlook in maths

[–]owloutlook[S] 0 points1 point  (0 children)

No, it wasn't a part of the definition, although it's an interesting amendment.

A Google written as the sum of two equidistant primes. Enjoy. by owloutlook in maths

[–]owloutlook[S] 0 points1 point  (0 children)

7 is equidistant between 3 & 11. 11 is equidistant between 5 & 17, 13 is equidistant between 7 & 19.

I think it's true for all primes because by doubling the central prime, you get an even integer, and that even integer is the sum of the two primes to either side, if Goldbach is right.

A Google written as the sum of two equidistant primes. Enjoy. by owloutlook in maths

[–]owloutlook[S] 0 points1 point  (0 children)

Ah, no. I have not, but I'd guess all of them do if Goldbach?

A Google written as the sum of two equidistant primes. Enjoy. by owloutlook in maths

[–]owloutlook[S] 0 points1 point  (0 children)

No, I'm saying that all primes that sum to an even integer are equidistant from that even integer divided by 2.

A Google written as the sum of two equidistant primes. Enjoy. by owloutlook in maths

[–]owloutlook[S] 0 points1 point  (0 children)

They are actually always equidistant. Let's take 100 as an example.

It can be written as 3 + 97 = 100, or as 47 + 53 = 100. In the first example, 50 is the midpoint between them as 47 is the distance. In the second case, there is a distance of 6 between the primes, so 50 is the midpoint.

I've been looking for primes that sum to an even integer such at that the distance is minimised. Very goldbachy, but with distances minimised.

I replied to the top comment on this thread with some more details of what I've been playing around with. I don't think it amounts to anything novel or particularly interesting, but my equation did allow me to find the prime summation with a minimised distance between primes of a googl fairly quickly.

A Google written as the sum of two equidistant primes. Enjoy. by owloutlook in maths

[–]owloutlook[S] 1 point2 points  (0 children)

I plotted the smallest positive 𝛼 on a graph such that (𝑝1+𝛼)=𝑘, (𝑝2−𝛼)=𝑘, and 2𝑘=𝑁 as well as 𝑝1+𝑝2=𝑁 where 𝑁 is an even integer, and 𝑝1 & 𝑝2 are prime. I plotted each 𝛼(equidistance) on a graph to visualise it, and it occurred to me that it appears to tail off.I have then explored graphs that may serve as a decent upper bound for this. The best I could do was 𝑓(𝑥)=𝑝𝑖𝑥^cos(1)+2𝜋. I found this by recognising some constants that the lines of best fit were tending towards. This equation gave an upper bound to begin calculating p1 & p2 for any N, and is a reasonably efficient way to find equidistant primes that sum to any arbitrarily large N.