Problems & Puzzles: Puzzles

Puzzle 985. Consecutive primes paired to consecutive powers of 2. JM Bergot sent the following nice puzzle   Find sequences of consecutive primes such that paired and added to consecutive powers of two sum each to a prime.   Carlos Rivera got this sequence of 8 terms:   1309757957 + 2 = 1309757959 1309757959 + 4 = 1309757963 ... 1309758067 + 256 = 1309758323 Q. Send a larger sequence.

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Contributions came from Giovanni Resta, Adam Stinchcombe, Jan van Delden; all from the week 11-17, January, 2020.

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Giovanni wrote:

9 terms from 814442926307,
10 terms from 516821428572467, and
11 terms from 1947120901401587.
 

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Adam wrote;

All I could so far is tie your 256 record with the prime 265517458637.

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Jan wrote:

Given a sequence of consecutive primes p[n] we have the condition that q[n]=p[n]+2^n must be prime, n>=1, as well.

If we study p[1],p[2],p[3] we may observe that we must have: p[2]=q[1], q[2]=p[3]. Given p[3] we also know where q[3] must be.
This sequence of primes: p[1],p[2],p[3],q[3] modulo 30 defines p[1]=17 mod 30.

What’s more, it is even true that we have p[4]=q[3]. If there is a prime p[4] between p[3] and q[3] it necessarily defines a prime q[4] that is divisible by 5.

I used the relations above to speed up my routine. However I was quite surprised by the results. I studied p[n+1]-p[n] and found that all solutions with length greater than or equal to 8 also have: p[5]=q[4]. Apparently the consecutive differences must be:

 

2,4,8,16

 

I didn’t study this, but if true it would speed up the routine.

 

Maybe some other contributor has an explication for this experimental fact. My first solution with length 9 is:

814442926307 - 814442926309
814442926309 - 814442926313
814442926313 - 814442926321
814442926321 - 814442926337
814442926337 - 814442926369
814442926369 - 814442926433
814442926373 - 814442926501
814442926387 - 814442926643
814442926397 - 814442926909

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