Problems & Puzzles: Puzzles

Puzzle 924. Follow up to Puzzle 923

Here we will study a special case of two of the sequences studied in Puzzle 922:

b) 2*P-nextprime(P)

d) 2*P-previousprime(P)

In each case we will ask only for recursive sequence such that the produced terms are consecutive primes.

 

Example: For b) 2*P-nextprime(P)

 

K=3: 263, 257, 251, because these are decreasing-consecutive and

 

263 is the starting prime
2*263-269=259
2*257-263=251

 

My largest sequence of this type is:

K=5: 121174931, ..., 121174811

 

Examples, for d) 2*P-previousprime(P):

 

K=4: 9843049, 9843079,9843109, 9843139 because these are increasing-consecutive and

 

9843049 is the starting prime

2*9843049-9843019=9843079

2*9843079-9843049=9843109

2*9843109-9843079=9843139

 

My largest sequence of this type is:

K=5: 121174841, ..., 121174961

 

Q. Can you send your largest sequence for each of these two defined above?

 

 

Jan van Delden wrote on May 2018:

The primes for both sequences of length k are part of a sequence of length k+1 and are called CPAP (consecutive primes in arithmetic progression).
We have 2p[n]-p[n+1]=p[n-1] or 2p[n]-p[n-1]=p[n+1], which can both be transformed to: p[n]-p[n-1]=p[n+1]-p[n], or the primegap is constant.

 

The difference in the two questions is the starting point in the CPAP. If one has a CPAP with length k equal to q[0],,q[k-1], than the answer to your first question is q[k-2] and to your second question q[1].

 

The largest known CPAP can be found at http://oeis.org/A033290 and is submitted by Jens Kruse Andersen and has length 10, with gap 210.
So from this list one can derive:

 

First question:      p=100996972469714247637786655587969840329509324689190041803603417758904341703348882159067229719+8*210

 

Second question: p=100996972469714247637786655587969840329509324689190041803603417758904341703348882159067229719+1*210

 

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