Problems & Puzzles: Puzzles

 Puzzle 761. On certain consecutive primes p & q Vic Bold sent the following nice puzzle:   Here we ask for chain of pairs of consecutive primes p & q such that p+q hits the center of another pair of consecutive primes and so on, for k steps.   Example: p=13 & q=17 makes k=4:   13+17=(29+31)/2 29+31=(59+61)/2 (59+61)=(113+127)/2 113+127=(239-241)/2 Carlos Rivera's preliminary search found an example for k=7 steps starting from the pair of consecutive primes 975027887 & 975027913.   Q. Send your largest k example.

Contributions came from J. K. Andersen and Giovanni Resta.

BTW, I must confess that it seems that this puzzle likes very much to me because I have posted similar puzzles several times before. See Puzzle 119 and Puzzle 260.

But this time the pair of primes required in each step of the sequence need not to be twin primes.

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Jens wrote:

Jaroslaw Wroblewski found the longest known BiTwin chain in 2008:
1873321386459914635*13# * 2^n +/- 1, n=1,...,9

This gives a chain with k=8 for starting primes
1873321386459914635*13# * 2^1 +/- 1 = 112511682470782472978100 +/- 1.
It doesn't extend to a longer chain for our problem.

He also found a BiTwin chain which is one shorter:
53780914344409189*13#*2^n +/-1, n=0,...,7 This one does extend to give us another k=8. The last primes are not twins.

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Giovanni wrote:

The smallest pair of consecutive primes for k=8
is (448866202747, 448866202793).

The smallest for k=9 is (891737206043, 891737206117).
In this case the last consecutive pair in the chain
is (456569449512949, 456569449512971).

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