Problems & Puzzles: Puzzles

Puzzle 619. Odd midpoints JM Bergot sent the following puzzle. The four consecutive primes 199, 211, 223, and 227 have midpoints 205, 217, 225, all odd numbers. Q.  Send your largest group of consecutive primes having all odd midpoints?

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Contributions cane from J.K. Andersen, Torbjörn Alm, Emmanuel Vantieghem, Hakan Summakoğlu

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J. K. Andersen wrote:

Consecutive primes having all odd midpoints is the same as consecutive
primes having the same residue modulo 4. Puzzle 539 is about this.
The longest known sequence is at
http://users.cybercity.dk/~dsl522332/math/congruent-primes.htm#mod4
It is 33 primes starting with 3278744415797.

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Emmanuel wrote:

A few minutes ago, I reallized that puzzle 619 is puzzle 539 without the condition that the length  n  of the sequence must be greater than  log(smallest prime).

Indeed if two consecutive primes p, q  have odd midpoint, it follows that  p+q = 2 (mod 4).  And this is possible only if  p  and  q  are in the same residu class modulo 4.

So, I think it's allmost impossible for me to do better than J. K. Andersen and M. Hassler.

 

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Torbjörn wrote:

This is the smallest Titanic solution

10^1000 + 46227, +46339, +52423, +55831

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Hakan wrote:

I found a group of 24 consecutive primes.All of these primes of form 4k+1.

1113443017, 1113443029, 1113443053, 1113443069, 1113443077, 1113443137, 1113443141, 1113443153, 1113443173, 1113443189, 1113443209, 1113443249, 1113443269, 1113443293, 1113443329, 1113443341, 1113443369, 1113443381, 1113443389, 1113443473, 1113443521, 1113443533, 1113443549, 1113443561.