Problems & Puzzles: Puzzles

Puzzle 423. K Consecutive even numbers such that... JM Bergot sent one more nice puzzle: Find K consecutive even numbers, E, such that for each E, the sum of its prime factors, S, is prime and (E-S) is prime too. Bergot found the smallest pair K=2:{10 & 12} C. Rivera found the smallest such K-tuples for K=3 & 4: K=3:{453436, 453438, 453440} K=4:{657488682, 657488684, 657488686, 657488688} Q1. Find the smallest K-tuplet, for K>4.

---

Farideh Firoozbakht wrote:

I think we aren't able to find the smallest k-tuplet for a k, k>4 even for k=5.

A 5-tuplet occurs when we find two consecutive 4-tuplets such that the difference between the first numbers of them is 2 but the next two 4-tuplet are {1991428338, 1991428340,1991428342,1991428344} & {2965167404, 2965167406, 2965167408, 2965167410} so according to your solution for k=4 and my findings the first two terms of the sequence of such differences are 1991428338-657488682=1333939656 &  2965167404-1991428338=973739066.  Hence it seems that the difference d=2 occurs for large numbers that we aren't able to find them.

***

J. K. Andersen wrote (May 08):

Puzzle 423

The smallest 5-tuplet is
K=5:{569255532972, 569255532974, 569255532976, 569255532978, 569255532980}

It is the only below 10^12. In a 4-tuplet there are 8 numbers which must be prime. In order to extend it to a 5-tuplet there are only 2 additional numbers which must also be prime. I counted 210 4-tuplets below 10^12, and expected 1 or 2 5-tuplets.

***

---