Problems & Puzzles: Puzzles

Puzzle 1061 Consecutive numbers and the quantity of distinct prime factors Metin Sariyar sent the following nice puzzle: Find the smallest number k such that number of distinct prime divisors of exactly n numbers from k to k+n-1 increase by 1, while it decreases by 1 for their reversals. The first 3 terms, for n=2, 3 & 4, I found are: 9, 13683, 1697257, ...?   n k because 2 9 9 , 10 : 1, 2 9, 01: 1, 0 3 13683 13683, 13684, 13685: 2, 3, 4 38631, 48631, 58631: 3, 2, 1 4 1697257 1697257, 1697258, 1697259, 1697260: 1, 2, 3, 4 7527961, 8527961, 9527961, 0627961: 4, 3, 2, 1 5 ? ?

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During the week1-5 November, 2021, contribution came from Oscar Volpatti

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Oscar wrote:

for n=5 the smallest number is k=8570100457.

 

I searched up to 3*10^10 only finding some more sequences with n=5 for the following k values:

 

10745845567

11076970173

11880356581

12076586761

14153664781

14448518431

16078740037

16079048371

17228392641

17522170381

18689539312

18709379881

19178015062

20428193041

21642933841

21654820741

22165958161

23889820327

29735850861

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