Problems & Puzzles: Puzzles

Puzzle 1062 Consecutive numbers and the quantity of distinct prime factors-II Metin Sariyar sent the following nice puzzle: The number of distinct prime divisors of numbers from 103 to the next prime 107 are respectively; 1,2,3,2,1. What is the smallest prime with this property with number of distinct prime divisors from p(k) to the next prime p(k+1) :1,2,3,4,3,2,1  The first two terms of this sequence are; 5, 103,..?  Example:  1) 5 has 1, 6 has 2 and 7 has 1 distinct prime divisors: 1-2-1 2) the distinct prime divisors of 103, 104, 105, 106, 107 are respectively: 1,2,3,2,1 3) ?  (With 1,2,3,4,3,2,1) Q. I ask the next term.

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During the week 6-12 Nov. 2021, contributions came from Shyam Sunder Gupta, Giorgos Kalogeropoulos, Oscar Volpatti, Emmanel Vantieghem

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Shyam wrote:

There is no set up to 10^10 such that the number of distinct prime divisors of seven consecutivenumbers are respectively; 1,2,3,4,3,2,1. However there are number of sets such that the number ofdistinct prime divisors of five consecutive numbers is respectively; 1, 2, 3, 2, 1.

For example numbers starting from 103,163,193,313,397,463,1093,1279,1303,2473,5413,5503,5923,7603,7723 etc.

There are number of sets such that the number of distinct prime divisors of seven consecutive numbers is respectively; 1, 3, 4, 5, 4, 3, 1. The starting number of first three such sets is 908597, 1856027 and 2690377.

On investigating numbers up to 10^10, it is found that there are sets such that the number of distinct prime divisors of nine consecutive numbers is respectively; 1, 3, 4, 5, 6, 5, 4, 3, 1. The starting numbers of first four such sets are 5195681741, 7629752531, 8822273741 and 9544602581.

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Giorgos wrote:

The prime p can be either of the form 6n+1 or 6n-1.

If it is 6n+1 then we ask for 6n+6 to have 2 distinct prime factors. This means that 6n+6 = 2^x*3^y for x,y>0.

If it is 6n-1 then we ask for 6n to have 2 distinct prime factors. This means that 6n = 2^x*3^y.

So, in both cases we only need to search numbers of the form q = 2^x*3^y and see if omega of

{q-5,q-4,q-3,q-2,q-1,q,q+1} = {1,2,3,4,3,2,1} and for the second case {q-1,q,q+1,q+2,q+3,q+4,q+5} = {1,2,3,4,3,2,1}.

A quick search didn't return any results. I don't think such a number exist but I cannot prove it is impossible.

 

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Oscar wrote:

I didn't find neither third term nor fourth term so far.

The fifth term is prime 42059287693, with pattern 1,2,3,4,5,6,5,4,3,2,1 up to next prime 42059287703

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Emmanuel wrote:

Let  F(n)  denote the number of different prime factors of  n.
 

Suppose there exist a number  n  such that

     F(n) = 1, F(n+1) = 2, F(n+2) = 3, F(n+3) = 4, F(n+4) = 3, F(n+5) = 2, F(n+6) =1    (***).

Then it is clear that one of the numbers  n+1, n+3  or  n+5  is divisible by  6.
Obviously it cannot be  n+3.  So, it should be  n+1  or  n+5.

But since  F(n+1) = F(n+5) = 2, either  n+1  or  n+5  should be on the form  (2^u)*(3^v)  with  u, v > 0.
 

I was able to check if the conditions  (***)  are fulfilled for  n  < 10^60  and of the form  (2^u)*(3^v) - 1  or  -5.

No such configuration was found.
 

Bigger values faced me with heavy factorizations, so I stopped the computations.

 

But I think it is not impossible to find a solution.

 

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