Problems & Puzzles: Puzzles

Puzzle 334. Farideh & the 2004 year.

Farideh Firoozbakht proposed* the following puzzle: 

2004 =2^2*3*167 and the three numbers 2*2004-1,3*2004-1& 167*2004-1 are prime numbers.

Q1. Find a large number m with the above property,i.e., if p is a prime divisor of m then m*p-1 is prime. (I found  a 100-digit such number.)

We define a(n) as the smallest number m such that m has n  distinct prime divisors and m has the above property. The first 7 terms of {a(n)} are: 2,6,30,420,32550,410970,55137810

Q2. Can you find more terms of this sequence ?

Similar property: n=295855560 ,n=2^3*3^2*5*7*11*13*821 and k*295855560+1 for k=2,3,5,7,11,13& 821 are prime numbers.

Q3. Find a large number like n (number of distinct prime divisors be arbitrary).

We define similarly b(n) as the smallest number m such that m has n distinct prime divisors and if p is a prime divisor of m then m*p+1 is prime. The first 7 terms of the sequence {b(n)} are: 2,6,354,210,43860,463980,189583590

Q4. Can you find more terms of this sequence?

(on February, 2004, so sorry for the delay because I'm publishing this on October 2005!)

Contributions came from Anton Vrba, J. K. Andersen, Fred Schneider and Farideh Firoozbakht:

Anton wrote:

Q1 : (smile) m=2^25964950 immediately comes to mind, which has the one prime divisor p=2 and p*m-1 has 7816230 digits and is the largest known prime number credited to Dr Martin Novak and GIMPS
Q2 : The 8th term could be a(8)=1109*31*17*11*7*5*3*2 = 1350063330. This is the smallest number that is square free. I have checked for non square free numbers with factors 2^i i=2..6, 3^j=2..4, 5^2 and 5^3, 7^2 and 7^3, 11^2 and their combinations leading me to believe that 1350063330 is the next term of the series
Q4 : The 8th term is b(8)=577*503*19*11*7*5*3*2=12738238590 such the p*m+1 is prime. Again I have checked for non square free numbers smaller than b(8).

J. K. Andersen wrote:

With 1 prime factor: m = 2^25964950.
2*m-1 = 2^25964951-1 is the largest known prime. It has 7816230 digits.
This method works for all Mersenne primes 2^p-1 with m = 2^(p-1).

With 2 prime factors: m = 4830107*2^3321.
2*m-1 and 4830107*m-1 are both titanic primes, found and proved with PrimeForm/Gw.

Fred wrote:

For puzzle #334's question #2, I could find the minimum answer for n=8. It's 1350063330=2*3*5*7*11*17*31*1109

Farideh wrote:

I have found a(8), a(9), b(8) & b(9). b(8) is not 12738238590 (given by Anton).


On March 2, 2023, Michael Branicky wrote:

I have found a(10) = 9396949341780 and a(11) = 6805591029957720, and also b(10) = 60958204112190 and b(11) = 3337686588286050


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