Problems & Puzzles: Puzzles

Puzzle 1180  A puzzle by V. F. Izquierdo V. F. Izquierdo sent the following puzzle: Find the 2 smallest consecutive primes, so that all the composite numbers between them have the same number of different prime factors, discounting multiplicities.   My table:   #Factors  1st Prime     2nd Prime Diff(2nd-1st) 1        3               5     2 2        5               7     2 3        29              31     2 4        419             421     2 5        2309         2311      2 6        43889        43891      2 7        1138829        1138831     2   Example: with 6 different factors: All composite numbers between 43889 and 43891 (only number 43890 is there) have 6 prime factors (2,3.5,7,11 and 19)   Only twin prime numbers appear in this table. Q1 Can there be other prime numbers that are not twin prime numbers? Q2. Expand the table.

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From 14-20 July, 2024 contributions came from Michael Branicky, Giorgos Kalogeropoulos, Paul Cleary, Adam Stinchcombe

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Michael wrote:

Restricting to twin primes, this has been studied in OEIS A075590, which records the

"Smallest number with n distinct prime divisors which is the average of a twin prime pair".

Using the b-file there, table entries (or their upper bounds) can be completed up to n = 50.
 

Any lower solutions would require at least k, k+1, and k+2 to share the same number of

distinct prime divisors (between consecutive primes that have a difference >= 4).

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Giorgos wrote:

Q1. Although there can be prime numbers that are not twin primes, it is highly improbable that the least number with n factors is not between twin primes.

For example if we choose not to include twin primes then the sequence would be

19 -> 2
643 -> 3
51427 -> 4
8083633 -> 5
 

We see that the primes get large very quickly and it is more probable to find a smaller twin prime.

I conjecture that all the terms will be twin primes
 

Q2. Next terms are

17160989 -> 8 prime factors (twin prime)
 

300690389 -> 9 prime factors (twin prime)
 

If my conjecture is true, then next term is

15651726089 -> 10 prime factors (twin prime) 

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Paul wrote:

Q1. I didn't find any up to the prime 15651728089.
 

Q2. 

No Factors First Prime         Second Prime

8.               17160989           17160991

9.               300690389         300690391

10.             15651726089    15651726091

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Adam wrote:

I obtain the following data (number of factors, first prime, last prime): (7,1138829, 1138831),(8,17160989, 17160991), (9,300690389, 300690391).  All first occurrences were twin primes.  I then required the gap to be larger than 2 and found (factors,first prime): (2,19),(3,643),(4,51427),(5,8083633).  All the gap sizes were equal to 4.  I figured it was a probability concept: if you have the probability that a number in a given range has exactly k prime divisors is p, your guess for the probability of 3 consecutive such numbers would be p^3.  So twin primes should happen first (being easiest), gaps of 4 next (harder), then 6 (even harder), etc.  I tested the probability hypothesis and found it failed.  For consecutive numbers with exactly 3 primes divisors, the gap size of 10 occurs before the gap size of 8.  Data (gap size, ending prime, start prime) were [10, 91823, 91813] & [8, 278119, 278111].

 

On a lark, I went looking for 17 prime divisors and looked for twin primes on both sides of such numbers and found 5171596727996514846030 which is the product of the primes {2,3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 53, 59, 71, 73}, not guaranteed to be smallest.

 

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