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.

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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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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