Problems & Puzzles: Puzzles

Puzzle 240.  Consecutive  numbers and consecutive prime factors

Here we will ask for pairs of consecutive numbers (n, n+1) such that each is composed only by consecutive prime factors.

As a matter of fact I have found four examples:


Example: 4199=13*17*19, 4200=2^3*3*5^2*7


1. Can you find the next 3 examples?

2. Can you find a moderate large example, let's say 20 digits?

3. Can you find a pair of these where the primes involved are not repeated in each number?


Jon Wharf, Adam Stinchcombe, Enoch Haga, Faride Firoozbakht, Eric Brier and J. K. Andersen solved Q1.

Wharf, Firoozbakht, Brier and Andersen observed the relation of these kind of asked pairs and the twin primes as a basis to construct a smart approach in order to get larger examples and, consequently, solved Q2 by far.

Nobody has found an example for the case Q3, so probably there is not any one.


Q1: Here are the first pairs below 100000, as reported by several of the solvers of this puzzle.

(35, 36)
(143, 144)
(323, 324)
(384, 385)*
(539, 540)*
(899, 900)
(2430, 2431)*
(3599, 3600)
(4199, 4200)*
(4374, 4375)*
(5183, 5184)
(11663, 11664)
(22499, 22500)
(32399, 32400)
(36863, 36864)
(57599, 57600)
(72899, 72900)

* given as examples in this puzzle
** found by the solvers

Q2: What is the relation on the pairs (n, n+1) and the twin primes, mentioned above?

Let p and p+2 twin primes; so if n=p(p+2) then n+1=(p+1)^2. From start n is a product of consecutive primes (because p & p+2 are twin and necessarily consecutive primes), so all we need is that p+1 (an even and composite quantity) can be factorized as a product of consecutive primes.

While not all the known solutions are of this type (all pairs in the list above marked with an asterisk are not of this type), this model provides a basis in order to hunt systematically the asked (n , n+1) pairs of large size for this puzzle

Before presenting the large solutions gotten I would like to say that Wharf and Andersen neatly noticed that after (4374, 4375) they are all of the 'twin-prime + square' type (Why?... See at the bottom the new questions emerging of this)

Large solutions: In order to get large solutions you should use the twin-prime + square model. In order to guarantee that (p+1) is a product of consecutive primes you have several options as: p=k*q# -1 or p=k-q! - 1, and so on, all related with the 'minus one" primorial or factorial functions, using a k multiplier appropriate, that is to say a multiplier k having its largest prime factor less that q.

J. K. Andersen sent the largest one (5138 digits)

I made my own search with PrimeForm/GW which found and proved the 2569-digit twins 2379752375*6000#+/-1. 6000# is the product of all primes below 6000, and 2379752375 = 5^3*7^2*11^2*13^2*19. The solution n=x^2-1 has 5138 digits.

Others sent by Andersen (I checked Chris Caldwell's prime database and found two large twin primes x+/-1 satisfying the requirement for x) were:

2846!4+/-1 with 2151 digits was found in 1992 by Harvey Dubner. 2846!4 is the multifactorial 2846*2842*2838*...*2 which contains all prime factors <= 1423

5775*2^5907+/-1 with 1782 digits was found in 1998 by Bradford Brown with Yves Gallot's Proth.exe. 5775 = 3*5^2*7*11 which is fine for our purpose.

Faride sent the following one:

2*3*5^995*7*11*...*541*547+/-1. digits=1835

Jon Wharf sent this one:

Titanic pair: ( (950*1193#)^2-1, (950*1193#)^2 ). Proved 950*1193# - 1 and 950*1193# + 1 with PrimeForm.

Adam Stinchcombe wrote:

I just wanted to add that I found some 200+ digit examples:

n=2^2*3^21*5^19*7^22*11^19*13^20*17^19, n^2 215 digits
n=2^8*3^11*5^20*7^22*(11*13817)^19, n^2 218 digits

Q3. In order to get this target you should use the twin-pair + square model together with the primorial minus one option. This is what two solvers wrote:


Assume n & n+1 be product of consecutive primes and be squarefree ,since one of them is even ,thus n or n+1 must be of the form p#.

If n be even then n=p# and n+1=p#+1 ,but for n < 163, p#+1 isn't product of consecutive primes (I checked ).

If n is odd then n+1=p# and n=p#-1 ,also for n < 163 , p#-1 isn't product of consecutive primes.

Thus if such n exist n must be greater than 163#-2.163#-2= 5766152219975951659023630035336134306565384015606066319856068808.

I think there is no such n.


The even number of n,n+1 must be p# for some prime p.

Then we need p#-1 or p#+1 to have consecutive distinct prime factors. It seems unlikely to me and a computer search to 3803# with 1624 digits gave no solutions.

My search used two methods to eliminate m=p#-1 or m=p#+1:

1) Trial factoring first searched for a small factor. If the smallest prime factor of m is found then the following prime must also be a factor to give us a chance. This never happened.

2) If m is assumed to be the product of k consecutive primes then the next prime after the k'th root of m must divide m to give us a chance. This was tried for k=2,3,... until the root became below the trial factor limit from 1). A factor of m was never found. The gmp library was used for prp testing to find the next prime. The prime/prp was not proved so very theoretically a factor could have been missed.


Rising questions

  • Is (4374, 4375)** the last of a finite set of solutions NOT of the twin-pair + square type?
  • Can you give a reason for this or find more examples?
  • In particular can you find one more example as (4199, 4200) where n=4199 is a product of three consecutive primes?




Records   |  Conjectures  |  Problems  |  Puzzles