Problems & Puzzles: Problems

Problem 56.  p1*p2**pk = q1+q2++qk

JM Bergot made me notice the following problem:

Let's seek for solutions to:

p1*p2**pk = q1+q2++qk .......... (A)

where sets {p1-pk} & {q1-qk} are two sets of consecutive primes.

For k=odd solutions have been reported by Jim Ferry for K=3 to 25. 

Example. For k=3, the smallest solution is:

 29*31*37 = 11083 + 11087 + 11093

For k=even, due to parity restriction, p1 must be 2. Then solutions should be unique when they exist.

It seems that up to this moment no one single solution has been found for k=even

Q1. Can you find one solution to (A) for k=even?


Contributions came from Jan van Delden and J. K. Andersen.


van Delden wrote:

I computed the minimal absolute difference between the sum and the product-term until n=240 (where the product consists of 639 decimals). I also computed the average gap between the consecutive primes, q1..qn, defining the sum. There are just a few values of n where this difference is smaller than the average gap (let alone zero):

n=64; pn=311; difference=108 avg. gap=296.10
n=66; pn=317; difference=110 avg. gap=308.46
n=98; pn=521; difference=212 avg. gap=382.95
n=112;pn=613; difference=262 avg. gap=552.34
n=150;pn=863; difference=670 avg. gap=729.18
n=230;pn=1451;difference=688 avg. gap=1205.59

This suggests that finding a solution gets less and less probable with increasing n, although not improbable.


Andersen wrote:

I found no solutions in a search to k=200 using prp's for sums with k>40.
The sums could theoretically change if some of the prp's are composite.
A PARI/GP prp test with no known composite pseudoprime was used.

I expect around 0.08% chance of a solution existing.
This is based on the following heuristics.
p1*p2*...*pk = pk#.
We want k consecutive primes around pk#/k with sum pk#.
The average distance between primes near pk#/k is log(pk#/k).
For large k this is roughly log(pk#) which is known to be around pk.
Consider the largest sum below pk# of k consecutive primes.
The next sum replaces the first of those k primes by a prime k places
further along, so the sum is expected to be around k*pk larger.
If k is even then the sum is even and the expected chance of hitting
exactly the even number pk# is 2/(k*pk).
If k is odd then the sum is odd and the chance is 0.
The average expected chance over both odd and even k is 1/(k*pk).
The expected number of solutions for k = a to b can then be
approximated by the sum of 1/(k*pk) for k = a to b.
This sum is convergent when b is infinite so the expected number
of solutions is finite.
The sum from 201 to infinite is approximately 0.0008, so this is the
total number of solutions I expect, given the search limit of 200.
The expected chance of at least one solution becomes around 0.08%.



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