Problems & Puzzles: Problems

Problem 51.  A kind of prime fractions

Jan van Delden raised to my attention a problem posed by a friend of him Mr. Arnout Jaspers, problem that has been published in the Dutch magazine Natuur Wetenschap en Techniek, page 79. Jasper has authorized to bring now his problem here.

The Delden version from the Jaspers original question asks for solutions to this kind of fractions, 1/Π(pi) = Σ (si/pi), where pi are distinct primes and si=+1 or -1, only.

Example: 1/(3*5*7*53*107)=1/3-1/5-1/7+1/53-1/107, n=5

Jan van Delden claims that:

For n=2 the sign pattern (+1,-1) will give only solutions for p[1]=2

For n=3 the sign pattern (+1,-1,-1) and (-1,+1,+1) will give solutions for p[1]=2; the latter is the only one for p[1]>2

For n=4 the sign pattern (+1,-1,-1,+1), (-1,+1,+1,-1), (-1,+1,+1,+1), (+1,-1,-1,-1) will give only solutions for p[1]=2

While in the puzzle already published several structured questions are posed, here I will ask one single question.

Q. Find all solutions ( I will organize them here by n value, by author & date, as they keep coming)


Delden wrote (30 Set. 07):

Some of my results.
There are no solutions with p[1]>2 for n even.
If n even the sign pattern can be split up in 2-terms of the form (+1,+1), (+1,-1), (-1,-1), or (-1,+1).
After forming a combined fraction for each of these 2-terms, the numerator will be even. Combining all  these 2-terms will therefore have an even numerator, which can't be 1.
The only 2-term is [2, -3] (where I included the sign in the solution).
If n=3 the number of solutions with fixed p[1]=p are finite, in fact if p>2 the upperbound is pi(2p)-pi(p) and if p=2 there are 2 solutions.
If p[1]=2 then we have the only solutions [2,-3,-7] and [-2,3,5].
A 3-term  with sign pattern (-1,+1,+1) with (p,q,r) as denominator with p<q<r will have r=(pq-1)/(q-p). Solving r>q gives q<p+sqrt(p^2-1)<2p. So (p,q) defines r and r>q fixes the number of q allowed. Since p<q<2p the upperbound is as given if p>2. If p=2 the only possible solution is q=3, this gives r=5. (The sign pattern (+1,-1,-1), only solutions for p=2, gives q=3, r=7 in a similar manner).
Furthermore r/p is uniform decreasing as a function of q, so the maximum for r is attained for the smallest q possible, so for p=2 we get r/p=2.5. If p>2 then we are looking for twin primes to achieve a maximum for r (resulting in r=p+(p-1)(p+1)/2), hence the maximum value of r will be O(p^2/2).
There's a simple way to eliminate a lot of sign patterns from the 2^n possible patterns.
Assume that the p[i] are in ascending order: p[i]<p[j] if i<j.
The patterns (+1,+1) and (+1,-1) will result in a + contribution to the total sum.
The patterns (-1,-1) and (-1,+1) will result in a - contribution to the total sum.
[In fact the signs used can be further apart then 1 position]
Assume our total sum is split in two parts (and we have n>2), the first part being a/b the second c/d after combining the unit fractions.
Then we have ad+bc=1, since b and d are (both) a product of primes these terms are positive. We get necessarily that sign(a) and sign(b) are opposite.
[If they are both positive then ad+bc>1, both negative then ad+bc<0].
The above rule should hold up using all possible splits for any pattern.
So for instance (n=3):
(+1,+1,+1) can't be split in a form with a sign change, therefore not allowed
(+1,+1,-1)  can be split as (+1,+1)(-1) giving (+1,-1) a sign change. But also (+1)(+1,-1) giving (+1,+1)  no sign change, so not allowed!
                This can ofcourse be seen differently by combining the first +1 with the last -1 giving (+1,+1) directly.
(+1,-1,+1)  Not allowed, same reason
(+1,-1,-1)   both splits result in (+1,-1) hence allowed.
The other patterns start with -1, use symmetry, so also (-1,+1,+1) allowed.
[You may verify that (-1,+1,+1) is the only pattern with p[1]>2]
Of the above sign-patterns we also see that (+1,+1,+1) ,(+1,+1,-1) and (+1,-1,+1) give a + contribution to the total sum. The opposite patterns give a - contribution and are therefore also not allowed. This contribution can be used if testing for splits for n>3 of size 3. The splits for n=3 that are admissible for n=3 are inconclusive if used as patterns in n>3!


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