Problems & Puzzles: Conjectures

Conjecture 43. Three consecutive primes

Werner D. Sand sent the following conjecture from his own invention:

Let p, q, r be three consecutive prime numbers, p<q<r. Then 1/p < 1/q + 1/r.

Question: Prove it or show it false.


Contributions came from Farideh Firoozbakht, J. K. Andersen, Jean Brette, Fred Schneider, Andrew Rupinski & Joseph L. Pe. At the end this 'conjecture' resulted easy to solve, nevertheless the proof came in distinct envelopes, as you will see.

Farideh, Andersen, Fred and Andrew sent proofs based in the same concept: there at least two primes between p & 2p:

For p>5 we get pi(2p) - pi(p) >= 2,a result by Ramanujan,

This means there are at least 2 primes in [p+1,...,2p].

If p, q, r are consecutive primes then both q and r must be in [p+1,...,2p].
q < 2p implies 1/(2p) < 1/q.
r < 2p implies 1/(2p) < 1/r.
1/p = 1/(2p) + 1/(2p) for any value of p.
Combining this gives the wanted 1/p = 1/(2p) + 1/(2p) < 1/q + 1/r.

Pe's attack came from a distinct side:

Conjecture #43 is true. Here is a proof:

The conclusion follows from these two propositions:

(1) Let a(n) be a sequence of positive integers such that
a(n+1) < phi*a(n) for all n; here phi = (1+sqrt(5))/2 is the golden ratio. Then 1/a(n) < 1/a(n+1) + 1/a(n+2).

(2) For any positive integer n > 48, there is a prime between n and (9/8)*n. (See the article on Bertrand's Postulate in "Prime Numbers" by D. Wells".)

Because of (2) above, the primes > 48 satisfy the condition of (1), and so 1/p < 1/q + 1/r for consecutive primes p < q < r larger than 48. But it is easily checked that the conclusion holds also for primes < 48. This shows that conjecture # 43 is true.

It only remains to prove proposition (1) above.
Proof of (1): From the hypothesis, we have
a(n+1) < phi*a(n)
a(n+2) < phi*a(n+1) < (phi^2)*a(n),

so that

1/(phi*a(n)) < 1/a(n+1)
1/((phi^2)*a(n)) < 1/a(n+2).

Adding corresponding sides of the previous two inequalities,

(1/a(n))*[1/phi + 1/(phi^2)] < 1/a(n+1) + 1/a(n+2).

Since phi^2 - phi - 1 = 0, we get 1/phi + 1/(phi^2) = 1, so that the expression in square brackets in the previous inequality equals 1.

1/a(n) < 1/a(n+1) + 1/a(n+2),

as required.

Brette argued from this side:

In fact , q<3p/2 suffices, as follow:

The  conjecture 43 is true.

The Prime pages  give some results on gaps betweeen consecutive primes.

Let g(p) be the number of composites between p and the next prime :  

if p and q are consecutive  primes, we have     q = p + g(p) + 1.

 In the 4,  Bounds on g(p), one can find an upper bound for  g(p), after Nagura (1952) g(pn) < (1/5) pn for n > 9  , where    pn  is the nth prime    
So, for   p > 23,   and    q    the next prime, we have     q < p + (p/5) + 1
or  q < p + (p/2) = 3p/2, which is true for all primes p>7

Taking the inverse, we have  (1/q) > 2/3p

Since  q   and    r   are also consecutive, we have  r < 3q/2 < 9p/4  and   (1/r) > 4/9p. So   (1/q) + (1/r) > 6/9p + 4/9p = 10/9p > 1/p.



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