Problems & Puzzles: Puzzles

Puzzle 315. pn => pn-i + pi

Sebastian Martin Ruiz, poses the following puzzle:
Prove that:

pn => pn-i + pi

for all 1<= i <=n-1, for all n=>3.



Contributions came from J. K. Andersen, Faride Firoozbakht, Luis Rodríguez & Rudolph Knjzek. Well, it seems that we have more problems than solutions now! See below why:

Andersen wrote:

This is equivalent to the second Hardy-Littlewood conjecture:

pi(x) + pi(y) >= pi(x+y).

It appears likely to be false but finding a counter example seems completely infeasible. Thomas J Engelsma has found many admissible prime tuplet patterns which will each give infinitely many counter examples if the k-tuple conjecture is true. See
For a discussion of the difficulty in finding such a counter example with one of the patterns, see

Faride wrote:
It seems that if n>2, the following stronger relation is true.

Max{ prime(i)+prime(n-i) | 1<=i<=n-1 } = prime(1)+prime(n-1) (*) (the maximum happens where i=1)

But at this moment I don't know how we can prove it.

The requested relation is a trivial result of (*) (d(n)=prime(n)-prime(n-1)>=2).

Luis Rodríguez wrote:

Starting from this two theorems of Dusart: "Autour de la function qui compte le nombre des nombres premiers". These de doctorat 1998

Theorem 1.5 Pag 26
p(n) > n[log(n) + logLog(n) -1] ; n > 1

Theorem 1.8 Pag. 32
p(n) < n[log(n) + logLog(n)-.9484]; n>39017

We can demonstrate that p(n) >=P(n-i)+p(i) provided that i/n >= 0.01

As the theorem 1.8 exige that n > 39017 or
p(n) > 467473 it is necessary first to check the conjecture from n = 39017 to 19508 and i from 1 to 19508. I checked that and it passed the test.

Making k = (1-i/n) and j = i/n ; k+j = 1

We will take the minimum for the left term that is: P(n) = n[log(n) + loglog(n) - 1] and the mximum for the right term, that is:
p(n-i)= p(kn)= kn(log(kn)+loglog(kn)-.9484]
p(i) = p(jn)= jn[log(jn)+loglog(jn)-.9484]

If the inequality holds up the conjecture is true.
Canceling n in the two members and developing the algebra we arrive to:
loglog(n) > 0.0516 + [klog(k)+jlog(j)] +

But loglog(n) > kloglog(kn)+ jloglog(jn) and I can cancel the two quantities without to destroy the inequality.
0 > 0.0516 + [klog(k) + jlog(j)]
This last sum results less than -0.056 provided j >= 0.01 that is i/n >= .01


Knjzek wrote:


The number of primes in the open interval (0,p(k)) is k-1.
The open interval (p(n),p(n-k)) contains also k-1 primes.
If no open interval (a,b) of same length as (0,p(k)), b-a = p(k), contains more than k-1 primes, thenp(n) - p(n-k) => p(k) and p(n) => p(k) + p(n-k)

Therefore it is equivalent to proof that pi(b) - pi(a) =< pi(b-a).
If P is the Product of all primes less or equal than sqrt(b) and d divides P we can sieve out the composites (Lagrange) and have to proof:
Sum over d (µ(d)*(int(b/d)-int(a/d))) =< Sum over d (µ(d)*int((b-a)/d)).
Assuming that int(b/d)-int(a/d) = int((b-a)/d) + r with r = 0 or 1, this seems to be true.


The 29/5/05 I asked this to JKA:

Perhaps today I'm a little bit sleep. Can you tell me in detail why do you affirm that the Sebastian Conjecture "is equivalent to the second Hardy-Littlewood conjecture"?

As a matter of fact the H&L conjecture is:

pi(x+y)<=pi(x) + pi(y), where pi is the prime counting function. 

By his side, the Sebastian's conjecture is:

p(n) =>p(n-i)+p(i), where p(x) is the x-th prime.

This Sunday morning for me is not as clear as a week before the equivalence between these two statements.

Two days later Luis Rodríguez wrote:

I don't understand what means Andersen by "equivalent", because in Sebastian's p(n) means the primes of rank n, but in Hardy's Pi(n) means the number of primes less than n.
Besides if we replace in Hardy's x+y = n x = n-i , y = i ; for the case i = n/2 results the false inequality: Pi(n) >= Pi(n/2) + Pi(n/2)

This is the JKA reply to both of us:

Let p(n) = n'th prime, and pi(x) = number of primes <=x.

I have been asked to explain in detail the connection to the
second Hardy-Littlewood conjecture:
(0) pi(x) + pi(y) >= pi(x+y), for x,y >= 2.

Consider the slightly weaker variant:
(1) If x or y is prime, then pi(x) + pi(y) >= pi(x+y), for x,y >= 2.

Ruiz asks us to prove (without knowing whether it's true):
(2) p(n) >= p(n-i) + p(i), for n >= 3.

(1) and (2) are either both true or both false. This is proved by showing that a counter example to one of them gives a counter example to the other.

First assume (1) is false with counter example:
pi(x) + pi(y) < pi(x+y), where x is prime (y is symmetric).
Let n = pi(x+y). Then p(n) <= x + y.
Let i satisfy x = p(n-i). Then pi(x) = n-i.
By assumption pi(y) < pi(x+y) - pi(x) = n - (n-i) = i.
pi(y) < i implies y < p(i). We now have:
p(n) <= x + y = p(n-i) + y < p(n-i) + p(i).
This is the wanted counter example to (2).

Now assume (2) is false with counter example:
p(n) < p(n-i) + p(i).
Let x = p(n-i). Let y = p(n) - p(n-i). Then x+y = p(n).
y = p(n) - p(n-i) < p(i) by assumption, so pi(y) < i. We now have:
pi(x+y) = pi(p(n)) = n = (n-i) + i = pi(x) + i > pi(x) + pi(y).
x is prime so this is the wanted counter example to (1).
That completes the proof.

Are (1) and (2) true or false? I think false, but nobody has been able to prove it.  says for example:
There is an admissible 459-tuple of width 3241.
(An even smaller width may exist).
If the k-tuple conjecture is true (I and most other think so),
then there is a prime 459-tuple of width 3241, i.e. a solution to:
p(n) - p(n-458) = 3240.
For comparison, p(458) = 3251. Let i = 458 and we get:
p(n) = p(n-i) + 3240 < p(n-i) + 3251 = p(n-i) + p(i).
This would be a counter example to (2).

A counter example to (0) is not necessarily a counter example to (2).
It appears conceivable (but unlikely in my opinion) that (2) is true and the second Hardy-Littlewood conjecture is false.




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