Problems & Puzzles: Puzzles

Puzzle 997. m - n + s - Floor[m*s/n]=1 Sebastián Martín Ruiz wrote on April 11, 2020:   1) Demonstrate that  m - n + s - Floor[m*s/n]=1 for all positive integers s<n<m iff m-s<Sqrt[m] .   2) Verify if Sqrt[m] is the largest bound for m-s that assures the equality above be satisfied.   3 )Demonstrate that for tor three consecutive primes and n>1 other that 3 & 8: Prime[n + 2] - Prime[n + 1] + Prime[n] - Floor[Prime[n + 2]Prime[n]/Prime[n + 1]] = 1 (for this puropose, once proved 1, is enough to prove that p(n+2)-p(n)<Sqrt[p(n+2)])   4)Proved 3, demonstrate: p(n+2)<(1/(p(n+1)-p(n))+1)p(n+1)

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Contributions came from Oscar Volpatti, Emmanuel Vantieghem, Jan van Delden and Sebastián Martín Ruiz. All of them during the week 18-25 April, 2020.

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Oscar wrote:

I have no complete answer this week, question 3 was beyond my reach.
 
Questions 1 and 2.
For convenience, rearrange the given equality as  floor(m*s/n) = m-n+s-1;
I claim that it holds for all positive integers  s<n<m  iff  m-s <= sqrt(4*m+1)-1,
essentially improving the bound  m-s < sqrt(m) by a constant factor 2.


If  m-s <= 1, there are no integers n with  s<n<m; let us assume  m-s >= 2.
From the definition of floor function, we obtain the following system of inequalities:
(a)  m*s/n >= m-n+s-1;
(b)  m*s/n < m-n+s.
Multiply by n (strictly positive) and rearrange:
(a)  n^2 - (m+s-1)*n + m*s >= 0;
(b)  n^2 - (m+s)*n + m*s < 0;
Let us call the quadratic polynomials Pa(n) and Pb(n) respectively.


Pb(n) has two distinct integer roots s and m, so inequality (b) holds for s<n<m, precisely the range of interest.
Therefore, the starting equality holds in such range iff inequality (a) holds too.


Pa(n) has discriminant  Da = (m+s-1)^2 - 4*m*s;
I will prove that (a) holds for all integers s<n<m (actually, for all positive integers n) iff  Da <= 1.
We need to distinguish two cases.


I) Da  is even,  m+s  is odd,  m-s >= 3.
Pa(n) has a unique minimum at the integer point n1 = (m+s-1)/2, with Pa(n1) = -Da/4 and s<n1<m;
iff Da<=0, inequality (a) holds for n1, and for any other positive integer n.


II) Da  is odd,  m+s  is even,  m-s >= 2.
Now n1 is not an integer; Pa(n) has two minima at the closest integer points n0 = (m+s)/2 and n2 = (m+s-2)/2 = n0-1,
with Pa(n0) = Pa(n2) = (1-Da)/4  and s<n0<m (but s<n2<m holds too only if m-s>=4);
iff Da<=1, inequality (a) holds for n0, and for any other positive integer n.


So, inequality  Da <= 1  actually covers both parity cases.
Rewrite it as  s^2 - 2*(m+1)*s + m*(m-2) <= 0,  and solve for s:
m+1-sqrt(4*m+1) <= s <= m+1+sqrt(4*m+1),
or, after rearranging:
-sqrt(4*m+1)-1 <= m-s <= sqrt(4*m+1)-1.
But only the second inequality is meaningful, as we worked in the range m-s >= 2.

Question 3.
As done before, we can rewrite the given equality as a system of two inequalities:
(a)  p(n+2)*p(n)/p(n+1) > p(n+2)-p(n+1)+p(n)-1;
(b)  p(n+2)*p(n)/p(n+1) < p(n+2)-p(n+1)+p(n).
Now both inequalities are strict, because p(n+1) doesn't divide the product p(n+2)*p(n) exactly.
Again, (b) holds iff  p(n)<p(n+1)<p(n+2),  a true statement for every n.
Multiply both sides of (a) by p(n+1), then rearrange:
(p(n+2)-p(n+1))*(p(n+1)-p(n)) < p(n+1).
Assuming the conjectured Oppermann's bound on prime gaps:
p(n+1)-p(n) < sqrt(p(n))  for every n > 30,
we obtain (for n> 30) the following chain of inequalities:
(p(n+2)-p(n+1))*(p(n+1)-p(n)) < sqrt(p(n+1))*sqrt(p(n)) < p(n+1). 
It is easy to check that, in the range n <= 30, (a) fails only for n=3 and for n=8.


Question 4.
The given inequality can be obtained from previous inequality (a), by solving for p(n+2).

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Emmanuel wrote:

About Q1.
Take two positive integers  s, m  such that s < m
Consider the function  f(x) = m - x + s - m*s / x  for  x  between  s  and  m  (s < x < m).
With a littlebit of elementary calculus we can show that  f(x)  reaches its maximum where  f '(x) = 0, i.e. : when  x0 = Sqrt(m*s).

We then have :
   f(x0) = m - Sqrt(m*s) + s - Sqrt(m*s) = m + s - 2Sqrt(m)Sqrt(s) = (Sqrt(m) - Sqrt(s))^2.
Now, if we want to have  m - n + s -Floor[m*s / n]  to be equal to  1  for all integers  n  between  s  and  m
we should have that  m - x + s - m*s / x  is a number N  that satisfies  0 < N <= 1.
Thus, that should be  Sqrt(m) - Sqrt(s) <= 1.
This is much more than  m - Sqrt(m) < s.
 
For the rest, the condition  Sqrt(p(n+2)) - Sqrt(p(n) < 1  would imply that there is always a prime in an interval  [ x^2, (x+1)^2 ].
As far as I know this is still a conjecture.
 

So, I think all subsequent inequalities are also conjectural.

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Jan wrote:

Q2

 

For all n, with s<n<m and s,n,m integer we are given: m-n+s-floor[m*s/n]=1

 

Write: Floor[m*s/n]=m*s/n-a with a in [0,1)

 

We thus obtain: m-n+s-m*s/n=1-a=b, with b in (0,1]

 

Multiply with n and simplify: (m-n)(n-s)/n=b

 

The function f(x,s,m)=(m-x)(x-s)/x is 0 iff x=m or x=s.  Its maximum value is attained for x=sqrt(ms).

Given s<n<m we won’t have x=s or x=m, thus b>0. We want this maximum value b to be less than or equal to 1 for all values of n.

 

We therefore need:

 

(m-sqrt(ms))(sqrt(ms)-s)/ sqrt(ms) = (sqrt(m)-sqrt(s))^2<=1 or sqrt(m)-sqrt(s)<=1 (since m>s).

 

If we multiply both sides of the last equation with sqrt(m)+sqrt(s) we arrive at:

 

m-s<=sqrt(m)+sqrt(s)

 

The reverse of this statement is that if we have m-s<=sqrt(m)+sqrt(s) we have:

 

m-n+s-Floor[m*s/n]=1 for all n with: s<n<m.

 

The proof just reverses every step and is left as an exercise.

 

That means that the given condition in Q1: m-s<sqrt(m) is too strict. It is sufficient, but not necessary.

 

Notes:

The equation: m-n+s-Floor[m*s/n]=1 with s<n<m may have particular solutions n even if m-s>sqrt(m)+sqrt(s), but it can’t have solutions for all n in the given interval.
This might happen if m-n or n-s is relatively small.

 

Q3.

 

It is very tempting to plug in: s=p[n], n=p[n+1], m=p[n+2]. But we can’t use our previous result. In order to be able to do that we definitely need to prove that we have:

 

p[n+2]-p[n] < sqrt(p[n+2])+sqrt(p[n])

 

But I don’t see how to prove that this condition applies. It might for large n, but even that is just a conjecture nowadays: Opperman’s conjecture, which is a statement regarding an average prime gap, for large n:

 

p[n+2]-p[n+1]<sqrt(p[n+1]) and

p[n+1]-p[n] < sqrt(p[n])

 

Add the two ‘inequalities’ and arrive at:

 

p[n+2]-p[n]<sqrt(p[n+1])+sqrt(p(n))<sqrt(p[n+2])+sqrt(p[n])

 

So it seems to fit nicely, but there are two drawbacks. One Opperman’s is a conjecture and not a theorem. And two the conjecture is about an average prime gap and not a prime gap. This does not mean that the given equality we wish to use is not true; as stated before it might also happen that m-n or n-s are small enough, or to put it differently, at least one of the two prime gaps needs to be relatively small.

 

Q4.

 

If we plug in the same values for s,n,m into (m-n)(n-s)/n<=b with b=1 and rewrite we arrive at:

 

p[n+2]<=p[n+1] (1+1/(p[n+1]-p[n]))

 

Note the possibility of equality above. And again, this statement might very well be true, with a few exceptions, but I don’t see how to prove the condition stated in my answer of Q3.

 

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Sebastián wrote:

He hecho la demostración del apartado 1:

 

Floor[ms/n]=m-n+s-1

es equivalente a probar:

m-n+s-1<=ms/n<m-n+s

 

ms<mn-n^2+sn

0<mn-n^2+sn-ms

0<n(m-n)-s(m-n)

0<(n-s)(m-n) los cual es obvio pues s<n<m

 

por otro lado

 

mn-n^2+sn-n<=ms

mn-n^2+sn-ms<=n

(m-n)(n-s)<=n

 

la cota correcta sería (m-s)<Sqrt[n] en lugar de Sqrt[m]

 

por lo que se tendría (m-n)(n-s)<(m-s)^2<=n

 

y así estarían probadas la dos desigualdades.

 

Para el recíproco buscamos un contraejemplo con (m-s)>Sqrt[n]

no he encontrado ninguno poniendo como cota Sqrt[n]+1, tampoco para Sqrt[n]+2, he encontrado  contraejemplos para Sqrt[2]+3 

 

m=5 n=2 s=1 

5-2+1-Floor[5*1/2]=4-2=2          5-1<=Sqrt[2]+3

 

por lo que la mejor cota sería Sqrt[n]+2 
 

pero esto último no lo puedo asegurar.

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