Problems & Puzzles:
Conjectures
Conjecture 53. Floor
function & consecutive primes
Patrick Capelle sent the following nice
conjecture & questions:
If p,
q, r are consecutive prime numbers such that p > q > r and if
r is different from
2,3,5,7,11,13,19,23,31,47,83,89,113,199,1327,
then floor((p*q)/r) = p + q  r.
Questions :
1. Can you prove this
conjecture or find a counterexample ?
2. Do you have some comments
about it ?
Contributions came from Giovanni Resta,
Phil Carmody, Kermite Rose, Bernardo Boncompagni, Anton Vrba & Farideh
Firoozbakht.
***
Giovanni wrote:
The conjecture is probably true, but
I think
that it is very difficult to prove it.
p*q/r (which is never an integer) can be rewritten as
p+qr + (g1^2+g1*g2)/r,
where g1=qr and q2=pq are the two gaps between the
consecutive primes r, q, p (a little strange the
choice to name them in decreasing order...)
Hence Floor[p*q/r] is equal to p+qr
if (g1^2+g1*g2)/r is smaller than 1.
Even if we forget about the term g1*g2, we
are left with r > g1^2 and this condition
is dangerously similar to the
Legendre's conjecture that between n^2 and (n+1)^2
there is always a prime number (and thus the gap
between a number x and the next prime does not
exceed the square root of x, more or less).
So it seems to me that Capelle's conjecture is probably
true as well, since it is very improbable
that so large gaps exist.
***
Phil wrote:
floor((p*q)/r) = p+qr is the same
as stating
pr+qrrr+s=pq, 0<=s<r
s=pqpr+rrqr
=p(qr)+r(rq)
=(pr)*(qr)
=(gap1+gap2)*(gap1)
for gap1, gap2 = next two prime gaps after r.
This is asymptotically equivalent to prime gaps being o(sqrt).
A similar expression could be derived for gap1*gap2<q
***
Kermite wrote:
This conjecture is almost certainly
true.
It is because it is equivalent to the statement that
(pr) * (qr) < r:
r < q < p
floor((p*q)/r) = p + q  r.
r divides (p * q) with quotient (p + q  r) and remainder s, where s
is < r.
p * q = r * (p + q  r) + s where 0 <= s < r
p * q = r * p + r * ( q  r) + s
p * q  r * p = r * (qr) + s
p * (q  r) = r * (qr) + s
p * (qr)  r * (qr) = s
(p  r) * (qr) = s
Since p > r and q > r, s is > 0.
0 < (pr ) * ( q  r) < r
All the steps in the derivation of
(p  r) * (qr) = s
are reversible, so therefore the equivalence is proved.
As r becomes larger, it becomes more
and more likely that (pr) *
(qr) would be less than r:
Bernardo wrote:
Be q=ra and p=rab. Then the
conjecture can be rewritten as
floor(r2ab+(a^2+ab)/r)=r2ab => a^2+ab<r. So if p, q and r are
sufficiently close (as the larger prime numbers tend to be), the
conjecture always holds. Actually, they don't even have to be prime!
They just need to be close enough.
Anton wrote:
Let q=p+a and r=p+b
then
p*q/r = p*(p+a)/(p+b)
and the Taylor
expansion p*(p+a)/(p+b) is
p(ba) + b(ba)/p
– b^2(ba)/p^2 + b^3(ba).p^3  ….
The first term
p(ba) or p+ab = p+qr is subject of this conjecture, so the second
term needs to be greater than one to disprove what is conjectured.
Analysing the second term b(ba)/p>1or expressing this in terms of the
prime gaps (g1+g2)g2 > p. Assuming g1 = g2 =g then g>sqrt(p/2), or if
g1=2 then g2>sqrt(p+1)1, or if g2=2 then g1>2p2, which all are far
greater what Cramer and Wolf conjectured concerning maximal prime gaps.
Capelle has
formulated a conjecture around the series expansion of p*q/r = p+qr
when reduced to integers, p, q, r being successive primes, and noted the
few exceptions for small p.
Farideh wrote:
Answer to Q1. I'm sure that this conjecture is true
but I can't prove it.
Answer to Q2: This conjecture is a result of
conjecture 30.
Proof : I used from the following result of the
conjecture 30.
If p & p+d are two consecutive primes then d < 3*log(p)^2.
(*)
Let q=r+d & p=q+d' so p*q/r = (r+d)*(r+d+d')/r =
(r^2+(2d+d')r+(d^2+dd'))/r
= r+(2d+d')+(d^2+dd')/r = (r+d+d')+(r+d)r+(r+(d^2+dd')/r
= p+qr+(d^2+dd')/r
hence, floor(p*q/r) = p+qr+floor((d^2+dd')/r) and we
must only prove that
floor((d^2+dd')/r) = 0 or d^2+d*d' < r.
Now according to (*) we have,
d < 3*log(r)^2
d' < 3*log(q)^2 = 3*log(r+d)^2 < 3*log(r+3*log(r)^2)^2
So d^2+d*d' < 9*log(r)^4+9*log(r)^2*log(r+3*log(r)^2)^2,
but for every number
r, r > 201770.565 we have f(r) = 9*log(r)^4+9*log(r)^2*log(r+3*log(r)^2)^2
< r.
Hence if r is a prime greater than 201770 then
d^2+d*d' < r and floor(p*q/r) = p+qr.
So we must only check the truth of conjecture for
primes r, where r<201770.
Between primes r, r < 201770 the equality
floor(p*q/r) = p+qr only for
r = 2, 3, 5, 7, 11, 13, 19, 23, 31, 47, 83,
89, 113, 199 & 1327 doesn't hold
and the proof is complete.
***
Capelle added (5/5/07):
As promised in my email of 23 april 2007, i propose here my other
results.
Since last time, i found some new identities, conjecturally. I
think that they have never been proposed before. I checked all the
equations mentioned, to avoid mistakes with the symbols.
If p, q, r are consecutive prime numbers such that p > q > r >
1327, then, conjecturally:
1) floor(p*q/r) = p + q  r.
2) floor(q*r/p) = q + r  p.
3) floor(r*p/q) = r + p  q  1.
4) floor(p*q/r + q*r/p) = 2q.
5) floor(p*q/r + r*p/q) = 2p.
6) floor(q*r/p + r*p/q) = 2r.
7) floor(p*q/r  r*p/q) = 2*(q  r).
8) floor(q*r/p  r*p/q) = 2*(q  p).
9) floor(p*q/r + q*r/p + r*p/q) = p + q + r.
We can obtain some interesting results by
combination among the three
first equations (to compare, for six of
them, with the conjectures 4 to 9) :
10) floor(p*q/r) + floor(q*r/p) = 2q.
11) floor(p*q/r) + floor(r*p/q) = 2p  1.
12) floor(q*r/p) + floor(r*p/q) = 2r  1.
13) floor(p*q/r)  floor(q*r/p) = 2(p  r).
14) floor(p*q/r)  floor(r*p/q) = 2(q  r) + 1.
15) floor(q*r/p)  floor(r*p/q) = 2(q  p) +
1.
16) floor(p*q/r) + floor(q*r/p) + floor(r*p/q) =
p + q + r  1.
17) floor(p*q/r) * floor(q*r/p) = q^2  (p 
r)^2.
18) floor(p*q/r) * (floor(r*p/q) + 1) = p^2  (q
 r)^2.
19) floor(q*r/p) * (floor(r*p/q) + 1) = r^2  (p
 q)^2.
By comparison of the equations 4 and 10 :
floor(p*q/r + q*r/p) = floor(p*q/r) + floor(q*r/p).
By comparison of the equations 5, 6 and 13 :
floor(p*q/r + r*p/q)  floor(q*r/p + r*p/q) = floor(p*q/r) 
floor(q*r/p).
By comparison of the equations 4, 6 and 7 :
floor(p*q/r  r*p/q) = floor(p*q/r + q*r/p)  floor(q*r/p
+ r*p/q).
By comparison of the equations 4, 5 and 8 :
floor(q*r/p  r*p/q) = floor(p*q/r + q*r/p)  floor(p*q/r
+ r*p/q).
By comparison of the equations 4, 5, 6, 9 :
floor(p*q/r + q*r/p) + floor(p*q/r + r*p/q) + floor(q*r/p
+ r*p/q)
= 2 * floor(p*q/r + q*r/p + r*p/q).
Question (to complete the first list of 9
equations) :
Can you propose a simple equation with floor(p*q/r 
q*r/p) ?
***
Werner D. Sand wrote:
conjecture 53 (like conjecture
52) does not only apply to prime numbers, but to positive integers
under a certain condition. Let be a>b>c the integers with d=bc and
e=ab, then the condition is: c > d^2+de. If a,b,c are consecutive
prime numbers, we call them p,q,r, and the condition is r > d^2+de.
The exceptions mentioned in the conjecture can be calculated by this
inequation. The essential statement of conjecture 53 is that all
prime triples > 1327 fulfill the condition.
***
