Problems & Puzzles: Conjectures

Conjecture 51 : Two approximate relationships about prime numbers

Patrick Devlin sent the following two approximate relationships & assertions about them, in a communication entitled "ON THE DISTRIBUTION OF THE PRIMES"

1. (p*Ln(p)) ~ ∑(n); p=2, 3, 5, 7, 11,...P; n=1,2,3,4,...P
2.
(Ln(p)) ~ P; p=2, 3, 5, 7, 11,...P

Assertions:
1. The above relationships hold true for p (2 to Infinity) and n (1 to Infinity).
2. They can be easily shown to be correct and equivalent to the Gauss prime
number conjecture (which has now been proven) and so in some sense they are
a re-statement of that conjecture.
3. They provide an interesting tool for further investigation into the
precise distribution of the prime numbers.
4. Unlike Euler's beautiful product identity for the zeta function which while being
correct, is always out of phase (ie there is no specific point where both
sides of the equation agree) these relationships hold true from the outset
while being in phase with each other.
5. The error ratio diminishes to zero as p tends to infinity.
6. Relationship number 1, is very interesting because it suggests that the
primes act as a moment balancing the moment exerted by the integers.
7. Relationship number 2 is interesting because it relates the nth prime to
all of the preceding primes.

A couple of comments [by CR] before the questions:

a) The Devlin's first relationship may be written in a simpler way as:

1'. (p*Ln(p)) ~ P2/2;  p=2, 3, 5, 7, 11,...P

b) Regarding the 5th statement please see a pair of graphs that I have produced: 1, 2.

Questions.

1. Please comment the Devlin's assertions

2. Do you devise a formal deduction of the Devlin's approximate relationships?

3. Do you devise an interesting application of the Devlin's approximate relationships?


Contributions came from Luis Rodríguez and Anton Vrba

***

Rodríguez wrote:

The first conjecture seems to me very interesting. I have not seen that before. I think it is better to present the conjecture in this form:

                  Lim Σ [p log(p), p= 2 to P] = P(P + 1 ) / 2

Because the n is currently reserved for the rank of a prime p. I myself have that confusion at a first lecture.

The second "conjecture" is [related to] a [known] theorem. Ross and Schoenfeld have fixed its bounds:
               
               If  Theta(p) =
Σ [log(p), p= 2 to p], then:

                p.(1 - 1/log(p)) <  Theta(p)  < p.(1 + 1/2log(p))


From this the Prime Numbers Theorem can be deducted. (See John Williams, Les Nombres Premiers . Hermann -Paris). Littlewood demonstrated that the maximum oscillation that theta(p) can attain is of order sqr(p). logloglog(p). That means that the Ross bounds are too much ample.

 

Anton wrote:

Investigating above one can add the generalized case of Patrick’s relation as 

3. (p^r*Ln(p)) ~ (n^r); p=2, 3, 5, 7, 11,...P; n=1,2,3,4,...P

I have calculated with r as big as 500 and above relation holds as well as for  r<1

My comment to Patrick’s assertion is (i) True for 1, 2, 5 and 7 (ii) False for 3, (iii) No comment for 4 and 6 and finally (iv) Prime Number Theory already states what is commented as True.

The above comment and Patrick’s relation can be simply demonstrated as follows:

Considering that Prime Number Theory (PNT) tells us the average gap of all primes less than P are Ln(P), therefore presume a prime P_i  then the next prime by PNT is expected at P_i+Ln(P_i). Now lets define a recursive logarithmic series as follows:

a1. q_1 = x,  for any x>1
a2. q_i = q_(i-1) + Log_a(q_(i-1))      

and the so obtained q_i  should resemble the series of prime numbers if the base of the logarithm a=e, where e is the famous constant 2.718… By calculation let q_1=2  then q_(10^6)=15479073 and Prime[10^6]=15485863  that is a  0.044% error!

I have discovered or re-discovered that if we define c as follows

a3. c= (q_i^r*Log_b(q_i) ) / (n^r)  ;  for  i=1,2,3,....k ; n=1,2,3,….Integer(q_k)

then

a4.  a = b^c  if  k ≈ infinity for any r, a, b and x.  

This leads me to believe the only logical reason for Patrick’s relation (1. and  2.) is because they are similar to the case a=b=e and r=1 and 0 in a1 to a4 above. Hence, all above relationships, in my opinion, add no further insight other than what is already known by the PNT, to the distribution of primes  as Patrick claims with assertion 3.

During the analysis when I started investigating the effect of the bases of the log terms, I was utterly amazed to find the relation a4 a=b^c, which reinforces my believe that this relation is true, the error gets extremely small very fast.

Now I pose the following questions:

Q1.  Is the property of recursive logarithmic series a1 - a4 discovered or re-discovered? If re-discovered are there any references?
Q2.  If discovered how to prove it other than by computational methods.

*** 

Werner Sand wrote (30/11/06) a short note on this issues:

Further approximate relationships see www.primzahlen.de Theorie/Referenten: "Eine neue Methode zur Berechnung der Anzahl der Primzahlen unter einer gegebenen Größe" by Werner D. Sand.

I have asked to Mr. Sand for an English version of his pdf article, published in German. Hopefully soon we may have it available... Two days latter he wrote and sent the following summary of his article:

Assertion: 

Σ(p, p=2 to x) ~ π(x²) or
Σ(p, p=2 to sqrt(x)) ~ π(x),

where p = prime and x = real, π = number of primes

Generalization:

Σ(p^r, p=2 to x) ~ π(x^(r+1)) or
Σ(p^r, p=2 to x^(1/(r+1))) ~ π(x)

where r = real >= 0 (r=0 means counting the primes) 

Proof for r = m = integer:

General formula for the sum of powers of natural numbers:

Σ(k^m, k=1 to n) = (n^(m+1))/(m+1) + (n^m)/2 + O(n^(m-1),

where the second and third summand can be omitted without effect on the order:

Σ(k^m, k=1 to n) ~ (n^(m+1))/(m+1)

Instead of n natural numbers we choose the π(n) ~ n/ln n prime numbers, which means we divide the equation by ln n:

Σ(p^m, p=2 to n) ~ ((n^(m+1))/(m+1)) * (1/ln n) =  (n^(m+1))/ln(n^(m+1)) = π(n^(m+1)), q.e.d.

Proof for r = real:

Instead of sum choose integral: int(x^r) = (x^(r+1))/(r+1). Then proof like above.

Let

S1 be π(x) ~ Σ(p, p=2 to sqrt(x)) and

S2 be π(x) ~ Σ(p², p=2 to x^(1/3)).

Then the quality of S1 or S2  is comparable to Legendre: π(x) ~ x/(ln x-1), in some cases better.

I didn't check yet Si with i >=3.

See this new article from W. Sand (April, 09)

***

As you all can observe, the approximate equations published by Sand are distinct to the submitted by Patrick Devlin, because the argument of the summing operator is distinct.

But share with them that they admit - conjecturally - a generalization to powers of the terms added, equal to the generalization that Anton made to the Devlin's approximate equations.

In the Devlin's generalized equations the arguments are p^r.Ln(p) and Ln(p) while in the Sand's equations the argument is p^r.

Let's give a comparison for the three equations without generalizations, in the following manner:
 
Author Argument of the Σ operator Limits of the sum Value approximated by Σ
Devlin p.ln(p) 2, P P2/2
Devlin ln(p) 2, P P
Sand p 2, P π(P2)

Do you like these results? Have they been suggested before? Are they useful?

***

Sebastián Martín Ruiz sent (Dec. 7, 2006) the following link related:

***

Anton Vrba wrote (Dec. 8, 06):

Author

Argument of the Σ operator

Limits of the sum

Value approximated by Σ

Deviation at π(P)=105

Devlin

p ln(p)

2, P

D1 = P2/2

0.060%

Devlin

1 ln(p)

2, P

D0 = P

0.094%

Sand

p

2, P

S1 = π(P2)

0.061%

Sand

1

2,P

S0 = π(P)

exact, counting

Vrba

p/ln(p)

2, P

V1= π(P)2/2

8.01%

Vrba

1/ln(p)

2, P

V0 = π(P)2/P

1.07%

After seeing the first three entries in above table the last three entries had to follow.  Note that is the number of primes squared in last two lines

Note the following relationship D1 /D0 =  V1 /V0 = P/2 and extended over Sands relationships  then

(2 D1 )/(P D0 ) =  (2 S1 )/(P S0 ) =  (2 V1 )/(P V0 ) = 1

The easy proof for S1/S0 was forwarded by Werner Sand: The Prime Number Theory states that number of primes equals  x / ln x hence

S1 /S0 = π(P2)/ π(P) ~ (P²/ln P²) / (P/ln P) = (P²*ln P)/(2ln P*P) = P/2.

Using the above, the following relationship between primes and prime counting function has now been identified

(2 π(P2)) / ( π(P) P) = 1

Example: Calculating for the first 105 primes above evaluates to 0.958 and substituting known values for P2=1022, then P=1011 and calculating π(P) with Mathematica  the above evaluates to 0.978.

Concerning the high deviation of V1 it is to be noted that V1 sums the counting function as defined in the PNT and we know that x/ln(x) consistently gives a lower value for the number of primes than the actual value (see the graph on Wolfram’s site) and thus this consistent lower value integrates into the larger deviation.

***

Werner Sand contributes with another approximate relationship, in this case based on the definition of Merten's constant.

Complement of Anton Vrba's table: 

Author  

Argument of the Σ operator

Limits

Value approximated

Sand/Mertens

1/p

2,x

lnln(x)

x is the upper limit of the running variable p (p<=x). x is real and may be prime or not.

***

Later Werner Sand added two contributions:

A. Another complement to the table:

Assumption: Σ (ln p)/p (p=2…x) ~ ln п(x) 

Proof:

Σ (ln p)/p = (ln 2)/2 + (ln 3)/3 + (ln 5)/5 + …+ (ln x)/x =

1/(2/(ln 2)) + 1/(3/(ln 3)) + 1/(5/(ln 5)) + ... + 1/(x/(ln x)) = (PNT)

1/1 + 1/2 + 1/3 + 1/4 + ... + 1/(x/(ln x)) ~ (PNT)

1/1 + 1/2 + 1/3 + 1/4 + ... 1/п(x) ~ (Euler)

ln п(x)), q.e.d.

B. The approximate relations are not necessarily one-to-one. E.g.

Σ (ln p)r/p (p=2…x) ~ (ln x)r/r    (*)

Then if r=1

Σ (ln p)/p (p=2…x) ~ ln x or

Σ (ln p)/p (p=2…x) ~ ln п(x) (as shown in A) (**)

(*) Just by trial. No proof yet. It's a conjecture. Let R = Σ {[(ln p)^r]/p}/{[(ln x)^r]/r}. I have tested R with MuPAD for  r from 0.5 to 1000 incl. ratios and irratios as sqrt(2) and for p up to 32.452.843  resp. 2*10^6 steps. R is sometimes oscillating, so I cannot state with absolute certainty that R is going to 1. But I am nearly sure. I would be glad if someone else could calculate.

(**) x is indeed unequal pi(x), but nevertheless ln x ~ ln pi(x). Proof:
[ln pi(x)]/ln x ~ [ln(x/ln x)]/ln x = [(ln x - ln ln x)/ln x = 1-(ln lnx)/ln x].
(ln x)/(ln ln x) is approximately the number of primes up to ln x, which
means goes to infinity with growing x. Hence the reciprocal (ln ln x)/ln x
goes to zero, and the above ratio goes to 1, q.e.d.

You can check that by pocket computer: lim [(ln x)/(ln pi(x))] ~ [lim(ln x)/(ln(x/ln x))]. Just insert a few values of x, e.g. x=10, x= 100, x=1000 and so on.

***

Anton Vrba wrote (May 09)

Regarding Werner's new conjecture Σ (ln p)r/p (p=2…x) ~ (ln x)r/r - the following:

I have done some plots. The filename indicates the r value.

Most interesting is the lower bound value of the ratio Σ (ln p)r/p / (ln x)r/r as can be seen for r>>

I think for x->infinity the ratio 1 will be attained for all r>0, his so called oscillations are getting asymptotically smaller as can be seen from these plots.

 


 

 

Records   |  Conjectures  |  Problems  |  Puzzles