![Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.10, No.3, September 2023
DOI : 10.5121/mathsj.2023.10301 1
A POSITIVE INTEGER 𝑵 SUCH THAT
𝒑𝒏 + 𝒑𝒏+𝟑 ~ 𝒑𝒏+𝟏 + 𝒑𝒏+𝟐
FOR ALL 𝒏 ≥ 𝑵
Pham Minh Duc, Luong Van Tam
VNU University of Science, 334, Nguyen Trai, Thanh Xuan Trung Ward, Thanh Xuan
District, Hanoi, Vietnam
ABSTRACT
According to Bertrand's postulate,we have 𝑝𝑛 + 𝑝𝑛 ≥ 𝑝𝑛+1. Is it true that for all 𝑛 > 1 then 𝑝𝑛−1 + 𝑝𝑛 ≥
𝑝𝑛+1? Then 𝑝𝑛 + 𝑝𝑛+3 > 𝑝𝑛+1 + 𝑝𝑛+2where 𝑛 ≥ 𝑁, 𝑁 is a large enough value?
KEYWORDS
Bertrand’s Postulate, Rosser's theorem, L’Hospital Rule, prime number.
1. INTRODUCTION
In 1845, Bertrand conjectured what became known as Bertrand’s postulate: twice any prime strictly
exceeds the next prime. Tchebichef presented his proof of Bertrand’s postulate in 1850 and published it in
1852. It is now sometimes called the Bertrand-Chebyshev theorem. Surprisingly, a stronger statement
seems not to be well known, but is elementary to prove: The sum of any two consecutive primes strictly
exceeds the next prime, except for the only equality 2 + 3 = 5. After I conjectured and proved this
statement independently, a very helpful referee pointed out that Ishikawa published this result in 1934
(with a different proof). This observation is a special case of a much more general result, Theorem 1, that
is also elementary to prove (given the prime number theorem), and perhaps not previously noticed:
If𝑝𝑛 denotes the 𝑛th prime, 𝑛 = 1, 2, 3, … with 𝑝1 = 2, 𝑝2 = 3, 𝑝3 = 5, …, and then there exists a positive
integer 𝑁 such that 𝑝𝑛 + 𝑝𝑛+3 ~ 𝑝𝑛+1 + 𝑝𝑛+2for ll 𝑛 ≥ 𝑁. We give the following result.
2. MAIN RESULT
Theorem 1. If𝑛 are nonnegative integers, then there exists a large enough positive integer 𝑁 such
that, for all 𝑛 ≥ 𝑁, 𝑝𝑛 + 𝑝𝑛+3 ~ 𝑝𝑛+1 + 𝑝𝑛+2.
Applying Rosser's theorem for all 𝑛 ≥ 6, we have
𝑛(ln𝑛 + lnln𝑛 − 1) < 𝑝𝑛 < 𝑛(ln𝑛 + lnln𝑛)
(𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1) − 1] < 𝑝𝑛+1 < (𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1)]
(𝑛 + 2)[ln(𝑛 + 2) + lnln(𝑛 + 2) − 1] < 𝑝𝑛+2 < (𝑛 + 2)[ln(𝑛 + 2) + lnln(𝑛 + 2)]
(𝑛 + 3)[ln(𝑛 + 3) + lnln(𝑛 + 3) − 1] < 𝑝𝑛+3 < (𝑛 + 3)[ln(𝑛 + 3) + lnln(𝑛 + 3)]
Consider the expression](https://image.slidesharecdn.com/10323mathsj01-231012122649-f34478eb/85/A-Positive-Integer-Such-That-For-All-1-320.jpg)
![Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.10, No.3, September 2023
2
𝐴 =
𝑛(ln𝑛 + lnln𝑛 − 1) + (𝑛 + 3)[ln(𝑛 + 3) + lnln(𝑛 + 3) − 1]
(𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1)] + (𝑛 + 2)[ln(𝑛 + 2) + lnln(𝑛 + 2)]
We consider the following limit
𝐵 = lim
𝑛→+∞
ln𝑛 + lnln𝑛 +
1
ln𝑛
+ ln(𝑛 + 3) + lnln(𝑛 + 3) +
1
ln(𝑛+3)
ln(𝑛 + 1) + lnln(𝑛 + 1) + 2 +
1
ln(𝑛+1)
+ ln(𝑛 + 2) + lnln(𝑛 + 2) +
1
ln(𝑛+2)
Taking the ln of the numerator and denominator and applying L’Hospital Rule gives
lim
𝑛→+∞
𝑛(ln𝑛 + lnln𝑛 − 1)
= lim
𝑛→+∞
ln𝑛 + lnln𝑛 − 1 + 𝑛 (
1
𝑛
+
1
𝑛
ln𝑛
)
= lim
𝑛→+∞
ln𝑛 + lnln𝑛 +
1
ln𝑛
lim
𝑛→+∞
𝑛 [ln(𝑛 + 3) + lnln(𝑛 + 3) − 1]
= lim
𝑛→+∞
ln(𝑛 + 3) + lnln(𝑛 + 3) − 1 + (𝑛 + 3) (
1
𝑛 + 3
+
1
𝑛+3
ln(𝑛 + 3)
)
= lim
𝑛→+∞
ln(𝑛 + 3) + lnln(𝑛 + 3) +
1
ln(𝑛 + 3)
lim
𝑛→+∞
𝑛 [ln(𝑛 + 1) + lnln(𝑛 + 1)]
= lim
𝑛→+∞
ln(𝑛 + 1) + lnln(𝑛 + 1) + (𝑛 + 1) (
1
𝑛 + 1
+
1
𝑛+1
ln(𝑛 + 1)
)
= lim
𝑛→+∞
ln(𝑛 + 1) + lnln(𝑛 + 1) +
1
ln(n + 1)
+ 1
lim
𝑛→+∞
𝑛 [ln(𝑛 + 2) + lnln(𝑛 + 2)]
= lim
𝑛→+∞
ln(𝑛 + 2) + lnln(𝑛 + 2) + (𝑛 + 2) (
1
𝑛 + 2
+
1
𝑛+2
ln(𝑛 + 2)
)
= lim
𝑛→+∞
ln(𝑛 + 2) + lnln(𝑛 + 2) +
1
ln(n + 2)
+ 1
Then we see
𝐵 = lim
𝑛→+∞
ln𝑛 + lnln𝑛 +
1
ln𝑛
+ ln(𝑛 + 3) + lnln(𝑛 + 3) +
1
ln(𝑛+3)
ln(𝑛 + 1) + lnln(𝑛 + 1) + 2 +
1
ln(𝑛+1)
+ ln(𝑛 + 2) + lnln(𝑛 + 2) +
1
ln(𝑛+2)](https://image.slidesharecdn.com/10323mathsj01-231012122649-f34478eb/85/A-Positive-Integer-Such-That-For-All-2-320.jpg)
![Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.10, No.3, September 2023
3
When 𝑛 → +∞ then
𝐵 = lim
𝑛→+∞
ln𝑛 + ln(𝑛 + 3)
ln(𝑛 + 1) + ln(𝑛 + 2)
= lim
𝑛→+∞
ln(𝑛2
+ 3𝑛)
ln(𝑛2 + 3𝑛 + 2)
= 1
(Because 𝑛2
+ 3𝑛 ~ 𝑛2
+ 3𝑛 + 2, for 𝑛 → +∞)
Or, for 𝑛 ≥ 𝑁, 𝑁 is a large enough positive integer, then𝐴 ~ 1,
𝑛(ln𝑛 + lnln𝑛 − 1) + (𝑛 + 3)[ln(𝑛 + 3) + lnln(𝑛 + 3) − 1]
(𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1)] + (𝑛 + 2)[ln(𝑛 + 2) + lnln(𝑛 + 2)]
~ 1
It turns out,
𝑝𝑛+𝑝𝑛+3
𝑝𝑛+1+𝑝𝑛+2
~1, or have mean is 𝑝𝑛 + 𝑝𝑛+3 ~ 𝑝𝑛+1 + 𝑝𝑛+2.
3. CONCLUSION
In short, that is also elementary to prove (given the prime number theorem), and perhaps not
previously noticed: If pn denotes the nth prime, n = 1,2,3,... with p1 = 2, p2 = 3, p3 = 5,..., and
then there exists a positive integer N such that pn + pn+3 ~ pn+1 + pn+2 for 11n ≥ N. But can not
confirm that pn + pn+3 = pn+1 + pn+2 where n ≥ N, N is a large enough value?
ACKNOWLEDGEMENT
I thank VNU University of Science for accompanying me.
REFERENCES
[1] J. Bertrand, “M´emoire sur le nombre de valeurs que peut prendre une fonction quand on y permute
les lettres qu’elle renferme”, Journal de l’Ecole Royale Polytechnique Cahier 30, 18 (1845), 123–
140.
[2] P. Tchebichef, “M´emoire sur les nombres premiers”, Journal de math´ematiques pures et
appliqu´ees series 1, 17 (1852), 366–390.
[3] H. Ishikawa, “1934 Ueber die Verteilung der Primzahlen”, Science Reports of the Tokyo Bunrika
Daigaku, Section A 2 (1934) (31/32), 27-40.
[4] J. B. Rosser, and L. Schoenfeld, Lowell, “Approximate formulas for some functions of prime
numbers”, Illinois Journal of Mathematics 6 (1962) (1), 64-94.
[5] R. P. Brent, J. van de Lune, H. J. J. te Riele, and D. T. Winter, On the zeros of the Riemann zeta
function in the critical strip. II, Math. Comp. 39 (1982), no. 160, 681–688.
[6] M. Cipolla, La determinazione assintotica dell’𝑛𝑖𝑚𝑜
numero primo, Matematiche Napoli 3 (1902),
132-166.
[7] J. van de Lune, H. J. J. te Riele, and D. T. Winter, On the zeros of the Riemann zeta function in the
critical strip. IV, Math. Comp. 46 (1986), no. 174, 667–681.
[8] Jean-Pierre Massias and Guy Robin, Bornes effectives pour certaines fonctions concernant les
nombres premiers, J. Théor. Nombres Bordeaux 8 (1996), no. 1, 215–242.
[9] Guy Robin, Estimation de la fonction de Tchebychef 𝜃 sur le 𝑘-ième nombre premier et grandes
valeurs de la fonction 𝑛 log 𝑛 nombre de diviseurs premiers de 𝑛, Acta Arith. 42 (1983), no. 4, 367–
389.
[10] J. B. Rosser, The n-th prime is greater than 𝑛 log 𝑛, Proc. London Math. Soc. (2) 45 (1939), 21-44.
[11] Leonard Eugene Dickson, New First Course in the Theory of Equations, John Wiley & Sons, Inc.,
New York, 1939.
[12] J. Barkley Rosser and Lowell Schoenfeld, Approximate formulas for some functions of prime
numbers, Illinois J. Math. 6 (1962), 64–94.
[13] Wilhelm Wirtinger, Über eine Minimalaufgabe im Gebiete der analytischen Funktionen von
mehreren Veränderlichen, Monatsh. Math. Phys. 47 (1939), 426–431.](https://image.slidesharecdn.com/10323mathsj01-231012122649-f34478eb/85/A-Positive-Integer-Such-That-For-All-3-320.jpg)
![Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.10, No.3, September 2023
4
[14] Wilhelm Wirtinger, Über eine Minimalaufgabe im Gebiete der analytischen Funktionen von
mehreren Veränderlichen, Monatsh. Math. Phys. 47 (1939), 426–431.
[15] Pierre Dusart (1999). "The kth prime is greater than k(log k + log log k−1) for k ≥ 2". Mathematics of
Computation. 68 (225): 411–415.](https://image.slidesharecdn.com/10323mathsj01-231012122649-f34478eb/85/A-Positive-Integer-Such-That-For-All-4-320.jpg)
A Positive Integer 𝑵 Such That 𝒑𝒏 + 𝒑𝒏+𝟑 ~ 𝒑𝒏+𝟏 + 𝒑𝒏+𝟐 For All 𝒏 ≥ 𝑵
- 1. Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.10, No.3, September 2023 DOI : 10.5121/mathsj.2023.10301 1 A POSITIVE INTEGER 𝑵 SUCH THAT 𝒑𝒏 + 𝒑𝒏+𝟑 ~ 𝒑𝒏+𝟏 + 𝒑𝒏+𝟐 FOR ALL 𝒏 ≥ 𝑵 Pham Minh Duc, Luong Van Tam VNU University of Science, 334, Nguyen Trai, Thanh Xuan Trung Ward, Thanh Xuan District, Hanoi, Vietnam ABSTRACT According to Bertrand's postulate,we have 𝑝𝑛 + 𝑝𝑛 ≥ 𝑝𝑛+1. Is it true that for all 𝑛 > 1 then 𝑝𝑛−1 + 𝑝𝑛 ≥ 𝑝𝑛+1? Then 𝑝𝑛 + 𝑝𝑛+3 > 𝑝𝑛+1 + 𝑝𝑛+2where 𝑛 ≥ 𝑁, 𝑁 is a large enough value? KEYWORDS Bertrand’s Postulate, Rosser's theorem, L’Hospital Rule, prime number. 1. INTRODUCTION In 1845, Bertrand conjectured what became known as Bertrand’s postulate: twice any prime strictly exceeds the next prime. Tchebichef presented his proof of Bertrand’s postulate in 1850 and published it in 1852. It is now sometimes called the Bertrand-Chebyshev theorem. Surprisingly, a stronger statement seems not to be well known, but is elementary to prove: The sum of any two consecutive primes strictly exceeds the next prime, except for the only equality 2 + 3 = 5. After I conjectured and proved this statement independently, a very helpful referee pointed out that Ishikawa published this result in 1934 (with a different proof). This observation is a special case of a much more general result, Theorem 1, that is also elementary to prove (given the prime number theorem), and perhaps not previously noticed: If𝑝𝑛 denotes the 𝑛th prime, 𝑛 = 1, 2, 3, … with 𝑝1 = 2, 𝑝2 = 3, 𝑝3 = 5, …, and then there exists a positive integer 𝑁 such that 𝑝𝑛 + 𝑝𝑛+3 ~ 𝑝𝑛+1 + 𝑝𝑛+2for ll 𝑛 ≥ 𝑁. We give the following result. 2. MAIN RESULT Theorem 1. If𝑛 are nonnegative integers, then there exists a large enough positive integer 𝑁 such that, for all 𝑛 ≥ 𝑁, 𝑝𝑛 + 𝑝𝑛+3 ~ 𝑝𝑛+1 + 𝑝𝑛+2. Applying Rosser's theorem for all 𝑛 ≥ 6, we have 𝑛(ln𝑛 + lnln𝑛 − 1) < 𝑝𝑛 < 𝑛(ln𝑛 + lnln𝑛) (𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1) − 1] < 𝑝𝑛+1 < (𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1)] (𝑛 + 2)[ln(𝑛 + 2) + lnln(𝑛 + 2) − 1] < 𝑝𝑛+2 < (𝑛 + 2)[ln(𝑛 + 2) + lnln(𝑛 + 2)] (𝑛 + 3)[ln(𝑛 + 3) + lnln(𝑛 + 3) − 1] < 𝑝𝑛+3 < (𝑛 + 3)[ln(𝑛 + 3) + lnln(𝑛 + 3)] Consider the expression
- 2. Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.10, No.3, September 2023 2 𝐴 = 𝑛(ln𝑛 + lnln𝑛 − 1) + (𝑛 + 3)[ln(𝑛 + 3) + lnln(𝑛 + 3) − 1] (𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1)] + (𝑛 + 2)[ln(𝑛 + 2) + lnln(𝑛 + 2)] We consider the following limit 𝐵 = lim 𝑛→+∞ ln𝑛 + lnln𝑛 + 1 ln𝑛 + ln(𝑛 + 3) + lnln(𝑛 + 3) + 1 ln(𝑛+3) ln(𝑛 + 1) + lnln(𝑛 + 1) + 2 + 1 ln(𝑛+1) + ln(𝑛 + 2) + lnln(𝑛 + 2) + 1 ln(𝑛+2) Taking the ln of the numerator and denominator and applying L’Hospital Rule gives lim 𝑛→+∞ 𝑛(ln𝑛 + lnln𝑛 − 1) = lim 𝑛→+∞ ln𝑛 + lnln𝑛 − 1 + 𝑛 ( 1 𝑛 + 1 𝑛 ln𝑛 ) = lim 𝑛→+∞ ln𝑛 + lnln𝑛 + 1 ln𝑛 lim 𝑛→+∞ 𝑛 [ln(𝑛 + 3) + lnln(𝑛 + 3) − 1] = lim 𝑛→+∞ ln(𝑛 + 3) + lnln(𝑛 + 3) − 1 + (𝑛 + 3) ( 1 𝑛 + 3 + 1 𝑛+3 ln(𝑛 + 3) ) = lim 𝑛→+∞ ln(𝑛 + 3) + lnln(𝑛 + 3) + 1 ln(𝑛 + 3) lim 𝑛→+∞ 𝑛 [ln(𝑛 + 1) + lnln(𝑛 + 1)] = lim 𝑛→+∞ ln(𝑛 + 1) + lnln(𝑛 + 1) + (𝑛 + 1) ( 1 𝑛 + 1 + 1 𝑛+1 ln(𝑛 + 1) ) = lim 𝑛→+∞ ln(𝑛 + 1) + lnln(𝑛 + 1) + 1 ln(n + 1) + 1 lim 𝑛→+∞ 𝑛 [ln(𝑛 + 2) + lnln(𝑛 + 2)] = lim 𝑛→+∞ ln(𝑛 + 2) + lnln(𝑛 + 2) + (𝑛 + 2) ( 1 𝑛 + 2 + 1 𝑛+2 ln(𝑛 + 2) ) = lim 𝑛→+∞ ln(𝑛 + 2) + lnln(𝑛 + 2) + 1 ln(n + 2) + 1 Then we see 𝐵 = lim 𝑛→+∞ ln𝑛 + lnln𝑛 + 1 ln𝑛 + ln(𝑛 + 3) + lnln(𝑛 + 3) + 1 ln(𝑛+3) ln(𝑛 + 1) + lnln(𝑛 + 1) + 2 + 1 ln(𝑛+1) + ln(𝑛 + 2) + lnln(𝑛 + 2) + 1 ln(𝑛+2)
- 3. Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.10, No.3, September 2023 3 When 𝑛 → +∞ then 𝐵 = lim 𝑛→+∞ ln𝑛 + ln(𝑛 + 3) ln(𝑛 + 1) + ln(𝑛 + 2) = lim 𝑛→+∞ ln(𝑛2 + 3𝑛) ln(𝑛2 + 3𝑛 + 2) = 1 (Because 𝑛2 + 3𝑛 ~ 𝑛2 + 3𝑛 + 2, for 𝑛 → +∞) Or, for 𝑛 ≥ 𝑁, 𝑁 is a large enough positive integer, then𝐴 ~ 1, 𝑛(ln𝑛 + lnln𝑛 − 1) + (𝑛 + 3)[ln(𝑛 + 3) + lnln(𝑛 + 3) − 1] (𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1)] + (𝑛 + 2)[ln(𝑛 + 2) + lnln(𝑛 + 2)] ~ 1 It turns out, 𝑝𝑛+𝑝𝑛+3 𝑝𝑛+1+𝑝𝑛+2 ~1, or have mean is 𝑝𝑛 + 𝑝𝑛+3 ~ 𝑝𝑛+1 + 𝑝𝑛+2. 3. CONCLUSION In short, that is also elementary to prove (given the prime number theorem), and perhaps not previously noticed: If pn denotes the nth prime, n = 1,2,3,... with p1 = 2, p2 = 3, p3 = 5,..., and then there exists a positive integer N such that pn + pn+3 ~ pn+1 + pn+2 for 11n ≥ N. But can not confirm that pn + pn+3 = pn+1 + pn+2 where n ≥ N, N is a large enough value? ACKNOWLEDGEMENT I thank VNU University of Science for accompanying me. REFERENCES [1] J. Bertrand, “M´emoire sur le nombre de valeurs que peut prendre une fonction quand on y permute les lettres qu’elle renferme”, Journal de l’Ecole Royale Polytechnique Cahier 30, 18 (1845), 123– 140. [2] P. Tchebichef, “M´emoire sur les nombres premiers”, Journal de math´ematiques pures et appliqu´ees series 1, 17 (1852), 366–390. [3] H. Ishikawa, “1934 Ueber die Verteilung der Primzahlen”, Science Reports of the Tokyo Bunrika Daigaku, Section A 2 (1934) (31/32), 27-40. [4] J. B. Rosser, and L. Schoenfeld, Lowell, “Approximate formulas for some functions of prime numbers”, Illinois Journal of Mathematics 6 (1962) (1), 64-94. [5] R. P. Brent, J. van de Lune, H. J. J. te Riele, and D. T. Winter, On the zeros of the Riemann zeta function in the critical strip. II, Math. Comp. 39 (1982), no. 160, 681–688. [6] M. Cipolla, La determinazione assintotica dell’𝑛𝑖𝑚𝑜 numero primo, Matematiche Napoli 3 (1902), 132-166. [7] J. van de Lune, H. J. J. te Riele, and D. T. Winter, On the zeros of the Riemann zeta function in the critical strip. IV, Math. Comp. 46 (1986), no. 174, 667–681. [8] Jean-Pierre Massias and Guy Robin, Bornes effectives pour certaines fonctions concernant les nombres premiers, J. Théor. Nombres Bordeaux 8 (1996), no. 1, 215–242. [9] Guy Robin, Estimation de la fonction de Tchebychef 𝜃 sur le 𝑘-ième nombre premier et grandes valeurs de la fonction 𝑛 log 𝑛 nombre de diviseurs premiers de 𝑛, Acta Arith. 42 (1983), no. 4, 367– 389. [10] J. B. Rosser, The n-th prime is greater than 𝑛 log 𝑛, Proc. London Math. Soc. (2) 45 (1939), 21-44. [11] Leonard Eugene Dickson, New First Course in the Theory of Equations, John Wiley & Sons, Inc., New York, 1939. [12] J. Barkley Rosser and Lowell Schoenfeld, Approximate formulas for some functions of prime numbers, Illinois J. Math. 6 (1962), 64–94. [13] Wilhelm Wirtinger, Über eine Minimalaufgabe im Gebiete der analytischen Funktionen von mehreren Veränderlichen, Monatsh. Math. Phys. 47 (1939), 426–431.
- 4. Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.10, No.3, September 2023 4 [14] Wilhelm Wirtinger, Über eine Minimalaufgabe im Gebiete der analytischen Funktionen von mehreren Veränderlichen, Monatsh. Math. Phys. 47 (1939), 426–431. [15] Pierre Dusart (1999). "The kth prime is greater than k(log k + log log k−1) for k ≥ 2". Mathematics of Computation. 68 (225): 411–415.