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Tackling Andrica's Conjecture - Part 3

, 27 May 2013 CPOL
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Tackling Andrica's conjecture - Part 3


Here is an interesting, but slightly detached from the previous two articles, result ...

Let's look at the following two sequences:

$$a_{n}=\sqrt{p_{n+1}} - \sqrt{p_{n}}$$ $$b_{n}=ln\left ( \frac{1+\sqrt{p_{n+1}}}
{1+\sqrt{p_{n}}} \right )^{1+\sqrt{p_{n+1}}}$$

Here is a short Python code to visualise the sequences:

import math
import matplotlib.pyplot as plt

primes = []

def isPrime(n):
    l = int(math.sqrt(n)) + 1
    for i in xrange(2,l):
        if (n % i) == 0:
            return False
    return True

def calculateLog(sqrt_p1, sqrt_p2):
    sqrt_1_p2 = 1.0 + sqrt_p2
    r = math.log(sqrt_1_p2/(1.0 + sqrt_p1))
    return r * sqrt_1_p2

N = 1500000

print "populate primes ..."
for i in xrange(2, N):
    if isPrime(i):

sqrt_diff = [] # sqrt diffs
diff = []      # simple diffs
log_calcs = [] # log calcs
x = []
for i in xrange(1, len(primes)):
    sqrt_p2 = math.sqrt(primes[i])
    sqrt_p1 = math.sqrt(primes[i-1])
    sqrt_diff.append(sqrt_p2 - sqrt_p1)
    diff.append(primes[i] - primes[i-1])
    log_calcs.append(calculateLog(sqrt_p1, sqrt_p2))

for i in xrange(len(sqrt_diff)):
    print sqrt_diff[i]," = s(",primes[i+1],") - s(",primes[i],")"

plt.plot(x, sqrt_diff)
plt.plot(x, log_calcs)
plt.hist(diff, 1000)

And here is how both sequences look like ($a_{n}$ the first and $b_{n}$ the second):

Quite asymptotic, aren't they? Indeed they are ...

Lemma 3. $$\sqrt{p_{n+1}} - \sqrt{p_{n}} \leq ln\left ( \frac{1+\sqrt{p_{n+1}}}
{1+\sqrt{p_{n}}} \right )^{1+\sqrt{p_{n+1}}} \leq \left ( \frac{1+\sqrt{p_{n+1}}}
{1+\sqrt{p_{n}}} \right )\cdot \left ( \sqrt{p_{n+1}} - \sqrt{p_{n}} \right )$$

Let's look at this function:

$f_{6}(x)=\frac{\sqrt{p_{n+1}}}{1+x\cdot \sqrt{p_{n+1}}}$. Obviously, ${ln\left
 ( 1+x\cdot \sqrt{p_{n+1}} \right )}'=f_{6}(x)$. As a result $$\int_{\sqrt{\frac{p_
{n}}{p_{n+1}}}}^{1} f_{6}\left ( x \right )dx = ln\left ( 1+x\cdot \sqrt{p_{n+1}}
 \right )|_{\sqrt{\frac{p_{n}}{p_{n+1}}}}^{1}=ln\left ( \frac{1+\sqrt{p_{n+1}}}
{1+\sqrt{p_{n}}} \right )$$

According to Mean Value Theorem,

$\exists \mu \in \left (\sqrt{\frac{p_{n}}{p_{n+1}}} ,1 \right )$ such that: $$\int_
{\sqrt{\frac{p_{n}}{p_{n+1}}}}^{1} f_{6}\left ( x \right )dx = f_{6}\left ( \mu 
 \right )\cdot \left ( 1- \sqrt{\frac{p_{n}}{p_{n+1}}} \right )$$

Putting all together:

$$ln\left ( \frac{1+\sqrt{p_{n+1}}}{1+\sqrt{p_{n}}} \right ) = \frac{\sqrt{p_{n+1}}
-\sqrt{p_{n}}}{1+\mu \cdot \sqrt{p_{n+1}}}$$


$$\sqrt{\frac{p_{n}}{p_{n+1}}}< \mu < 1 \Rightarrow 1+\sqrt{p_{n}}< 1+\mu \cdot \sqrt
{p_{n+1}} < 1+\sqrt{p_{n+1}} $$

And we get:

$$\frac{\sqrt{p_{n+1}}-\sqrt{p_{n}}}{1+\sqrt{p_{n+1}}}\leq ln\left ( \frac{1+\sqrt{p_
{n+1}}}{1+\sqrt{p_{n}}} \right )\leq \frac{\sqrt{p_{n+1}}-\sqrt{p_{n}}}{1+\sqrt{p_

which proves this lemma.


$\Delta_{n}=\sqrt{p_{n+1}}-\sqrt{p_{n}}$, this becomes: $$\frac{\Delta_{n}}{1+\sqrt{p_
{n+1}}}\leq ln\left ( 1 + \frac{\Delta_{n}}{1+\sqrt{p_{n}}} \right )\leq \frac
{\Delta_{n}}{1+\sqrt{p_{n}}}$$ or $$\frac{1+\sqrt{p_{n}}}{1+\sqrt{p_{n+1}}}\leq
 ln\left ( 1 + \frac{\Delta_{n}}{1+\sqrt{p_{n}}} \right )^{\frac{1+\sqrt{p_{n}}}
{\Delta_{n}}}\leq 1$$

Is this result of any use? I don't know yet, but it looks like:

$$\left ( 1 + \frac{\Delta_{n}}{1+\sqrt{p_{n}}} \right )^{\frac{1+\sqrt{p_{n}}}
{\Delta_{n}}} \rightarrow e, n \to \infty$$


This article, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)


About the Author

Software Developer (Senior) Snappli Ltd.
United Kingdom United Kingdom
My name is Ruslan Ciurca. Currently I am a Software Developer at

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