# Can the number of terms of sequence $p_1$, $2p_1+1$, $2(2p_1+1)+1$, . . . be more than 3?

Can the number of terms of sequence $$p_1$$, $$2p_1+1$$, $$2(2p_1+1)+1$$, . . . be more than 3? Where all terms are prime.

For primes $$p_1\equiv 1 mod 3$$ we have:

$$2p_1+1\equiv (2+1) mod 3 \equiv 0 mod 3$$

So eligible primes to produce new primes must be of the form $$3k-1$$ such that:

$$p_1\equiv -1 mod 3$$$$p_2=2p_1+1 \equiv (-2+1=-1) mod 3$$

$$p_3=2(2p_2)+1\equiv -1 mod 3$$

And it will give numbers $$p_n\equiv -1 mod 3$$ if we continue.

So we expect number of terms being prime not limited.But I could not find number of terms more than 3. here are some examples:

$$p_1=11$$$$p_2=2\times 11+1=23$$$$p_3=2\times 23+1=47$$ $$p_4=2\times 47=95$$

$$p_1=41$$$$p_2=2\times 41+1=83$$$$2\times 83+1=167$$$$2\times 167+1=235$$

My conjecture is that the number of terms of such sequence can not be more than 3, Can some one gives a counter example?

• $5, 11, 23, 47$? They are related to Sophie-Germain/Safe Primes and Cunningham Chains. Oct 26, 2020 at 16:23
• Please do not edit your question in such a way that invalidates answers after they have been provided. Instead, if you have another question, use the "Ask Question" button to ask that question. Oct 26, 2020 at 21:47

Define sequence $$\{p_n\}_{n\in\mathbb{N}}$$ as follows: $$p_1=q$$ where $$q$$ is some prime number and $$p_{n+1}=2p_n+1$$ for $$n\geq 1$$. The question is whether there is a prime $$q$$ such that all $$p_n$$ are prime numbers (for all $$n\in\mathbb{N}$$). We claim that there is no such $$q$$.

Suppose the contrary. Firstly, we will obtain explicit formula for $$p_n$$. Indeed, note that $$p_{n+1}+1=2(p_n+1)$$ for $$n\geq 1$$, so $$p_{n+1}+1=2(p_n+1)=2^2(p_{n-1}+1)=\ldots=2^n(p_1+1)=(q+1)\cdot 2^n.$$ Hence, $$p_{n}=(q+1)\cdot 2^{n-1}-1$$ for $$n\geq 1$$.

Now if $$q>2$$, we will consider $$p_{q}$$. Note that $$p_{q}=(q+1)\cdot 2^{q-1}-1\equiv 2^{q-1}-1\equiv 0\pmod q$$ due to Fermat's Little Theorem. However, $$p_q$$ is prime, so $$p_q=q$$ but this impossible because $$q>1$$.

It remains to consider case $$q=2$$. In this case, $$p_n=3\cdot 2^{n-1}-1$$, so $$p_6=3\cdot 32-1=95=5\cdot 19$$ is not prime.

Thus, there is no such $$q$$ and such a sequence cannot be infinite.

• Minor error: the sequence for $q = 2$ goes $2, 5, 11, 23, 47, 95, \dots$; there's no 25. Oct 26, 2020 at 18:35
• Thanks, edited. Oct 26, 2020 at 21:00

The previous question was can the number of terms in the sequence be more than $$3$$:

The comment has answered the question by giving an example.

Out of curiosity, I have written a simple code to print out some of such sequences:

import math
def checkprime(num):
if num > 1:
for i in range(2,int(math.sqrt(num))+1):
if (num % i) == 0:
return False
break
else:
return True

print([2, 5, 11, 23, 47])
print([5, 11, 23, 47])
for i in range(29, 10000,30):
tmp = []
j = i
while checkprime(j):
tmp.append(j)
j = 2*j + 1
if len(tmp) > 3:
print(tmp)

Here are some of the prime sequences satisfying the condition:

[2, 5, 11, 23, 47]
[5, 11, 23, 47]
[89, 179, 359, 719, 1439, 2879]
[179, 359, 719, 1439, 2879]
[359, 719, 1439, 2879]
[509, 1019, 2039, 4079]
[1229, 2459, 4919, 9839]
[1409, 2819, 5639, 11279]
[2699, 5399, 10799, 21599]
[3539, 7079, 14159, 28319]
[6449, 12899, 25799, 51599]
• Note that primes ending in the digits $1,3,7$ necessarily start a chain that yields a number ending in $5$, which is not prime and terminates the chain. Longer chains are possible only for primes ending in the digit $9$. Oct 26, 2020 at 17:15
• thanks for the input, from the chinese remainder theorem, besides the chain starting from $2$ and $5$, the rest must begins with $29 \pmod{30}$ then. Oct 26, 2020 at 17:31