Let \(a_{0}=p\), where \(p\) is a prime number and

\(a_{n+1}=2a_{n}+1\), for \(n=0,1,2...\)

Does there exist a value of \(p\) such that the sequence consists entirely of prime numbers.

Let \(a_{0}=p\), where \(p\) is a prime number and

\(a_{n+1}=2a_{n}+1\), for \(n=0,1,2...\)

Does there exist a value of \(p\) such that the sequence consists entirely of prime numbers.

No vote yet

2 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestYou can surely have sequence of limited numbers which follows your desired pattern as if a(0) = 2, a(1) = 5, a(2) = 11, a(3) = 23, a(4) = 47. But can't obtain infinite sequence as it will break down as soon as unit's digit come to be 5.

As you can observe that prime numbers (except 2 & 5) has their units digits as 1, 3, 7, 9. Now suppose you choose some prime with their unit's digit as 1, 3, 7, 9.

a(0) = 31, a(1) = 63 [breakdown but sometimes number ending with 3 can be a prime number]

a(0) = 53, a(1) = 107, a(2) = 215 [sure breakdown as once you got a number a(n) ending with 7 then the number a(n+1) is ending with 5 and surely composite number.]

You can see examples 7 and 9 on your own according to this observation.

Thank You! – Shubham Kumar · 3 years, 8 months ago

Log in to reply

Thinking differently!!! Good I appreciate your thinking. – Shubham Kumar · 3 years, 8 months ago

Log in to reply