Let $a(n)$ be A106483 (i.e., primes $p$ such that $2p^2-1$ is also prime).

Let $b(n)$ be an integer sequence such that $b(n) = B$ after the whole transformation where we start with $A = n$, $B = 1$, $C$ and then for $i$ from $1$ to $n-1$ apply $C := \gcd(A, B)$, $A := A + 2n - C$, $B := B + C$.

I conjecture that odd number $k$ belongs to $a(n)$ iff $b(2k+1)=6k$.

Here is the PARI/GP program to check it numerically:

b(n) = my(A = n, B = 1); for(i=1, n-1, C = gcd(A, B); A += 2*n - C; B += C); B
my(x=3, z=3); for(k=1,299, while(!(isprime(x) && isprime(2*x^2-1)), x++); while(!(z%2 && b(2*z+1)==6*z), z++); print(x==z); x++; z++;);

Is there a way to prove it?

• Let $a(n)$ be A106483 (i.e., primes $p$ such that $2p^2-1$ is also prime).

• Let $b(n)$ be an integer sequence such that $b(n) = B$ after the whole transformation where we start with $A = n$, $B = 1$, $C$ and then for $i$ from $1$ to $n-1$ apply $C := \gcd(A, B)$, $A := A + 2n - C$, $B := B + C$.

I conjecture that odd number $k$ belongs to $a(n)$ iff $b(2k+1)=6k$.

Here is the PARI/GP program to check it numerically:

b(n) = my(A = n, B = 1, C); for(i=1, n-1, C = gcd(A, B); A += 2*n - C; B += C); B
my(x=3, z=3); for(k=1,299, while(!(isprime(x) && isprime(2*x^2-1)), x++); while(!(z%2 && b(2*z+1)==6*z), z++); print(x==z); x++; z++;);

Conjecture was verified with the given program up to $a(1000)=58901$ with no counterexamples.

Is there a way to prove it?