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  • Let $a(n)$ be A057326 (i.e., first member of a prime triple in a $2p-1$ progression).
  • Let $b(n) = B$ after $n-1$ iterations where we start with $A=n, B=1$ and for $i$ from $1$ to $n-1$ simultaneously apply $$ A:=A+\gcd(A+B, A-n+i), B := B + \gcd(A, B+i-1). $$

I conjecture that $k$ is an odd term of $a(n)$ iff $k>18$ and $b(k)=k$.

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

b(n) = my(A = n, B = 1, C); for(i=1, n-1, C = B; [A, B] = [A + gcd(A+B, A-n+i), B + gcd(A, B+i-1)]; if(B-C>1, break)); B
my(x=1, z=1); for(k=1, 299, while(!(x%2 && isprime(x) && isprime(2*x-1) && isprime(4*x-3)), x++); while(!(z>18 && z%2 && b(z)==z), z++); print(x==z); x++; z++;)

Conjecture was verified up to first $1000$ odd terms of $a(n)$ (i.e., up to $a(1001) = 799171$) with no counterexamples.

Is there a way to prove it?

Improve this question
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    $\begingroup$ Similar to mathoverflow.net/q/470505 ? $\endgroup$
    – Max Alekseyev
    Commented Oct 16 at 4:12
  • $\begingroup$ @MaxAlekseyev, thank you for comment! In this question, there are two different complicated pairs for $\gcd$ instead of one simple pair for $\gcd$ in the question you mentioned. $\endgroup$
    – Notamathematician
    Commented Oct 16 at 5:37

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