Are undulating prime numbers(literally, prime numbers that can be represented as abababab...aba, such as 919 or 131) endless or limited? Is this problem improvable?
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2See OEIS and wikipedia. – Dietrich Burde Jan 24 '24 at 21:13
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1There are only finitely many. See the argument and comment here – Brevan Ellefsen Jan 24 '24 at 21:14
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3@BrevanEllefsen: That argument doesn't apply here. In base $2$, the second digit (labelled $b$ in the present question) must be $0$, and that forces the undulating number to be all $1$s in base $4$. That's not the case here; the second digit in base $10$ is arbitrary, and thus the number isn't the repetition of a single digit in base $100$. The comment you refer to doesn't cover this case. – joriki Jan 24 '24 at 21:45
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@joriki the way that answer gets the formula is incorrect in base 10, but there still exist formulas of the same type in all dimensions that one can work out pretty immediately after seeing the base 2 version. As noted on Wikipedia, there are really only two cases for an undulating number with digits $A, B$: either it's $AB\frac{10^{2n}-1}{99}$ or it's $\frac{AB10^{2n+1}-BA}{99}$ (depending on the last digit). Once one sees these formulas, it's obvious to see how they generalize (and why they collapse to a single case in base 2). W+The argument I linked is the essential idea for case 1. – Brevan Ellefsen Jan 25 '24 at 11:18
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1@BrevanEllefsen: Are you saying that the form $\frac{AB10^{2n+1}-BA}{99}$ is also necessarily composite? If so, why? – joriki Jan 25 '24 at 11:25
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No, of course not. That's where the finitely many primes come from. I didn't mean to imply that the linked post is a complete solution (it's poorly written anyhow), merely that the idea reduces one of the cases in base 10 and is a starting point for the OP to try some things and show work. – Brevan Ellefsen Jan 25 '24 at 12:00
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2@BrevanEllefsen: But then from what are you concluding that there are only finitely many such primes? – joriki Jan 25 '24 at 12:52
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The case $1717171\cdots 7171$ , for example , gives a large proven prime. That does not speak for finite many such primes. The usual composite proofs (cover or algebraic/auriferuillan factors) apparently do not apply here. If $a=b=1$ is allowed (I guess it is not) , then there is a conjecture anyway that there are infinite many such primes. – Peter Jan 25 '24 at 13:39
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A proof of infiitie many primes of any such forms is , on the other hand , almost surely out of reach. We do not even know this for simpler expressions like $n^2+1$ – Peter Jan 25 '24 at 13:42
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an even larger (this time probable) prime – Peter Jan 25 '24 at 16:20
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@Peter: Could you give a reference for the conjecture for $a=b=1$? (I didn't find it in this thread.) – joriki Jan 25 '24 at 18:11