Why did this prime-sequence puzzle not work?





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While attacking a recent puzzle (whose solution ended up being entirely different from what I was trying), I was inspired to create a number-sequence puzzle with a sequence $(p_n)$ of primes where the secret rule is




$p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.




Unfortunately, I couldn't turn this into a reasonable puzzle. So I'm posting a meta-puzzle about it instead: can you see why this wouldn't make a good puzzle? Specifically, why couldn't I generate a good sequence to put in the question?










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  • 2




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    It was great to see the teamwork on this one, as people inched closer to the full realisation/solution in successive answers :-)
    $endgroup$
    – Rand al'Thor
    May 27 at 12:57


















5












$begingroup$


While attacking a recent puzzle (whose solution ended up being entirely different from what I was trying), I was inspired to create a number-sequence puzzle with a sequence $(p_n)$ of primes where the secret rule is




$p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.




Unfortunately, I couldn't turn this into a reasonable puzzle. So I'm posting a meta-puzzle about it instead: can you see why this wouldn't make a good puzzle? Specifically, why couldn't I generate a good sequence to put in the question?










share|improve this question









$endgroup$










  • 2




    $begingroup$
    It was great to see the teamwork on this one, as people inched closer to the full realisation/solution in successive answers :-)
    $endgroup$
    – Rand al'Thor
    May 27 at 12:57














5












5








5





$begingroup$


While attacking a recent puzzle (whose solution ended up being entirely different from what I was trying), I was inspired to create a number-sequence puzzle with a sequence $(p_n)$ of primes where the secret rule is




$p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.




Unfortunately, I couldn't turn this into a reasonable puzzle. So I'm posting a meta-puzzle about it instead: can you see why this wouldn't make a good puzzle? Specifically, why couldn't I generate a good sequence to put in the question?










share|improve this question









$endgroup$




While attacking a recent puzzle (whose solution ended up being entirely different from what I was trying), I was inspired to create a number-sequence puzzle with a sequence $(p_n)$ of primes where the secret rule is




$p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.




Unfortunately, I couldn't turn this into a reasonable puzzle. So I'm posting a meta-puzzle about it instead: can you see why this wouldn't make a good puzzle? Specifically, why couldn't I generate a good sequence to put in the question?







mathematics puzzle-creation number-theory






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asked May 27 at 12:00









Rand al'ThorRand al'Thor

76.5k15 gold badges252 silver badges505 bronze badges




76.5k15 gold badges252 silver badges505 bronze badges











  • 2




    $begingroup$
    It was great to see the teamwork on this one, as people inched closer to the full realisation/solution in successive answers :-)
    $endgroup$
    – Rand al'Thor
    May 27 at 12:57














  • 2




    $begingroup$
    It was great to see the teamwork on this one, as people inched closer to the full realisation/solution in successive answers :-)
    $endgroup$
    – Rand al'Thor
    May 27 at 12:57








2




2




$begingroup$
It was great to see the teamwork on this one, as people inched closer to the full realisation/solution in successive answers :-)
$endgroup$
– Rand al'Thor
May 27 at 12:57




$begingroup$
It was great to see the teamwork on this one, as people inched closer to the full realisation/solution in successive answers :-)
$endgroup$
– Rand al'Thor
May 27 at 12:57










4 Answers
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Theorem: The number $2^n+1$ is divisible by $3$ for any odd number $n$.
Proof: Set $n=2k+1$. Then modulo $3$ we have $$2^n+1 = 2^{2k+1}+1 = 2cdot 4^k+1 equiv 2cdot1^k+1 equiv 0 mod 3$$

which proves the theorem.


Since $2^n+1$ is odd for $n>0$, its lowest factor will be an odd prime. The sequence therefore will become 3 at the second step, and then remain $3$ afterwards.







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  • $begingroup$
    Yep, there we go. That's the key discovery.
    $endgroup$
    – Rand al'Thor
    May 27 at 12:55










  • $begingroup$
    Did you nip my proof? Well played!
    $endgroup$
    – El-Guest
    May 27 at 13:05










  • $begingroup$
    @El-Guest When I started writing my proof, I don't think you had cottoned on to the fact that the factor 3 occurs at all odd exponents, but by the time I submitted you had and were working on a proof yourself.
    $endgroup$
    – Jaap Scherphuis
    May 27 at 13:23










  • $begingroup$
    @JaapScherphuis I would’ve beaten you if I wasn’t so hell-bent on using induction haha! Well done on the modular arithmetic, your proof is much cleaner than mine. +1
    $endgroup$
    – El-Guest
    May 27 at 13:27



















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$p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.




Let’s try something:




Let $p_0 = 2$. Then $p_1$ is the lowest prime factor of $2^2 + 1 = 5$, so 5. Then $p_2$ is the lowest prime factor of $2^5 +!1 = 33$, so 3. Then $p_3$ is the lowest prime factor of $2^3 + 1 = 9$....which is 3 again! Looks like this sequence has an infinite repetition and converges to 3...




The question is,




Is this true for any $p_0$? The fact of the matter is that we have a countable set of $p_n$, because the number of $p_n$ must be less than the number of integers $2^x + 1$ there are — and this is one-to-one with the natural numbers $mathbb{N}$. It seems logical that at some point in the sequence of $p_n$, you’ll find a number divisible by 3. To avoid this, you’d essentially need to find a series of primes — quasi-Mersenne — where each continuing $2^{p_n} + 1$ is also prime. I don’t think that such a sequence exists...




Sample starting points:




$p_0 = 2$. 2, 5, 3, 3, 3, ...

(My favourite) $p_0 = 3$. 3, 3, 3, 3, 3, ...
$p_0 = 5$. 5, 3, 3, 3, 3, ...
$p_0 = 7$. 7, 3, 3, 3, 3, ... (because of 129)
$p_0 = 11$. 11, 3, 3, 3, 3, ... (because of 2049)....




Hmm....




Does every $p_0 > 2$ converge to 3 within one step? @WeatherVane found that this is at least the case for $3 leq p_0 leq 61$. This is essentially saying for prime $p$, does $3 | 2^p + 1$? Further, for any odd natural number 2m+1, does $3 | 2^{2m+1} + 1$? Let’s see...this is true iff $3 | 2^{2m+1} - 2 = 2( 2^{2m} - 1)$ iff $3 | 2^{2m} - 1$ iff $3 | 2^{2m} - 4$.... commences rambling...




Let’s use




Induction. We seem to have exhaustively proved base cases above. Then, assume that for some odd integer $n = 2m + 1$, $3 | 2^n + 1$. Then consider $p$ = 2m + 3, the next odd integer. Does $3 | 2^p + 1 = 2^{2m+3} + 1 = 4(2^{2m+1}) + 1$? Well, if $3 | 2^n + 1 = 2^{2m+1} + 1$, then $3 | 4(2^{2m+1} + 1)$. So $3 | 4(2^{2m+1}) + 4$. It must therefore also divide 3 less than that, and so $3 | 4(2^{2m+1}) + 1 = 2^{2m+3} + 1 = 2^p + 1$. We have therefore shown that the inductive hypothesis does indeed imply the inductive conclusion. By induction, therefore, $3|2^n+1$ for any positive odd integer $n$.




It naturally follows that




Any starting prime number greater than 2 is a positive odd integer, and so for any $p_0 > 2$, we must have $p_n = 3 forall n in mathbb{N}$. Since we have shown that $p_0 = 2$ converges to 3 within 2 steps, for any prime $p_0$, $p_n = 3 forall n geq 2$. This is why there is no sequence (not even the hypothetical prime ladder postulated above) which is suitable for puzzling, because every prime sequence (without restriction) converges to 3 infinitely. $square$







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  • $begingroup$
    Ah, you beat me by 37 seconds....
    $endgroup$
    – tom
    May 27 at 12:12






  • 4




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    Which is a prime number. How suitable.
    $endgroup$
    – Florian F
    May 27 at 12:18












  • $begingroup$
    Please see my amended answer having read yours.
    $endgroup$
    – Weather Vane
    May 27 at 12:43










  • $begingroup$
    Did you just throw it into a computer? Thanks for the computation — good to see that the hypothesis appears correct so far, at least @WeatherVane! Thanks!!
    $endgroup$
    – El-Guest
    May 27 at 12:45










  • $begingroup$
    Yes, the easy ones $ lt 2^{64}$. But actually, there is not always just one other prime factor.
    $endgroup$
    – Weather Vane
    May 27 at 12:46





















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I think




It looks as if you would get stuck at the number 3




because




often 3 is going to be the lowest prime factor and $2^3+1=9$ which again gives 3







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    2














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    Unsure... partial




    Because some $2^{x} + 1$ are prime numbers: no factors.

    For example $x = 8$




    Edit:




    The above wasn't well thought, because $8$ cannot be a lowest factor.


    But from @El-Guest's answer, I found that


    For every prime $3 le x le 61$, $2^{x} + 1$ is divisible by $3$.







    share|improve this answer











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      4 Answers
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      4 Answers
      4






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      14














      $begingroup$


      Theorem: The number $2^n+1$ is divisible by $3$ for any odd number $n$.
      Proof: Set $n=2k+1$. Then modulo $3$ we have $$2^n+1 = 2^{2k+1}+1 = 2cdot 4^k+1 equiv 2cdot1^k+1 equiv 0 mod 3$$

      which proves the theorem.


      Since $2^n+1$ is odd for $n>0$, its lowest factor will be an odd prime. The sequence therefore will become 3 at the second step, and then remain $3$ afterwards.







      share|improve this answer











      $endgroup$















      • $begingroup$
        Yep, there we go. That's the key discovery.
        $endgroup$
        – Rand al'Thor
        May 27 at 12:55










      • $begingroup$
        Did you nip my proof? Well played!
        $endgroup$
        – El-Guest
        May 27 at 13:05










      • $begingroup$
        @El-Guest When I started writing my proof, I don't think you had cottoned on to the fact that the factor 3 occurs at all odd exponents, but by the time I submitted you had and were working on a proof yourself.
        $endgroup$
        – Jaap Scherphuis
        May 27 at 13:23










      • $begingroup$
        @JaapScherphuis I would’ve beaten you if I wasn’t so hell-bent on using induction haha! Well done on the modular arithmetic, your proof is much cleaner than mine. +1
        $endgroup$
        – El-Guest
        May 27 at 13:27
















      14














      $begingroup$


      Theorem: The number $2^n+1$ is divisible by $3$ for any odd number $n$.
      Proof: Set $n=2k+1$. Then modulo $3$ we have $$2^n+1 = 2^{2k+1}+1 = 2cdot 4^k+1 equiv 2cdot1^k+1 equiv 0 mod 3$$

      which proves the theorem.


      Since $2^n+1$ is odd for $n>0$, its lowest factor will be an odd prime. The sequence therefore will become 3 at the second step, and then remain $3$ afterwards.







      share|improve this answer











      $endgroup$















      • $begingroup$
        Yep, there we go. That's the key discovery.
        $endgroup$
        – Rand al'Thor
        May 27 at 12:55










      • $begingroup$
        Did you nip my proof? Well played!
        $endgroup$
        – El-Guest
        May 27 at 13:05










      • $begingroup$
        @El-Guest When I started writing my proof, I don't think you had cottoned on to the fact that the factor 3 occurs at all odd exponents, but by the time I submitted you had and were working on a proof yourself.
        $endgroup$
        – Jaap Scherphuis
        May 27 at 13:23










      • $begingroup$
        @JaapScherphuis I would’ve beaten you if I wasn’t so hell-bent on using induction haha! Well done on the modular arithmetic, your proof is much cleaner than mine. +1
        $endgroup$
        – El-Guest
        May 27 at 13:27














      14














      14










      14







      $begingroup$


      Theorem: The number $2^n+1$ is divisible by $3$ for any odd number $n$.
      Proof: Set $n=2k+1$. Then modulo $3$ we have $$2^n+1 = 2^{2k+1}+1 = 2cdot 4^k+1 equiv 2cdot1^k+1 equiv 0 mod 3$$

      which proves the theorem.


      Since $2^n+1$ is odd for $n>0$, its lowest factor will be an odd prime. The sequence therefore will become 3 at the second step, and then remain $3$ afterwards.







      share|improve this answer











      $endgroup$




      Theorem: The number $2^n+1$ is divisible by $3$ for any odd number $n$.
      Proof: Set $n=2k+1$. Then modulo $3$ we have $$2^n+1 = 2^{2k+1}+1 = 2cdot 4^k+1 equiv 2cdot1^k+1 equiv 0 mod 3$$

      which proves the theorem.


      Since $2^n+1$ is odd for $n>0$, its lowest factor will be an odd prime. The sequence therefore will become 3 at the second step, and then remain $3$ afterwards.








      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited May 27 at 13:24

























      answered May 27 at 12:54









      Jaap ScherphuisJaap Scherphuis

      19.4k1 gold badge34 silver badges84 bronze badges




      19.4k1 gold badge34 silver badges84 bronze badges















      • $begingroup$
        Yep, there we go. That's the key discovery.
        $endgroup$
        – Rand al'Thor
        May 27 at 12:55










      • $begingroup$
        Did you nip my proof? Well played!
        $endgroup$
        – El-Guest
        May 27 at 13:05










      • $begingroup$
        @El-Guest When I started writing my proof, I don't think you had cottoned on to the fact that the factor 3 occurs at all odd exponents, but by the time I submitted you had and were working on a proof yourself.
        $endgroup$
        – Jaap Scherphuis
        May 27 at 13:23










      • $begingroup$
        @JaapScherphuis I would’ve beaten you if I wasn’t so hell-bent on using induction haha! Well done on the modular arithmetic, your proof is much cleaner than mine. +1
        $endgroup$
        – El-Guest
        May 27 at 13:27


















      • $begingroup$
        Yep, there we go. That's the key discovery.
        $endgroup$
        – Rand al'Thor
        May 27 at 12:55










      • $begingroup$
        Did you nip my proof? Well played!
        $endgroup$
        – El-Guest
        May 27 at 13:05










      • $begingroup$
        @El-Guest When I started writing my proof, I don't think you had cottoned on to the fact that the factor 3 occurs at all odd exponents, but by the time I submitted you had and were working on a proof yourself.
        $endgroup$
        – Jaap Scherphuis
        May 27 at 13:23










      • $begingroup$
        @JaapScherphuis I would’ve beaten you if I wasn’t so hell-bent on using induction haha! Well done on the modular arithmetic, your proof is much cleaner than mine. +1
        $endgroup$
        – El-Guest
        May 27 at 13:27
















      $begingroup$
      Yep, there we go. That's the key discovery.
      $endgroup$
      – Rand al'Thor
      May 27 at 12:55




      $begingroup$
      Yep, there we go. That's the key discovery.
      $endgroup$
      – Rand al'Thor
      May 27 at 12:55












      $begingroup$
      Did you nip my proof? Well played!
      $endgroup$
      – El-Guest
      May 27 at 13:05




      $begingroup$
      Did you nip my proof? Well played!
      $endgroup$
      – El-Guest
      May 27 at 13:05












      $begingroup$
      @El-Guest When I started writing my proof, I don't think you had cottoned on to the fact that the factor 3 occurs at all odd exponents, but by the time I submitted you had and were working on a proof yourself.
      $endgroup$
      – Jaap Scherphuis
      May 27 at 13:23




      $begingroup$
      @El-Guest When I started writing my proof, I don't think you had cottoned on to the fact that the factor 3 occurs at all odd exponents, but by the time I submitted you had and were working on a proof yourself.
      $endgroup$
      – Jaap Scherphuis
      May 27 at 13:23












      $begingroup$
      @JaapScherphuis I would’ve beaten you if I wasn’t so hell-bent on using induction haha! Well done on the modular arithmetic, your proof is much cleaner than mine. +1
      $endgroup$
      – El-Guest
      May 27 at 13:27




      $begingroup$
      @JaapScherphuis I would’ve beaten you if I wasn’t so hell-bent on using induction haha! Well done on the modular arithmetic, your proof is much cleaner than mine. +1
      $endgroup$
      – El-Guest
      May 27 at 13:27













      3














      $begingroup$


      $p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.




      Let’s try something:




      Let $p_0 = 2$. Then $p_1$ is the lowest prime factor of $2^2 + 1 = 5$, so 5. Then $p_2$ is the lowest prime factor of $2^5 +!1 = 33$, so 3. Then $p_3$ is the lowest prime factor of $2^3 + 1 = 9$....which is 3 again! Looks like this sequence has an infinite repetition and converges to 3...




      The question is,




      Is this true for any $p_0$? The fact of the matter is that we have a countable set of $p_n$, because the number of $p_n$ must be less than the number of integers $2^x + 1$ there are — and this is one-to-one with the natural numbers $mathbb{N}$. It seems logical that at some point in the sequence of $p_n$, you’ll find a number divisible by 3. To avoid this, you’d essentially need to find a series of primes — quasi-Mersenne — where each continuing $2^{p_n} + 1$ is also prime. I don’t think that such a sequence exists...




      Sample starting points:




      $p_0 = 2$. 2, 5, 3, 3, 3, ...

      (My favourite) $p_0 = 3$. 3, 3, 3, 3, 3, ...
      $p_0 = 5$. 5, 3, 3, 3, 3, ...
      $p_0 = 7$. 7, 3, 3, 3, 3, ... (because of 129)
      $p_0 = 11$. 11, 3, 3, 3, 3, ... (because of 2049)....




      Hmm....




      Does every $p_0 > 2$ converge to 3 within one step? @WeatherVane found that this is at least the case for $3 leq p_0 leq 61$. This is essentially saying for prime $p$, does $3 | 2^p + 1$? Further, for any odd natural number 2m+1, does $3 | 2^{2m+1} + 1$? Let’s see...this is true iff $3 | 2^{2m+1} - 2 = 2( 2^{2m} - 1)$ iff $3 | 2^{2m} - 1$ iff $3 | 2^{2m} - 4$.... commences rambling...




      Let’s use




      Induction. We seem to have exhaustively proved base cases above. Then, assume that for some odd integer $n = 2m + 1$, $3 | 2^n + 1$. Then consider $p$ = 2m + 3, the next odd integer. Does $3 | 2^p + 1 = 2^{2m+3} + 1 = 4(2^{2m+1}) + 1$? Well, if $3 | 2^n + 1 = 2^{2m+1} + 1$, then $3 | 4(2^{2m+1} + 1)$. So $3 | 4(2^{2m+1}) + 4$. It must therefore also divide 3 less than that, and so $3 | 4(2^{2m+1}) + 1 = 2^{2m+3} + 1 = 2^p + 1$. We have therefore shown that the inductive hypothesis does indeed imply the inductive conclusion. By induction, therefore, $3|2^n+1$ for any positive odd integer $n$.




      It naturally follows that




      Any starting prime number greater than 2 is a positive odd integer, and so for any $p_0 > 2$, we must have $p_n = 3 forall n in mathbb{N}$. Since we have shown that $p_0 = 2$ converges to 3 within 2 steps, for any prime $p_0$, $p_n = 3 forall n geq 2$. This is why there is no sequence (not even the hypothetical prime ladder postulated above) which is suitable for puzzling, because every prime sequence (without restriction) converges to 3 infinitely. $square$







      share|improve this answer











      $endgroup$















      • $begingroup$
        Ah, you beat me by 37 seconds....
        $endgroup$
        – tom
        May 27 at 12:12






      • 4




        $begingroup$
        Which is a prime number. How suitable.
        $endgroup$
        – Florian F
        May 27 at 12:18












      • $begingroup$
        Please see my amended answer having read yours.
        $endgroup$
        – Weather Vane
        May 27 at 12:43










      • $begingroup$
        Did you just throw it into a computer? Thanks for the computation — good to see that the hypothesis appears correct so far, at least @WeatherVane! Thanks!!
        $endgroup$
        – El-Guest
        May 27 at 12:45










      • $begingroup$
        Yes, the easy ones $ lt 2^{64}$. But actually, there is not always just one other prime factor.
        $endgroup$
        – Weather Vane
        May 27 at 12:46


















      3














      $begingroup$


      $p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.




      Let’s try something:




      Let $p_0 = 2$. Then $p_1$ is the lowest prime factor of $2^2 + 1 = 5$, so 5. Then $p_2$ is the lowest prime factor of $2^5 +!1 = 33$, so 3. Then $p_3$ is the lowest prime factor of $2^3 + 1 = 9$....which is 3 again! Looks like this sequence has an infinite repetition and converges to 3...




      The question is,




      Is this true for any $p_0$? The fact of the matter is that we have a countable set of $p_n$, because the number of $p_n$ must be less than the number of integers $2^x + 1$ there are — and this is one-to-one with the natural numbers $mathbb{N}$. It seems logical that at some point in the sequence of $p_n$, you’ll find a number divisible by 3. To avoid this, you’d essentially need to find a series of primes — quasi-Mersenne — where each continuing $2^{p_n} + 1$ is also prime. I don’t think that such a sequence exists...




      Sample starting points:




      $p_0 = 2$. 2, 5, 3, 3, 3, ...

      (My favourite) $p_0 = 3$. 3, 3, 3, 3, 3, ...
      $p_0 = 5$. 5, 3, 3, 3, 3, ...
      $p_0 = 7$. 7, 3, 3, 3, 3, ... (because of 129)
      $p_0 = 11$. 11, 3, 3, 3, 3, ... (because of 2049)....




      Hmm....




      Does every $p_0 > 2$ converge to 3 within one step? @WeatherVane found that this is at least the case for $3 leq p_0 leq 61$. This is essentially saying for prime $p$, does $3 | 2^p + 1$? Further, for any odd natural number 2m+1, does $3 | 2^{2m+1} + 1$? Let’s see...this is true iff $3 | 2^{2m+1} - 2 = 2( 2^{2m} - 1)$ iff $3 | 2^{2m} - 1$ iff $3 | 2^{2m} - 4$.... commences rambling...




      Let’s use




      Induction. We seem to have exhaustively proved base cases above. Then, assume that for some odd integer $n = 2m + 1$, $3 | 2^n + 1$. Then consider $p$ = 2m + 3, the next odd integer. Does $3 | 2^p + 1 = 2^{2m+3} + 1 = 4(2^{2m+1}) + 1$? Well, if $3 | 2^n + 1 = 2^{2m+1} + 1$, then $3 | 4(2^{2m+1} + 1)$. So $3 | 4(2^{2m+1}) + 4$. It must therefore also divide 3 less than that, and so $3 | 4(2^{2m+1}) + 1 = 2^{2m+3} + 1 = 2^p + 1$. We have therefore shown that the inductive hypothesis does indeed imply the inductive conclusion. By induction, therefore, $3|2^n+1$ for any positive odd integer $n$.




      It naturally follows that




      Any starting prime number greater than 2 is a positive odd integer, and so for any $p_0 > 2$, we must have $p_n = 3 forall n in mathbb{N}$. Since we have shown that $p_0 = 2$ converges to 3 within 2 steps, for any prime $p_0$, $p_n = 3 forall n geq 2$. This is why there is no sequence (not even the hypothetical prime ladder postulated above) which is suitable for puzzling, because every prime sequence (without restriction) converges to 3 infinitely. $square$







      share|improve this answer











      $endgroup$















      • $begingroup$
        Ah, you beat me by 37 seconds....
        $endgroup$
        – tom
        May 27 at 12:12






      • 4




        $begingroup$
        Which is a prime number. How suitable.
        $endgroup$
        – Florian F
        May 27 at 12:18












      • $begingroup$
        Please see my amended answer having read yours.
        $endgroup$
        – Weather Vane
        May 27 at 12:43










      • $begingroup$
        Did you just throw it into a computer? Thanks for the computation — good to see that the hypothesis appears correct so far, at least @WeatherVane! Thanks!!
        $endgroup$
        – El-Guest
        May 27 at 12:45










      • $begingroup$
        Yes, the easy ones $ lt 2^{64}$. But actually, there is not always just one other prime factor.
        $endgroup$
        – Weather Vane
        May 27 at 12:46
















      3














      3










      3







      $begingroup$


      $p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.




      Let’s try something:




      Let $p_0 = 2$. Then $p_1$ is the lowest prime factor of $2^2 + 1 = 5$, so 5. Then $p_2$ is the lowest prime factor of $2^5 +!1 = 33$, so 3. Then $p_3$ is the lowest prime factor of $2^3 + 1 = 9$....which is 3 again! Looks like this sequence has an infinite repetition and converges to 3...




      The question is,




      Is this true for any $p_0$? The fact of the matter is that we have a countable set of $p_n$, because the number of $p_n$ must be less than the number of integers $2^x + 1$ there are — and this is one-to-one with the natural numbers $mathbb{N}$. It seems logical that at some point in the sequence of $p_n$, you’ll find a number divisible by 3. To avoid this, you’d essentially need to find a series of primes — quasi-Mersenne — where each continuing $2^{p_n} + 1$ is also prime. I don’t think that such a sequence exists...




      Sample starting points:




      $p_0 = 2$. 2, 5, 3, 3, 3, ...

      (My favourite) $p_0 = 3$. 3, 3, 3, 3, 3, ...
      $p_0 = 5$. 5, 3, 3, 3, 3, ...
      $p_0 = 7$. 7, 3, 3, 3, 3, ... (because of 129)
      $p_0 = 11$. 11, 3, 3, 3, 3, ... (because of 2049)....




      Hmm....




      Does every $p_0 > 2$ converge to 3 within one step? @WeatherVane found that this is at least the case for $3 leq p_0 leq 61$. This is essentially saying for prime $p$, does $3 | 2^p + 1$? Further, for any odd natural number 2m+1, does $3 | 2^{2m+1} + 1$? Let’s see...this is true iff $3 | 2^{2m+1} - 2 = 2( 2^{2m} - 1)$ iff $3 | 2^{2m} - 1$ iff $3 | 2^{2m} - 4$.... commences rambling...




      Let’s use




      Induction. We seem to have exhaustively proved base cases above. Then, assume that for some odd integer $n = 2m + 1$, $3 | 2^n + 1$. Then consider $p$ = 2m + 3, the next odd integer. Does $3 | 2^p + 1 = 2^{2m+3} + 1 = 4(2^{2m+1}) + 1$? Well, if $3 | 2^n + 1 = 2^{2m+1} + 1$, then $3 | 4(2^{2m+1} + 1)$. So $3 | 4(2^{2m+1}) + 4$. It must therefore also divide 3 less than that, and so $3 | 4(2^{2m+1}) + 1 = 2^{2m+3} + 1 = 2^p + 1$. We have therefore shown that the inductive hypothesis does indeed imply the inductive conclusion. By induction, therefore, $3|2^n+1$ for any positive odd integer $n$.




      It naturally follows that




      Any starting prime number greater than 2 is a positive odd integer, and so for any $p_0 > 2$, we must have $p_n = 3 forall n in mathbb{N}$. Since we have shown that $p_0 = 2$ converges to 3 within 2 steps, for any prime $p_0$, $p_n = 3 forall n geq 2$. This is why there is no sequence (not even the hypothetical prime ladder postulated above) which is suitable for puzzling, because every prime sequence (without restriction) converges to 3 infinitely. $square$







      share|improve this answer











      $endgroup$




      $p_{n+1}$ is defined as the lowest prime factor of the number $2^{p_n}+1$.




      Let’s try something:




      Let $p_0 = 2$. Then $p_1$ is the lowest prime factor of $2^2 + 1 = 5$, so 5. Then $p_2$ is the lowest prime factor of $2^5 +!1 = 33$, so 3. Then $p_3$ is the lowest prime factor of $2^3 + 1 = 9$....which is 3 again! Looks like this sequence has an infinite repetition and converges to 3...




      The question is,




      Is this true for any $p_0$? The fact of the matter is that we have a countable set of $p_n$, because the number of $p_n$ must be less than the number of integers $2^x + 1$ there are — and this is one-to-one with the natural numbers $mathbb{N}$. It seems logical that at some point in the sequence of $p_n$, you’ll find a number divisible by 3. To avoid this, you’d essentially need to find a series of primes — quasi-Mersenne — where each continuing $2^{p_n} + 1$ is also prime. I don’t think that such a sequence exists...




      Sample starting points:




      $p_0 = 2$. 2, 5, 3, 3, 3, ...

      (My favourite) $p_0 = 3$. 3, 3, 3, 3, 3, ...
      $p_0 = 5$. 5, 3, 3, 3, 3, ...
      $p_0 = 7$. 7, 3, 3, 3, 3, ... (because of 129)
      $p_0 = 11$. 11, 3, 3, 3, 3, ... (because of 2049)....




      Hmm....




      Does every $p_0 > 2$ converge to 3 within one step? @WeatherVane found that this is at least the case for $3 leq p_0 leq 61$. This is essentially saying for prime $p$, does $3 | 2^p + 1$? Further, for any odd natural number 2m+1, does $3 | 2^{2m+1} + 1$? Let’s see...this is true iff $3 | 2^{2m+1} - 2 = 2( 2^{2m} - 1)$ iff $3 | 2^{2m} - 1$ iff $3 | 2^{2m} - 4$.... commences rambling...




      Let’s use




      Induction. We seem to have exhaustively proved base cases above. Then, assume that for some odd integer $n = 2m + 1$, $3 | 2^n + 1$. Then consider $p$ = 2m + 3, the next odd integer. Does $3 | 2^p + 1 = 2^{2m+3} + 1 = 4(2^{2m+1}) + 1$? Well, if $3 | 2^n + 1 = 2^{2m+1} + 1$, then $3 | 4(2^{2m+1} + 1)$. So $3 | 4(2^{2m+1}) + 4$. It must therefore also divide 3 less than that, and so $3 | 4(2^{2m+1}) + 1 = 2^{2m+3} + 1 = 2^p + 1$. We have therefore shown that the inductive hypothesis does indeed imply the inductive conclusion. By induction, therefore, $3|2^n+1$ for any positive odd integer $n$.




      It naturally follows that




      Any starting prime number greater than 2 is a positive odd integer, and so for any $p_0 > 2$, we must have $p_n = 3 forall n in mathbb{N}$. Since we have shown that $p_0 = 2$ converges to 3 within 2 steps, for any prime $p_0$, $p_n = 3 forall n geq 2$. This is why there is no sequence (not even the hypothetical prime ladder postulated above) which is suitable for puzzling, because every prime sequence (without restriction) converges to 3 infinitely. $square$








      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited May 27 at 13:04

























      answered May 27 at 12:10









      El-GuestEl-Guest

      26.4k3 gold badges63 silver badges109 bronze badges




      26.4k3 gold badges63 silver badges109 bronze badges















      • $begingroup$
        Ah, you beat me by 37 seconds....
        $endgroup$
        – tom
        May 27 at 12:12






      • 4




        $begingroup$
        Which is a prime number. How suitable.
        $endgroup$
        – Florian F
        May 27 at 12:18












      • $begingroup$
        Please see my amended answer having read yours.
        $endgroup$
        – Weather Vane
        May 27 at 12:43










      • $begingroup$
        Did you just throw it into a computer? Thanks for the computation — good to see that the hypothesis appears correct so far, at least @WeatherVane! Thanks!!
        $endgroup$
        – El-Guest
        May 27 at 12:45










      • $begingroup$
        Yes, the easy ones $ lt 2^{64}$. But actually, there is not always just one other prime factor.
        $endgroup$
        – Weather Vane
        May 27 at 12:46




















      • $begingroup$
        Ah, you beat me by 37 seconds....
        $endgroup$
        – tom
        May 27 at 12:12






      • 4




        $begingroup$
        Which is a prime number. How suitable.
        $endgroup$
        – Florian F
        May 27 at 12:18












      • $begingroup$
        Please see my amended answer having read yours.
        $endgroup$
        – Weather Vane
        May 27 at 12:43










      • $begingroup$
        Did you just throw it into a computer? Thanks for the computation — good to see that the hypothesis appears correct so far, at least @WeatherVane! Thanks!!
        $endgroup$
        – El-Guest
        May 27 at 12:45










      • $begingroup$
        Yes, the easy ones $ lt 2^{64}$. But actually, there is not always just one other prime factor.
        $endgroup$
        – Weather Vane
        May 27 at 12:46


















      $begingroup$
      Ah, you beat me by 37 seconds....
      $endgroup$
      – tom
      May 27 at 12:12




      $begingroup$
      Ah, you beat me by 37 seconds....
      $endgroup$
      – tom
      May 27 at 12:12




      4




      4




      $begingroup$
      Which is a prime number. How suitable.
      $endgroup$
      – Florian F
      May 27 at 12:18






      $begingroup$
      Which is a prime number. How suitable.
      $endgroup$
      – Florian F
      May 27 at 12:18














      $begingroup$
      Please see my amended answer having read yours.
      $endgroup$
      – Weather Vane
      May 27 at 12:43




      $begingroup$
      Please see my amended answer having read yours.
      $endgroup$
      – Weather Vane
      May 27 at 12:43












      $begingroup$
      Did you just throw it into a computer? Thanks for the computation — good to see that the hypothesis appears correct so far, at least @WeatherVane! Thanks!!
      $endgroup$
      – El-Guest
      May 27 at 12:45




      $begingroup$
      Did you just throw it into a computer? Thanks for the computation — good to see that the hypothesis appears correct so far, at least @WeatherVane! Thanks!!
      $endgroup$
      – El-Guest
      May 27 at 12:45












      $begingroup$
      Yes, the easy ones $ lt 2^{64}$. But actually, there is not always just one other prime factor.
      $endgroup$
      – Weather Vane
      May 27 at 12:46






      $begingroup$
      Yes, the easy ones $ lt 2^{64}$. But actually, there is not always just one other prime factor.
      $endgroup$
      – Weather Vane
      May 27 at 12:46













      2














      $begingroup$

      I think




      It looks as if you would get stuck at the number 3




      because




      often 3 is going to be the lowest prime factor and $2^3+1=9$ which again gives 3







      share|improve this answer









      $endgroup$




















        2














        $begingroup$

        I think




        It looks as if you would get stuck at the number 3




        because




        often 3 is going to be the lowest prime factor and $2^3+1=9$ which again gives 3







        share|improve this answer









        $endgroup$


















          2














          2










          2







          $begingroup$

          I think




          It looks as if you would get stuck at the number 3




          because




          often 3 is going to be the lowest prime factor and $2^3+1=9$ which again gives 3







          share|improve this answer









          $endgroup$



          I think




          It looks as if you would get stuck at the number 3




          because




          often 3 is going to be the lowest prime factor and $2^3+1=9$ which again gives 3








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered May 27 at 12:10









          tomtom

          2,1481 gold badge7 silver badges30 bronze badges




          2,1481 gold badge7 silver badges30 bronze badges


























              2














              $begingroup$

              Unsure... partial




              Because some $2^{x} + 1$ are prime numbers: no factors.

              For example $x = 8$




              Edit:




              The above wasn't well thought, because $8$ cannot be a lowest factor.


              But from @El-Guest's answer, I found that


              For every prime $3 le x le 61$, $2^{x} + 1$ is divisible by $3$.







              share|improve this answer











              $endgroup$




















                2














                $begingroup$

                Unsure... partial




                Because some $2^{x} + 1$ are prime numbers: no factors.

                For example $x = 8$




                Edit:




                The above wasn't well thought, because $8$ cannot be a lowest factor.


                But from @El-Guest's answer, I found that


                For every prime $3 le x le 61$, $2^{x} + 1$ is divisible by $3$.







                share|improve this answer











                $endgroup$


















                  2














                  2










                  2







                  $begingroup$

                  Unsure... partial




                  Because some $2^{x} + 1$ are prime numbers: no factors.

                  For example $x = 8$




                  Edit:




                  The above wasn't well thought, because $8$ cannot be a lowest factor.


                  But from @El-Guest's answer, I found that


                  For every prime $3 le x le 61$, $2^{x} + 1$ is divisible by $3$.







                  share|improve this answer











                  $endgroup$



                  Unsure... partial




                  Because some $2^{x} + 1$ are prime numbers: no factors.

                  For example $x = 8$




                  Edit:




                  The above wasn't well thought, because $8$ cannot be a lowest factor.


                  But from @El-Guest's answer, I found that


                  For every prime $3 le x le 61$, $2^{x} + 1$ is divisible by $3$.








                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited May 27 at 12:49

























                  answered May 27 at 12:10









                  Weather VaneWeather Vane

                  6,4061 gold badge4 silver badges26 bronze badges




                  6,4061 gold badge4 silver badges26 bronze badges


































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