Digits of the square root of primes












2












$begingroup$


Here is something interesting I discovered recently.




Let $S_n$ be the sum from the first decimal of $sqrt2+sqrt{p_n}$ to the $p_n$th decimal, where $p_n$ is the $n$th prime number. Then I conjecture that $S_{n+1}>S_n$ for all $n>1$.




Although I have no way of proving this and no feasible way to disprove this, I have a few comments that may be useful.




  • The first few values of $S_n$ for $n=2,3,4,5,6$ are $10,21,41,48,64$. Clearly these are increasing.


  • If we pick $p_{n+1}$ such that it is greater than a square number by a small amount, then it will have more zeroes, especially when $n$ is large. Therefore using this method there could be a higher chance that $S_{n+1}le S_n$ and thus disproving the conjecture. I have done so for $p_6=13$ and $p_7=17$, and for $p_{12}=37$ and $p_{11}=31$, but to no avail since $85=S_7>S_6=64$ and $149=S_{12}>S_{11}=123$.


  • I have a strong feeling that the conjecture is true, since every time we are adding more digits, and the chance of zeroes occurring that overcome the increasing sum is subjectively low.



Any advances on this are welcome!



If you do happen to find a value of $n$ that disprove the conjecture, please give a few more so that a pattern can hopefully be observed, if any.






prime-numbers recreational-mathematics radicals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your thought to have $p_{n+1}$ just greater than a square is a good one. It would also be good to have $p_n=p_{n+1}-2$, which means the square is a multiple of $36$ so you don't get so many digits added. This requires that the square be a multiple of $900$ because otherwise the square ends in $4$ or $6$ and one of the neighboring numbers is a multiple of $5$. I am sure there are people with code sitting around that can try millions or billions of cases easily.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 18:16






  • 2




    $begingroup$
    @RossMillikan counterexample found
    $endgroup$
    – Quintec
    Dec 30 '18 at 18:25






  • 2




    $begingroup$
    @Quintec: you should write it up as an answer. You just need to copy in the decimals and sums.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 18:36










  • $begingroup$
    @RossMillikan Done.
    $endgroup$
    – Quintec
    Dec 30 '18 at 18:55
















2












$begingroup$


Here is something interesting I discovered recently.




Let $S_n$ be the sum from the first decimal of $sqrt2+sqrt{p_n}$ to the $p_n$th decimal, where $p_n$ is the $n$th prime number. Then I conjecture that $S_{n+1}>S_n$ for all $n>1$.




Although I have no way of proving this and no feasible way to disprove this, I have a few comments that may be useful.




  • The first few values of $S_n$ for $n=2,3,4,5,6$ are $10,21,41,48,64$. Clearly these are increasing.


  • If we pick $p_{n+1}$ such that it is greater than a square number by a small amount, then it will have more zeroes, especially when $n$ is large. Therefore using this method there could be a higher chance that $S_{n+1}le S_n$ and thus disproving the conjecture. I have done so for $p_6=13$ and $p_7=17$, and for $p_{12}=37$ and $p_{11}=31$, but to no avail since $85=S_7>S_6=64$ and $149=S_{12}>S_{11}=123$.


  • I have a strong feeling that the conjecture is true, since every time we are adding more digits, and the chance of zeroes occurring that overcome the increasing sum is subjectively low.



Any advances on this are welcome!



If you do happen to find a value of $n$ that disprove the conjecture, please give a few more so that a pattern can hopefully be observed, if any.






prime-numbers recreational-mathematics radicals







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your thought to have $p_{n+1}$ just greater than a square is a good one. It would also be good to have $p_n=p_{n+1}-2$, which means the square is a multiple of $36$ so you don't get so many digits added. This requires that the square be a multiple of $900$ because otherwise the square ends in $4$ or $6$ and one of the neighboring numbers is a multiple of $5$. I am sure there are people with code sitting around that can try millions or billions of cases easily.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 18:16






  • 2




    $begingroup$
    @RossMillikan counterexample found
    $endgroup$
    – Quintec
    Dec 30 '18 at 18:25






  • 2




    $begingroup$
    @Quintec: you should write it up as an answer. You just need to copy in the decimals and sums.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 18:36










  • $begingroup$
    @RossMillikan Done.
    $endgroup$
    – Quintec
    Dec 30 '18 at 18:55














2












2








2





$begingroup$


Here is something interesting I discovered recently.




Let $S_n$ be the sum from the first decimal of $sqrt2+sqrt{p_n}$ to the $p_n$th decimal, where $p_n$ is the $n$th prime number. Then I conjecture that $S_{n+1}>S_n$ for all $n>1$.




Although I have no way of proving this and no feasible way to disprove this, I have a few comments that may be useful.




  • The first few values of $S_n$ for $n=2,3,4,5,6$ are $10,21,41,48,64$. Clearly these are increasing.


  • If we pick $p_{n+1}$ such that it is greater than a square number by a small amount, then it will have more zeroes, especially when $n$ is large. Therefore using this method there could be a higher chance that $S_{n+1}le S_n$ and thus disproving the conjecture. I have done so for $p_6=13$ and $p_7=17$, and for $p_{12}=37$ and $p_{11}=31$, but to no avail since $85=S_7>S_6=64$ and $149=S_{12}>S_{11}=123$.


  • I have a strong feeling that the conjecture is true, since every time we are adding more digits, and the chance of zeroes occurring that overcome the increasing sum is subjectively low.



Any advances on this are welcome!



If you do happen to find a value of $n$ that disprove the conjecture, please give a few more so that a pattern can hopefully be observed, if any.






prime-numbers recreational-mathematics radicals







share|cite|improve this question









$endgroup$




Here is something interesting I discovered recently.




Let $S_n$ be the sum from the first decimal of $sqrt2+sqrt{p_n}$ to the $p_n$th decimal, where $p_n$ is the $n$th prime number. Then I conjecture that $S_{n+1}>S_n$ for all $n>1$.




Although I have no way of proving this and no feasible way to disprove this, I have a few comments that may be useful.




  • The first few values of $S_n$ for $n=2,3,4,5,6$ are $10,21,41,48,64$. Clearly these are increasing.


  • If we pick $p_{n+1}$ such that it is greater than a square number by a small amount, then it will have more zeroes, especially when $n$ is large. Therefore using this method there could be a higher chance that $S_{n+1}le S_n$ and thus disproving the conjecture. I have done so for $p_6=13$ and $p_7=17$, and for $p_{12}=37$ and $p_{11}=31$, but to no avail since $85=S_7>S_6=64$ and $149=S_{12}>S_{11}=123$.


  • I have a strong feeling that the conjecture is true, since every time we are adding more digits, and the chance of zeroes occurring that overcome the increasing sum is subjectively low.



Any advances on this are welcome!



If you do happen to find a value of $n$ that disprove the conjecture, please give a few more so that a pattern can hopefully be observed, if any.






prime-numbers recreational-mathematics radicals




prime-numbers recreational-mathematics radicals






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 30 '18 at 17:40









TheSimpliFireTheSimpliFire

12.6k62360




12.6k62360












  • $begingroup$
    Your thought to have $p_{n+1}$ just greater than a square is a good one. It would also be good to have $p_n=p_{n+1}-2$, which means the square is a multiple of $36$ so you don't get so many digits added. This requires that the square be a multiple of $900$ because otherwise the square ends in $4$ or $6$ and one of the neighboring numbers is a multiple of $5$. I am sure there are people with code sitting around that can try millions or billions of cases easily.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 18:16






  • 2




    $begingroup$
    @RossMillikan counterexample found
    $endgroup$
    – Quintec
    Dec 30 '18 at 18:25






  • 2




    $begingroup$
    @Quintec: you should write it up as an answer. You just need to copy in the decimals and sums.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 18:36










  • $begingroup$
    @RossMillikan Done.
    $endgroup$
    – Quintec
    Dec 30 '18 at 18:55


















  • $begingroup$
    Your thought to have $p_{n+1}$ just greater than a square is a good one. It would also be good to have $p_n=p_{n+1}-2$, which means the square is a multiple of $36$ so you don't get so many digits added. This requires that the square be a multiple of $900$ because otherwise the square ends in $4$ or $6$ and one of the neighboring numbers is a multiple of $5$. I am sure there are people with code sitting around that can try millions or billions of cases easily.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 18:16






  • 2




    $begingroup$
    @RossMillikan counterexample found
    $endgroup$
    – Quintec
    Dec 30 '18 at 18:25






  • 2




    $begingroup$
    @Quintec: you should write it up as an answer. You just need to copy in the decimals and sums.
    $endgroup$
    – Ross Millikan
    Dec 30 '18 at 18:36










  • $begingroup$
    @RossMillikan Done.
    $endgroup$
    – Quintec
    Dec 30 '18 at 18:55
















$begingroup$
Your thought to have $p_{n+1}$ just greater than a square is a good one. It would also be good to have $p_n=p_{n+1}-2$, which means the square is a multiple of $36$ so you don't get so many digits added. This requires that the square be a multiple of $900$ because otherwise the square ends in $4$ or $6$ and one of the neighboring numbers is a multiple of $5$. I am sure there are people with code sitting around that can try millions or billions of cases easily.
$endgroup$
– Ross Millikan
Dec 30 '18 at 18:16




$begingroup$
Your thought to have $p_{n+1}$ just greater than a square is a good one. It would also be good to have $p_n=p_{n+1}-2$, which means the square is a multiple of $36$ so you don't get so many digits added. This requires that the square be a multiple of $900$ because otherwise the square ends in $4$ or $6$ and one of the neighboring numbers is a multiple of $5$. I am sure there are people with code sitting around that can try millions or billions of cases easily.
$endgroup$
– Ross Millikan
Dec 30 '18 at 18:16




2




2




$begingroup$
@RossMillikan counterexample found
$endgroup$
– Quintec
Dec 30 '18 at 18:25




$begingroup$
@RossMillikan counterexample found
$endgroup$
– Quintec
Dec 30 '18 at 18:25




2




2




$begingroup$
@Quintec: you should write it up as an answer. You just need to copy in the decimals and sums.
$endgroup$
– Ross Millikan
Dec 30 '18 at 18:36




$begingroup$
@Quintec: you should write it up as an answer. You just need to copy in the decimals and sums.
$endgroup$
– Ross Millikan
Dec 30 '18 at 18:36












$begingroup$
@RossMillikan Done.
$endgroup$
– Quintec
Dec 30 '18 at 18:55




$begingroup$
@RossMillikan Done.
$endgroup$
– Quintec
Dec 30 '18 at 18:55










1 Answer
1






active

oldest

votes


















9












$begingroup$

Checking all primes in Python, a counterexample is found at 19. Here are the first few sums:



$$
begin{array}{r|r}
p_n & S_n \
hline
2 & 10\
3 & 11 \
5 & 21 \
7 & 41 \
11 & 48 \
13 & 63 \
17 & 93 \
textbf{19} & textbf{72} \
end{array}
$$



As you can see, $72 notgt 93$.



Here is the source code with output up to the counterexample, and here is some more data that goes up to $p_n < 100$. The data includes $p_n$, $sqrt2+sqrt{p_n}$, and $S_n$. It also notes when another counterexample is found - as you can see, there are many.






share|cite|improve this answer









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    active

    oldest

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    9












    $begingroup$

    Checking all primes in Python, a counterexample is found at 19. Here are the first few sums:



    $$
    begin{array}{r|r}
    p_n & S_n \
    hline
    2 & 10\
    3 & 11 \
    5 & 21 \
    7 & 41 \
    11 & 48 \
    13 & 63 \
    17 & 93 \
    textbf{19} & textbf{72} \
    end{array}
    $$



    As you can see, $72 notgt 93$.



    Here is the source code with output up to the counterexample, and here is some more data that goes up to $p_n < 100$. The data includes $p_n$, $sqrt2+sqrt{p_n}$, and $S_n$. It also notes when another counterexample is found - as you can see, there are many.






    share|cite|improve this answer









    $endgroup$


















      9












      $begingroup$

      Checking all primes in Python, a counterexample is found at 19. Here are the first few sums:



      $$
      begin{array}{r|r}
      p_n & S_n \
      hline
      2 & 10\
      3 & 11 \
      5 & 21 \
      7 & 41 \
      11 & 48 \
      13 & 63 \
      17 & 93 \
      textbf{19} & textbf{72} \
      end{array}
      $$



      As you can see, $72 notgt 93$.



      Here is the source code with output up to the counterexample, and here is some more data that goes up to $p_n < 100$. The data includes $p_n$, $sqrt2+sqrt{p_n}$, and $S_n$. It also notes when another counterexample is found - as you can see, there are many.






      share|cite|improve this answer









      $endgroup$
















        9












        9








        9





        $begingroup$

        Checking all primes in Python, a counterexample is found at 19. Here are the first few sums:



        $$
        begin{array}{r|r}
        p_n & S_n \
        hline
        2 & 10\
        3 & 11 \
        5 & 21 \
        7 & 41 \
        11 & 48 \
        13 & 63 \
        17 & 93 \
        textbf{19} & textbf{72} \
        end{array}
        $$



        As you can see, $72 notgt 93$.



        Here is the source code with output up to the counterexample, and here is some more data that goes up to $p_n < 100$. The data includes $p_n$, $sqrt2+sqrt{p_n}$, and $S_n$. It also notes when another counterexample is found - as you can see, there are many.






        share|cite|improve this answer









        $endgroup$



        Checking all primes in Python, a counterexample is found at 19. Here are the first few sums:



        $$
        begin{array}{r|r}
        p_n & S_n \
        hline
        2 & 10\
        3 & 11 \
        5 & 21 \
        7 & 41 \
        11 & 48 \
        13 & 63 \
        17 & 93 \
        textbf{19} & textbf{72} \
        end{array}
        $$



        As you can see, $72 notgt 93$.



        Here is the source code with output up to the counterexample, and here is some more data that goes up to $p_n < 100$. The data includes $p_n$, $sqrt2+sqrt{p_n}$, and $S_n$. It also notes when another counterexample is found - as you can see, there are many.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 18:52









        QuintecQuintec

        22625




        22625






























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