For $n,k in {mathbb{Z}}^{+}$ (excluding $n=1$), does $frac{(n+k)!}{n!}$ ever equal $n!$











up vote
2
down vote

favorite
1












While investigating an integer sequence, I came across the following two OEIS entries:





  • A094331: Least k such that n! < (n+1)(n+2)(n+3)...(n+k).


  • A075357: a(n) = smallest k such that (n+1)(n+2)...(n+k) is just >= n!.


The generating rule for both of these sequences is basically identical, except A094331 uses $lt$ and A075357 uses $le$. This made me curious whether both sequences are actually identical (and I'm not the first; David Wasserman commented the same thing when he was adding more terms to A075357.) For the purposes of the OEIS, the sequences are technically different, because A094331(1) = 1 and A075357(1) = 0 (i.e. for $n=1$ and $k=0$, $frac{(n+k)!}{n!} = n!$.)



In approaching this problem, I first tried computational brute forcing. For values of $n$ from 2 to 1000000, $n ne k$. However, this approach is obviously limited. Since my ability in number theory is very weak, I was wondering if anyone with a greater knowledge of number theory may be able provide a definitive answer to this question.










share|cite|improve this question
























  • i'm confused. For $jne m$ then $frac {(n+j)!}{n!} ne frac {(n+m)!}{n1}$ so obviously that can't be true for all $k$ and furthermore $frac {(n+k)!}{n!} = n!implies (n+k)! = (n!)^2$ which can't ever be true for $n+k> 1$ as $m!$ is never a perfect square.
    – fleablood
    Nov 22 at 1:52










  • You mean $k = 0$ in your example, right?
    – anomaly
    Nov 22 at 1:53










  • Did you mean to ask does $frac{(n+k)!}{n!} = n!$ EVER true. As it stands it seems as though you are asking is it always true which of course it isn't.
    – fleablood
    Nov 22 at 2:03










  • @anomaly Yes, I've made an edit.
    – Graham
    Nov 22 at 3:11










  • @fleablood I've also made an edit regarding your comment.
    – Graham
    Nov 22 at 3:16















up vote
2
down vote

favorite
1












While investigating an integer sequence, I came across the following two OEIS entries:





  • A094331: Least k such that n! < (n+1)(n+2)(n+3)...(n+k).


  • A075357: a(n) = smallest k such that (n+1)(n+2)...(n+k) is just >= n!.


The generating rule for both of these sequences is basically identical, except A094331 uses $lt$ and A075357 uses $le$. This made me curious whether both sequences are actually identical (and I'm not the first; David Wasserman commented the same thing when he was adding more terms to A075357.) For the purposes of the OEIS, the sequences are technically different, because A094331(1) = 1 and A075357(1) = 0 (i.e. for $n=1$ and $k=0$, $frac{(n+k)!}{n!} = n!$.)



In approaching this problem, I first tried computational brute forcing. For values of $n$ from 2 to 1000000, $n ne k$. However, this approach is obviously limited. Since my ability in number theory is very weak, I was wondering if anyone with a greater knowledge of number theory may be able provide a definitive answer to this question.










share|cite|improve this question
























  • i'm confused. For $jne m$ then $frac {(n+j)!}{n!} ne frac {(n+m)!}{n1}$ so obviously that can't be true for all $k$ and furthermore $frac {(n+k)!}{n!} = n!implies (n+k)! = (n!)^2$ which can't ever be true for $n+k> 1$ as $m!$ is never a perfect square.
    – fleablood
    Nov 22 at 1:52










  • You mean $k = 0$ in your example, right?
    – anomaly
    Nov 22 at 1:53










  • Did you mean to ask does $frac{(n+k)!}{n!} = n!$ EVER true. As it stands it seems as though you are asking is it always true which of course it isn't.
    – fleablood
    Nov 22 at 2:03










  • @anomaly Yes, I've made an edit.
    – Graham
    Nov 22 at 3:11










  • @fleablood I've also made an edit regarding your comment.
    – Graham
    Nov 22 at 3:16













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





While investigating an integer sequence, I came across the following two OEIS entries:





  • A094331: Least k such that n! < (n+1)(n+2)(n+3)...(n+k).


  • A075357: a(n) = smallest k such that (n+1)(n+2)...(n+k) is just >= n!.


The generating rule for both of these sequences is basically identical, except A094331 uses $lt$ and A075357 uses $le$. This made me curious whether both sequences are actually identical (and I'm not the first; David Wasserman commented the same thing when he was adding more terms to A075357.) For the purposes of the OEIS, the sequences are technically different, because A094331(1) = 1 and A075357(1) = 0 (i.e. for $n=1$ and $k=0$, $frac{(n+k)!}{n!} = n!$.)



In approaching this problem, I first tried computational brute forcing. For values of $n$ from 2 to 1000000, $n ne k$. However, this approach is obviously limited. Since my ability in number theory is very weak, I was wondering if anyone with a greater knowledge of number theory may be able provide a definitive answer to this question.










share|cite|improve this question















While investigating an integer sequence, I came across the following two OEIS entries:





  • A094331: Least k such that n! < (n+1)(n+2)(n+3)...(n+k).


  • A075357: a(n) = smallest k such that (n+1)(n+2)...(n+k) is just >= n!.


The generating rule for both of these sequences is basically identical, except A094331 uses $lt$ and A075357 uses $le$. This made me curious whether both sequences are actually identical (and I'm not the first; David Wasserman commented the same thing when he was adding more terms to A075357.) For the purposes of the OEIS, the sequences are technically different, because A094331(1) = 1 and A075357(1) = 0 (i.e. for $n=1$ and $k=0$, $frac{(n+k)!}{n!} = n!$.)



In approaching this problem, I first tried computational brute forcing. For values of $n$ from 2 to 1000000, $n ne k$. However, this approach is obviously limited. Since my ability in number theory is very weak, I was wondering if anyone with a greater knowledge of number theory may be able provide a definitive answer to this question.







number-theory oeis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 3:11

























asked Nov 22 at 1:25









Graham

1157




1157












  • i'm confused. For $jne m$ then $frac {(n+j)!}{n!} ne frac {(n+m)!}{n1}$ so obviously that can't be true for all $k$ and furthermore $frac {(n+k)!}{n!} = n!implies (n+k)! = (n!)^2$ which can't ever be true for $n+k> 1$ as $m!$ is never a perfect square.
    – fleablood
    Nov 22 at 1:52










  • You mean $k = 0$ in your example, right?
    – anomaly
    Nov 22 at 1:53










  • Did you mean to ask does $frac{(n+k)!}{n!} = n!$ EVER true. As it stands it seems as though you are asking is it always true which of course it isn't.
    – fleablood
    Nov 22 at 2:03










  • @anomaly Yes, I've made an edit.
    – Graham
    Nov 22 at 3:11










  • @fleablood I've also made an edit regarding your comment.
    – Graham
    Nov 22 at 3:16


















  • i'm confused. For $jne m$ then $frac {(n+j)!}{n!} ne frac {(n+m)!}{n1}$ so obviously that can't be true for all $k$ and furthermore $frac {(n+k)!}{n!} = n!implies (n+k)! = (n!)^2$ which can't ever be true for $n+k> 1$ as $m!$ is never a perfect square.
    – fleablood
    Nov 22 at 1:52










  • You mean $k = 0$ in your example, right?
    – anomaly
    Nov 22 at 1:53










  • Did you mean to ask does $frac{(n+k)!}{n!} = n!$ EVER true. As it stands it seems as though you are asking is it always true which of course it isn't.
    – fleablood
    Nov 22 at 2:03










  • @anomaly Yes, I've made an edit.
    – Graham
    Nov 22 at 3:11










  • @fleablood I've also made an edit regarding your comment.
    – Graham
    Nov 22 at 3:16
















i'm confused. For $jne m$ then $frac {(n+j)!}{n!} ne frac {(n+m)!}{n1}$ so obviously that can't be true for all $k$ and furthermore $frac {(n+k)!}{n!} = n!implies (n+k)! = (n!)^2$ which can't ever be true for $n+k> 1$ as $m!$ is never a perfect square.
– fleablood
Nov 22 at 1:52




i'm confused. For $jne m$ then $frac {(n+j)!}{n!} ne frac {(n+m)!}{n1}$ so obviously that can't be true for all $k$ and furthermore $frac {(n+k)!}{n!} = n!implies (n+k)! = (n!)^2$ which can't ever be true for $n+k> 1$ as $m!$ is never a perfect square.
– fleablood
Nov 22 at 1:52












You mean $k = 0$ in your example, right?
– anomaly
Nov 22 at 1:53




You mean $k = 0$ in your example, right?
– anomaly
Nov 22 at 1:53












Did you mean to ask does $frac{(n+k)!}{n!} = n!$ EVER true. As it stands it seems as though you are asking is it always true which of course it isn't.
– fleablood
Nov 22 at 2:03




Did you mean to ask does $frac{(n+k)!}{n!} = n!$ EVER true. As it stands it seems as though you are asking is it always true which of course it isn't.
– fleablood
Nov 22 at 2:03












@anomaly Yes, I've made an edit.
– Graham
Nov 22 at 3:11




@anomaly Yes, I've made an edit.
– Graham
Nov 22 at 3:11












@fleablood I've also made an edit regarding your comment.
– Graham
Nov 22 at 3:16




@fleablood I've also made an edit regarding your comment.
– Graham
Nov 22 at 3:16










3 Answers
3






active

oldest

votes

















up vote
5
down vote



accepted










By a theorem of Chebyshev, there exists for any integer $n > 1$ a prime $p$ with $n < p < 2n$. Hence $(2n)!$ and $(2n+1)$! are not perfect squares, since they're divisible by $p$ but not $p^2$. It follows that $(n+k)!/n! not = n!$ for any integers $n, k > 0$.






share|cite|improve this answer




























    up vote
    1
    down vote













    With the exception of $N=1$, $N!$ is never a square. This is because, by Bertrand's Postulate, there is always a prime between $N$ and $lfloor N/2rfloor$ (to be precise, a prime $p$ satisfying $lfloor N/2rfloorlt ple N$), and such a prime can only divide $N!$ once. So if we take $N=n+k$ with positive integers $n$ and $k$, we have $Ngt1$ and so $N!=(n+k)!$ is not a square. In particular, it cannot be the case that $(n+k)!=(n!)^2$. Thus it is never the case that $(n+k)!/n!=n!$.



    It might be of interest to see if there is a proof that $N!$ is never the square of a factorial that doesn't rely on Bertrand's Postulate.






    share|cite|improve this answer




























      up vote
      1
      down vote













      The only cases where those two sequences would differ would be when $n! = frac {(n+k)!}{n!}$ or $(n+k)! = (n!)^2$ and that is the only time you can have $n! le frac {(n+k)!}{n!}$ and $n! < frac {(n+k)!}{n!}$ not both be true or not both be false.



      anomaly's answer explains why that can never for $n > 1$.



      I just want to add that if $n = 1$ then this is $(k+1)! = 1$ which is true only if $k =0$.



      Hence the sequences do differ at $n=1$. The first term of A075357 is $0$ while the term of A094331 is $1$. Otherwise the sequences are identical. But they do differ and $n=1$



      So for A075357 you want the least $k$ so that $1! < frac {(1+k)!}{1!}$ and that is $1! = frac {1!}{1!}$ and $1! < frac {2!}{1!}$ and that is when $1+ k = 2$ or $k =1$. But of A094331 you want the least $k$ so that $1! le frac {(1+k)!}{1!}$ and because $1! = frac {1!}{1!}$ that is $1+k =1$ or $k = 0$.



      As for any other $n > 1$ we have $frac {(n+k)!}{n!}ne n!$ we have $n! le frac {(n+k)!}{n!} iff n! < frac {(n+k)!}{n!}$ and the sequences are equal.






      share|cite|improve this answer























        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008628%2ffor-n-k-in-mathbbz-excluding-n-1-does-fracnkn-ever%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        5
        down vote



        accepted










        By a theorem of Chebyshev, there exists for any integer $n > 1$ a prime $p$ with $n < p < 2n$. Hence $(2n)!$ and $(2n+1)$! are not perfect squares, since they're divisible by $p$ but not $p^2$. It follows that $(n+k)!/n! not = n!$ for any integers $n, k > 0$.






        share|cite|improve this answer

























          up vote
          5
          down vote



          accepted










          By a theorem of Chebyshev, there exists for any integer $n > 1$ a prime $p$ with $n < p < 2n$. Hence $(2n)!$ and $(2n+1)$! are not perfect squares, since they're divisible by $p$ but not $p^2$. It follows that $(n+k)!/n! not = n!$ for any integers $n, k > 0$.






          share|cite|improve this answer























            up vote
            5
            down vote



            accepted







            up vote
            5
            down vote



            accepted






            By a theorem of Chebyshev, there exists for any integer $n > 1$ a prime $p$ with $n < p < 2n$. Hence $(2n)!$ and $(2n+1)$! are not perfect squares, since they're divisible by $p$ but not $p^2$. It follows that $(n+k)!/n! not = n!$ for any integers $n, k > 0$.






            share|cite|improve this answer












            By a theorem of Chebyshev, there exists for any integer $n > 1$ a prime $p$ with $n < p < 2n$. Hence $(2n)!$ and $(2n+1)$! are not perfect squares, since they're divisible by $p$ but not $p^2$. It follows that $(n+k)!/n! not = n!$ for any integers $n, k > 0$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 at 1:48









            anomaly

            17.2k42662




            17.2k42662






















                up vote
                1
                down vote













                With the exception of $N=1$, $N!$ is never a square. This is because, by Bertrand's Postulate, there is always a prime between $N$ and $lfloor N/2rfloor$ (to be precise, a prime $p$ satisfying $lfloor N/2rfloorlt ple N$), and such a prime can only divide $N!$ once. So if we take $N=n+k$ with positive integers $n$ and $k$, we have $Ngt1$ and so $N!=(n+k)!$ is not a square. In particular, it cannot be the case that $(n+k)!=(n!)^2$. Thus it is never the case that $(n+k)!/n!=n!$.



                It might be of interest to see if there is a proof that $N!$ is never the square of a factorial that doesn't rely on Bertrand's Postulate.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  With the exception of $N=1$, $N!$ is never a square. This is because, by Bertrand's Postulate, there is always a prime between $N$ and $lfloor N/2rfloor$ (to be precise, a prime $p$ satisfying $lfloor N/2rfloorlt ple N$), and such a prime can only divide $N!$ once. So if we take $N=n+k$ with positive integers $n$ and $k$, we have $Ngt1$ and so $N!=(n+k)!$ is not a square. In particular, it cannot be the case that $(n+k)!=(n!)^2$. Thus it is never the case that $(n+k)!/n!=n!$.



                  It might be of interest to see if there is a proof that $N!$ is never the square of a factorial that doesn't rely on Bertrand's Postulate.






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    With the exception of $N=1$, $N!$ is never a square. This is because, by Bertrand's Postulate, there is always a prime between $N$ and $lfloor N/2rfloor$ (to be precise, a prime $p$ satisfying $lfloor N/2rfloorlt ple N$), and such a prime can only divide $N!$ once. So if we take $N=n+k$ with positive integers $n$ and $k$, we have $Ngt1$ and so $N!=(n+k)!$ is not a square. In particular, it cannot be the case that $(n+k)!=(n!)^2$. Thus it is never the case that $(n+k)!/n!=n!$.



                    It might be of interest to see if there is a proof that $N!$ is never the square of a factorial that doesn't rely on Bertrand's Postulate.






                    share|cite|improve this answer












                    With the exception of $N=1$, $N!$ is never a square. This is because, by Bertrand's Postulate, there is always a prime between $N$ and $lfloor N/2rfloor$ (to be precise, a prime $p$ satisfying $lfloor N/2rfloorlt ple N$), and such a prime can only divide $N!$ once. So if we take $N=n+k$ with positive integers $n$ and $k$, we have $Ngt1$ and so $N!=(n+k)!$ is not a square. In particular, it cannot be the case that $(n+k)!=(n!)^2$. Thus it is never the case that $(n+k)!/n!=n!$.



                    It might be of interest to see if there is a proof that $N!$ is never the square of a factorial that doesn't rely on Bertrand's Postulate.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 22 at 2:16









                    Barry Cipra

                    58.6k653122




                    58.6k653122






















                        up vote
                        1
                        down vote













                        The only cases where those two sequences would differ would be when $n! = frac {(n+k)!}{n!}$ or $(n+k)! = (n!)^2$ and that is the only time you can have $n! le frac {(n+k)!}{n!}$ and $n! < frac {(n+k)!}{n!}$ not both be true or not both be false.



                        anomaly's answer explains why that can never for $n > 1$.



                        I just want to add that if $n = 1$ then this is $(k+1)! = 1$ which is true only if $k =0$.



                        Hence the sequences do differ at $n=1$. The first term of A075357 is $0$ while the term of A094331 is $1$. Otherwise the sequences are identical. But they do differ and $n=1$



                        So for A075357 you want the least $k$ so that $1! < frac {(1+k)!}{1!}$ and that is $1! = frac {1!}{1!}$ and $1! < frac {2!}{1!}$ and that is when $1+ k = 2$ or $k =1$. But of A094331 you want the least $k$ so that $1! le frac {(1+k)!}{1!}$ and because $1! = frac {1!}{1!}$ that is $1+k =1$ or $k = 0$.



                        As for any other $n > 1$ we have $frac {(n+k)!}{n!}ne n!$ we have $n! le frac {(n+k)!}{n!} iff n! < frac {(n+k)!}{n!}$ and the sequences are equal.






                        share|cite|improve this answer



























                          up vote
                          1
                          down vote













                          The only cases where those two sequences would differ would be when $n! = frac {(n+k)!}{n!}$ or $(n+k)! = (n!)^2$ and that is the only time you can have $n! le frac {(n+k)!}{n!}$ and $n! < frac {(n+k)!}{n!}$ not both be true or not both be false.



                          anomaly's answer explains why that can never for $n > 1$.



                          I just want to add that if $n = 1$ then this is $(k+1)! = 1$ which is true only if $k =0$.



                          Hence the sequences do differ at $n=1$. The first term of A075357 is $0$ while the term of A094331 is $1$. Otherwise the sequences are identical. But they do differ and $n=1$



                          So for A075357 you want the least $k$ so that $1! < frac {(1+k)!}{1!}$ and that is $1! = frac {1!}{1!}$ and $1! < frac {2!}{1!}$ and that is when $1+ k = 2$ or $k =1$. But of A094331 you want the least $k$ so that $1! le frac {(1+k)!}{1!}$ and because $1! = frac {1!}{1!}$ that is $1+k =1$ or $k = 0$.



                          As for any other $n > 1$ we have $frac {(n+k)!}{n!}ne n!$ we have $n! le frac {(n+k)!}{n!} iff n! < frac {(n+k)!}{n!}$ and the sequences are equal.






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            The only cases where those two sequences would differ would be when $n! = frac {(n+k)!}{n!}$ or $(n+k)! = (n!)^2$ and that is the only time you can have $n! le frac {(n+k)!}{n!}$ and $n! < frac {(n+k)!}{n!}$ not both be true or not both be false.



                            anomaly's answer explains why that can never for $n > 1$.



                            I just want to add that if $n = 1$ then this is $(k+1)! = 1$ which is true only if $k =0$.



                            Hence the sequences do differ at $n=1$. The first term of A075357 is $0$ while the term of A094331 is $1$. Otherwise the sequences are identical. But they do differ and $n=1$



                            So for A075357 you want the least $k$ so that $1! < frac {(1+k)!}{1!}$ and that is $1! = frac {1!}{1!}$ and $1! < frac {2!}{1!}$ and that is when $1+ k = 2$ or $k =1$. But of A094331 you want the least $k$ so that $1! le frac {(1+k)!}{1!}$ and because $1! = frac {1!}{1!}$ that is $1+k =1$ or $k = 0$.



                            As for any other $n > 1$ we have $frac {(n+k)!}{n!}ne n!$ we have $n! le frac {(n+k)!}{n!} iff n! < frac {(n+k)!}{n!}$ and the sequences are equal.






                            share|cite|improve this answer














                            The only cases where those two sequences would differ would be when $n! = frac {(n+k)!}{n!}$ or $(n+k)! = (n!)^2$ and that is the only time you can have $n! le frac {(n+k)!}{n!}$ and $n! < frac {(n+k)!}{n!}$ not both be true or not both be false.



                            anomaly's answer explains why that can never for $n > 1$.



                            I just want to add that if $n = 1$ then this is $(k+1)! = 1$ which is true only if $k =0$.



                            Hence the sequences do differ at $n=1$. The first term of A075357 is $0$ while the term of A094331 is $1$. Otherwise the sequences are identical. But they do differ and $n=1$



                            So for A075357 you want the least $k$ so that $1! < frac {(1+k)!}{1!}$ and that is $1! = frac {1!}{1!}$ and $1! < frac {2!}{1!}$ and that is when $1+ k = 2$ or $k =1$. But of A094331 you want the least $k$ so that $1! le frac {(1+k)!}{1!}$ and because $1! = frac {1!}{1!}$ that is $1+k =1$ or $k = 0$.



                            As for any other $n > 1$ we have $frac {(n+k)!}{n!}ne n!$ we have $n! le frac {(n+k)!}{n!} iff n! < frac {(n+k)!}{n!}$ and the sequences are equal.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Nov 22 at 2:33

























                            answered Nov 22 at 2:10









                            fleablood

                            67.5k22684




                            67.5k22684






























                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.





                                Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                Please pay close attention to the following guidance:


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008628%2ffor-n-k-in-mathbbz-excluding-n-1-does-fracnkn-ever%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                Popular posts from this blog

                                Ellipse (mathématiques)

                                Quarter-circle Tiles

                                Mont Emei