Reference Request for solution of Ramanujan Identities












2












$begingroup$


Where can I find the identities of Ramanujan concerning the Floor Function with its solution? Any site you can recomend to me? enter image description here






elementary-number-theory reference-request floor-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    science.ncue.edu.tw/journal/article/1-2-7.pdf
    $endgroup$
    – Karn Watcharasupat
    Jul 28 '18 at 14:49










  • $begingroup$
    (iii) Can be found here: math.stackexchange.com/questions/6087/…
    $endgroup$
    – Robert Z
    Jul 28 '18 at 15:09
















2












$begingroup$


Where can I find the identities of Ramanujan concerning the Floor Function with its solution? Any site you can recomend to me? enter image description here






elementary-number-theory reference-request floor-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    science.ncue.edu.tw/journal/article/1-2-7.pdf
    $endgroup$
    – Karn Watcharasupat
    Jul 28 '18 at 14:49










  • $begingroup$
    (iii) Can be found here: math.stackexchange.com/questions/6087/…
    $endgroup$
    – Robert Z
    Jul 28 '18 at 15:09














2












2








2


1



$begingroup$


Where can I find the identities of Ramanujan concerning the Floor Function with its solution? Any site you can recomend to me? enter image description here






elementary-number-theory reference-request floor-function







share|cite|improve this question









$endgroup$




Where can I find the identities of Ramanujan concerning the Floor Function with its solution? Any site you can recomend to me? enter image description here






elementary-number-theory reference-request floor-function




elementary-number-theory reference-request floor-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jul 28 '18 at 14:33









user573497user573497

16619




16619












  • $begingroup$
    science.ncue.edu.tw/journal/article/1-2-7.pdf
    $endgroup$
    – Karn Watcharasupat
    Jul 28 '18 at 14:49










  • $begingroup$
    (iii) Can be found here: math.stackexchange.com/questions/6087/…
    $endgroup$
    – Robert Z
    Jul 28 '18 at 15:09


















  • $begingroup$
    science.ncue.edu.tw/journal/article/1-2-7.pdf
    $endgroup$
    – Karn Watcharasupat
    Jul 28 '18 at 14:49










  • $begingroup$
    (iii) Can be found here: math.stackexchange.com/questions/6087/…
    $endgroup$
    – Robert Z
    Jul 28 '18 at 15:09
















$begingroup$
science.ncue.edu.tw/journal/article/1-2-7.pdf
$endgroup$
– Karn Watcharasupat
Jul 28 '18 at 14:49




$begingroup$
science.ncue.edu.tw/journal/article/1-2-7.pdf
$endgroup$
– Karn Watcharasupat
Jul 28 '18 at 14:49












$begingroup$
(iii) Can be found here: math.stackexchange.com/questions/6087/…
$endgroup$
– Robert Z
Jul 28 '18 at 15:09




$begingroup$
(iii) Can be found here: math.stackexchange.com/questions/6087/…
$endgroup$
– Robert Z
Jul 28 '18 at 15:09










2 Answers
2






active

oldest

votes


















2












$begingroup$

For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
&= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
&= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
end{eqnarray}$$
$$E_r=left{%
begin{array}{ll}
0, & r=0,1\
1, & r=2 \
2, & r=3 \
3, & r=4,5 \
end{array}%
right.$$
And the right side is
$$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
&= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
&= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
end{eqnarray}$$
$$F_r=left{%
begin{array}{ll}
0, & r=0,1\
1, & r=2 \
2, & r=3 \
3, & r=4,5 \
end{array}%
right.$$
So both sides are the same for all $r$ and we are done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why did you just let $n=6k+r$? Is this the general form of an positive integer?
    $endgroup$
    – user573497
    Jul 28 '18 at 15:56










  • $begingroup$
    That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
    $endgroup$
    – greedoid
    Jul 28 '18 at 16:08






  • 1




    $begingroup$
    oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
    $endgroup$
    – user573497
    Jul 28 '18 at 16:33






  • 1




    $begingroup$
    Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
    $endgroup$
    – user573497
    Jul 28 '18 at 16:36










  • $begingroup$
    I can understand now! Thanks for the feedback! :)
    $endgroup$
    – user573497
    Jul 28 '18 at 16:38



















0












$begingroup$

If $[x]$ denotes floor we have
$$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
$$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
$$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
and obviously
$$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
$$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
and also if $[x]$ denotes nearest integer we have
$$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$






share|cite|improve this answer











$endgroup$













    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',
    autoActivateHeartbeat: false,
    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%2f2865296%2freference-request-for-solution-of-ramanujan-identities%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
    &= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
    &= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
    end{eqnarray}$$
    $$E_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    And the right side is
    $$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
    &= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
    &= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
    end{eqnarray}$$
    $$F_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    So both sides are the same for all $r$ and we are done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why did you just let $n=6k+r$? Is this the general form of an positive integer?
      $endgroup$
      – user573497
      Jul 28 '18 at 15:56










    • $begingroup$
      That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
      $endgroup$
      – greedoid
      Jul 28 '18 at 16:08






    • 1




      $begingroup$
      oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
      $endgroup$
      – user573497
      Jul 28 '18 at 16:33






    • 1




      $begingroup$
      Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
      $endgroup$
      – user573497
      Jul 28 '18 at 16:36










    • $begingroup$
      I can understand now! Thanks for the feedback! :)
      $endgroup$
      – user573497
      Jul 28 '18 at 16:38
















    2












    $begingroup$

    For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
    &= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
    &= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
    end{eqnarray}$$
    $$E_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    And the right side is
    $$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
    &= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
    &= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
    end{eqnarray}$$
    $$F_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    So both sides are the same for all $r$ and we are done.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Why did you just let $n=6k+r$? Is this the general form of an positive integer?
      $endgroup$
      – user573497
      Jul 28 '18 at 15:56










    • $begingroup$
      That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
      $endgroup$
      – greedoid
      Jul 28 '18 at 16:08






    • 1




      $begingroup$
      oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
      $endgroup$
      – user573497
      Jul 28 '18 at 16:33






    • 1




      $begingroup$
      Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
      $endgroup$
      – user573497
      Jul 28 '18 at 16:36










    • $begingroup$
      I can understand now! Thanks for the feedback! :)
      $endgroup$
      – user573497
      Jul 28 '18 at 16:38














    2












    2








    2





    $begingroup$

    For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
    &= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
    &= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
    end{eqnarray}$$
    $$E_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    And the right side is
    $$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
    &= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
    &= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
    end{eqnarray}$$
    $$F_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    So both sides are the same for all $r$ and we are done.






    share|cite|improve this answer









    $endgroup$



    For (i), write $n=6k+r$ where $rin {0,1,...,5}$. Then left side is $$begin{eqnarray}Big[{nover 3}Big]+ Big[{n+2over 6}Big]+Big[{n+4over 6}Big] &= &Big[{6k+rover 3}Big]+ Big[{6k+r+2over 6}Big]+Big[{6k+r+4over 6}Big]\
    &= &2k+Big[{rover 3}Big]+ k+Big[{r+2over 6}Big]+k+Big[{r+4over 6}Big]\
    &= &4k+underbrace{Big[{rover 3}Big]+Big[{r+2over 6}Big]+Big[{r+4over 6}Big]}_{E_r}\
    end{eqnarray}$$
    $$E_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    And the right side is
    $$begin{eqnarray}Big[{nover 2}Big]+ Big[{n+3over 6}Big]&= &Big[{6k+rover 2}Big]+ Big[{6k+r+3over 6}Big]\
    &= &3k+Big[{rover 2}Big]+ k+Big[{r+3over 6}Big]\
    &= &4k+underbrace{Big[{rover 2}Big]+Big[{r+3over 6}Big]}_{F_r}\
    end{eqnarray}$$
    $$F_r=left{%
    begin{array}{ll}
    0, & r=0,1\
    1, & r=2 \
    2, & r=3 \
    3, & r=4,5 \
    end{array}%
    right.$$
    So both sides are the same for all $r$ and we are done.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jul 28 '18 at 14:51









    greedoidgreedoid

    40.1k114799




    40.1k114799












    • $begingroup$
      Why did you just let $n=6k+r$? Is this the general form of an positive integer?
      $endgroup$
      – user573497
      Jul 28 '18 at 15:56










    • $begingroup$
      That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
      $endgroup$
      – greedoid
      Jul 28 '18 at 16:08






    • 1




      $begingroup$
      oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
      $endgroup$
      – user573497
      Jul 28 '18 at 16:33






    • 1




      $begingroup$
      Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
      $endgroup$
      – user573497
      Jul 28 '18 at 16:36










    • $begingroup$
      I can understand now! Thanks for the feedback! :)
      $endgroup$
      – user573497
      Jul 28 '18 at 16:38


















    • $begingroup$
      Why did you just let $n=6k+r$? Is this the general form of an positive integer?
      $endgroup$
      – user573497
      Jul 28 '18 at 15:56










    • $begingroup$
      That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
      $endgroup$
      – greedoid
      Jul 28 '18 at 16:08






    • 1




      $begingroup$
      oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
      $endgroup$
      – user573497
      Jul 28 '18 at 16:33






    • 1




      $begingroup$
      Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
      $endgroup$
      – user573497
      Jul 28 '18 at 16:36










    • $begingroup$
      I can understand now! Thanks for the feedback! :)
      $endgroup$
      – user573497
      Jul 28 '18 at 16:38
















    $begingroup$
    Why did you just let $n=6k+r$? Is this the general form of an positive integer?
    $endgroup$
    – user573497
    Jul 28 '18 at 15:56




    $begingroup$
    Why did you just let $n=6k+r$? Is this the general form of an positive integer?
    $endgroup$
    – user573497
    Jul 28 '18 at 15:56












    $begingroup$
    That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
    $endgroup$
    – greedoid
    Jul 28 '18 at 16:08




    $begingroup$
    That is correct. If we had $[n/11]$ then I would write $n=11k+r$ and $0leq rleq 10$
    $endgroup$
    – greedoid
    Jul 28 '18 at 16:08




    1




    1




    $begingroup$
    oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
    $endgroup$
    – user573497
    Jul 28 '18 at 16:33




    $begingroup$
    oh, I get it! You did that for convenience since in the denominator there is a $6$. If it were some arbitrary number un the denominator like $m$, then I consider the integer $n=mk+r$ where 0leqrls
    $endgroup$
    – user573497
    Jul 28 '18 at 16:33




    1




    1




    $begingroup$
    Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
    $endgroup$
    – user573497
    Jul 28 '18 at 16:36




    $begingroup$
    Sorry, I couldn't edit the last comment. I tried to say that $n=mk+r$ where $r$ is an integer between $0$ and $m-1$.
    $endgroup$
    – user573497
    Jul 28 '18 at 16:36












    $begingroup$
    I can understand now! Thanks for the feedback! :)
    $endgroup$
    – user573497
    Jul 28 '18 at 16:38




    $begingroup$
    I can understand now! Thanks for the feedback! :)
    $endgroup$
    – user573497
    Jul 28 '18 at 16:38











    0












    $begingroup$

    If $[x]$ denotes floor we have
    $$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
    for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
    $$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
    $$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
    and obviously
    $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
    With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
    $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
    and also if $[x]$ denotes nearest integer we have
    $$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      If $[x]$ denotes floor we have
      $$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
      for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
      $$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
      $$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
      and obviously
      $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
      With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
      $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
      and also if $[x]$ denotes nearest integer we have
      $$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        If $[x]$ denotes floor we have
        $$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
        for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
        $$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
        $$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
        and obviously
        $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
        With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
        $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
        and also if $[x]$ denotes nearest integer we have
        $$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$






        share|cite|improve this answer











        $endgroup$



        If $[x]$ denotes floor we have
        $$sumlimits_{k=0}^{m-1}Big[{n+ks+tover ms}Big]=Big[{n+tover s}Big]$$
        for $ngeqslant0$, $m>0$, $s>tgeqslant0$, $n,m,s,t$ - integers. So
        $$Big[{nover 2}Big]=Big[{nover 6}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]$$
        $$Big[{nover 3}Big]=Big[{nover 6}Big]+Big[{n+3over 6}Big]$$
        and obviously
        $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{nover 2}Big]+Big[{n+3over 6}Big]$$
        With same conditions for $n,m,s,t$ if $[x]$ denotes ceiling we have
        $$Big[{nover 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$
        and also if $[x]$ denotes nearest integer we have
        $$Big[{n+2over 3}Big]+Big[{n+2over 6}Big]+Big[{n+4over 6}Big]=Big[{n+2over 2}Big]+Big[{n+3over 6}Big]$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 18:42

























        answered Jan 2 at 18:15









        user514787user514787

        715210




        715210






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2865296%2freference-request-for-solution-of-ramanujan-identities%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

            Human spaceflight

            Image
            Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
            Read more

            張江高科駅

            Image
            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
            Read more