How to find the nth term in the following sequence: 1,1,2,2,4,4,8,8,16,16












2












$begingroup$


I'm having difficulty in finding the formula for the sequence above, when I put this in wolframalpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.



I'm thinking I'll most likely need to have a condition for even numbers and another for non even numbers.



Any help would be highly appreciated.










share|cite|improve this question







New contributor




Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    23 mins ago


















2












$begingroup$


I'm having difficulty in finding the formula for the sequence above, when I put this in wolframalpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.



I'm thinking I'll most likely need to have a condition for even numbers and another for non even numbers.



Any help would be highly appreciated.










share|cite|improve this question







New contributor




Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    23 mins ago
















2












2








2





$begingroup$


I'm having difficulty in finding the formula for the sequence above, when I put this in wolframalpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.



I'm thinking I'll most likely need to have a condition for even numbers and another for non even numbers.



Any help would be highly appreciated.










share|cite|improve this question







New contributor




Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I'm having difficulty in finding the formula for the sequence above, when I put this in wolframalpha it gave me a rather complex formula which I'm not convinced even works properly but I'm sure there's a simple way to achieve this. I've searched for many similar sequences but couldn't find anything that helped me.



I'm thinking I'll most likely need to have a condition for even numbers and another for non even numbers.



Any help would be highly appreciated.







sequences-and-series






share|cite|improve this question







New contributor




Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 28 mins ago









AnonymousAnonymous

132




132




New contributor




Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Anonymous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    23 mins ago
















  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    23 mins ago










1




1




$begingroup$
How about using the floor function?
$endgroup$
– John. P
23 mins ago






$begingroup$
How about using the floor function?
$endgroup$
– John. P
23 mins ago












3 Answers
3






active

oldest

votes


















4












$begingroup$

These are just powers of two. So: $2^{lfloor n / 2rfloor}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
    $endgroup$
    – Anonymous
    11 mins ago










  • $begingroup$
    This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
    $endgroup$
    – Flowers
    8 mins ago



















1












$begingroup$

Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
$$a_n = 2 a_{n - 2}, qquad a_0 = a_1 = 1.$$
Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^{n - 2}$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt{2}$. So, the general solution is
$$a_n = A (sqrt{2})^n + B(-sqrt{2})^n = (sqrt{2})^n [A + B(-1)^n] .$$
Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.






share|cite









$endgroup$





















    0












    $begingroup$

    The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor frac{n}{2} rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.




    So, the sequence is given (for appropriate indexing) by $$color{#df0000}{boxed{a_n := 2^{lfloor n / 2 rfloor}}} .$$







    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
      });


      }
      });






      Anonymous is a new contributor. Be nice, and check out our Code of Conduct.










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3169109%2fhow-to-find-the-nth-term-in-the-following-sequence-1-1-2-2-4-4-8-8-16-16%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









      4












      $begingroup$

      These are just powers of two. So: $2^{lfloor n / 2rfloor}$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
        $endgroup$
        – Anonymous
        11 mins ago










      • $begingroup$
        This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
        $endgroup$
        – Flowers
        8 mins ago
















      4












      $begingroup$

      These are just powers of two. So: $2^{lfloor n / 2rfloor}$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
        $endgroup$
        – Anonymous
        11 mins ago










      • $begingroup$
        This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
        $endgroup$
        – Flowers
        8 mins ago














      4












      4








      4





      $begingroup$

      These are just powers of two. So: $2^{lfloor n / 2rfloor}$






      share|cite|improve this answer









      $endgroup$



      These are just powers of two. So: $2^{lfloor n / 2rfloor}$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 24 mins ago









      FlowersFlowers

      653410




      653410












      • $begingroup$
        I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
        $endgroup$
        – Anonymous
        11 mins ago










      • $begingroup$
        This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
        $endgroup$
        – Flowers
        8 mins ago


















      • $begingroup$
        I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
        $endgroup$
        – Anonymous
        11 mins ago










      • $begingroup$
        This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
        $endgroup$
        – Flowers
        8 mins ago
















      $begingroup$
      I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
      $endgroup$
      – Anonymous
      11 mins ago




      $begingroup$
      I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
      $endgroup$
      – Anonymous
      11 mins ago












      $begingroup$
      This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
      $endgroup$
      – Flowers
      8 mins ago




      $begingroup$
      This is assuming zero-indexing. So the first element is $n=0$. If you want it to be one-indexed then just subtract 1 from n in the formula.
      $endgroup$
      – Flowers
      8 mins ago











      1












      $begingroup$

      Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
      $$a_n = 2 a_{n - 2}, qquad a_0 = a_1 = 1.$$
      Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^{n - 2}$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt{2}$. So, the general solution is
      $$a_n = A (sqrt{2})^n + B(-sqrt{2})^n = (sqrt{2})^n [A + B(-1)^n] .$$
      Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.






      share|cite









      $endgroup$


















        1












        $begingroup$

        Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
        $$a_n = 2 a_{n - 2}, qquad a_0 = a_1 = 1.$$
        Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^{n - 2}$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt{2}$. So, the general solution is
        $$a_n = A (sqrt{2})^n + B(-sqrt{2})^n = (sqrt{2})^n [A + B(-1)^n] .$$
        Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.






        share|cite









        $endgroup$
















          1












          1








          1





          $begingroup$

          Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
          $$a_n = 2 a_{n - 2}, qquad a_0 = a_1 = 1.$$
          Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^{n - 2}$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt{2}$. So, the general solution is
          $$a_n = A (sqrt{2})^n + B(-sqrt{2})^n = (sqrt{2})^n [A + B(-1)^n] .$$
          Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.






          share|cite









          $endgroup$



          Alternatively, this is an example of a sequence where the $n$th term is a fixed linear combination of the immediately previous terms: We can write it as
          $$a_n = 2 a_{n - 2}, qquad a_0 = a_1 = 1.$$
          Using the ansatz $a_n = C r^n$ and substituting in the recursion formula gives $C r^n = 2 C r^{n - 2}$. Rearranging and clearing gives the characteristic equation $r^2 - 2 = 0$, whose solutions are $pm sqrt{2}$. So, the general solution is
          $$a_n = A (sqrt{2})^n + B(-sqrt{2})^n = (sqrt{2})^n [A + B(-1)^n] .$$
          Substituting the initial values $a_0 = a_1 = 1$ gives a linear system in the coefficients $A, B$.







          share|cite












          share|cite



          share|cite










          answered 8 mins ago









          TravisTravis

          63.8k769151




          63.8k769151























              0












              $begingroup$

              The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor frac{n}{2} rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.




              So, the sequence is given (for appropriate indexing) by $$color{#df0000}{boxed{a_n := 2^{lfloor n / 2 rfloor}}} .$$







              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor frac{n}{2} rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.




                So, the sequence is given (for appropriate indexing) by $$color{#df0000}{boxed{a_n := 2^{lfloor n / 2 rfloor}}} .$$







                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor frac{n}{2} rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.




                  So, the sequence is given (for appropriate indexing) by $$color{#df0000}{boxed{a_n := 2^{lfloor n / 2 rfloor}}} .$$







                  share|cite|improve this answer









                  $endgroup$



                  The sequence is the powers of two, each repeated twice. We can encode the latter feature using the quantity $lfloor frac{n}{2} rfloor$, which has values $0, 0, 1, 1, 2, 2, ldots$.




                  So, the sequence is given (for appropriate indexing) by $$color{#df0000}{boxed{a_n := 2^{lfloor n / 2 rfloor}}} .$$








                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 21 mins ago









                  TravisTravis

                  63.8k769151




                  63.8k769151






















                      Anonymous is a new contributor. Be nice, and check out our Code of Conduct.










                      draft saved

                      draft discarded


















                      Anonymous is a new contributor. Be nice, and check out our Code of Conduct.













                      Anonymous is a new contributor. Be nice, and check out our Code of Conduct.












                      Anonymous is a new contributor. Be nice, and check out our Code of Conduct.
















                      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%2f3169109%2fhow-to-find-the-nth-term-in-the-following-sequence-1-1-2-2-4-4-8-8-16-16%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

                      flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror

                      Mangá

                      Eduardo VII do Reino Unido