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












4












$begingroup$


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 noneven numbers.



Any help would be highly appreciated.






sequences-and-series







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  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    Mar 31 at 6:35


















4












$begingroup$


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 noneven numbers.



Any help would be highly appreciated.






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    Mar 31 at 6:35
















4












4








4


1



$begingroup$


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 noneven numbers.



Any help would be highly appreciated.






sequences-and-series







share|cite|improve this question











$endgroup$




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 noneven numbers.



Any help would be highly appreciated.






sequences-and-series




sequences-and-series






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share|cite|improve this question













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share|cite|improve this question








edited Mar 31 at 7:04









YuiTo Cheng

2,3144937




2,3144937










asked Mar 31 at 6:30









AnonymousAnonymous

233




233








  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    Mar 31 at 6:35
















  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    Mar 31 at 6:35










1




1




$begingroup$
How about using the floor function?
$endgroup$
– John. P
Mar 31 at 6:35






$begingroup$
How about using the floor function?
$endgroup$
– John. P
Mar 31 at 6:35












3 Answers
3






active

oldest

votes


















7












$begingroup$

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
    Mar 31 at 6:47






  • 1




    $begingroup$
    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
    Mar 31 at 6:50



















3












$begingroup$

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|improve this answer









$endgroup$





















    1












    $begingroup$

    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









    $endgroup$














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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      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
        Mar 31 at 6:47






      • 1




        $begingroup$
        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
        Mar 31 at 6:50
















      7












      $begingroup$

      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
        Mar 31 at 6:47






      • 1




        $begingroup$
        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
        Mar 31 at 6:50














      7












      7








      7





      $begingroup$

      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 Mar 31 at 6:34









      FlowersFlowers

      683410




      683410












      • $begingroup$
        I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
        $endgroup$
        – Anonymous
        Mar 31 at 6:47






      • 1




        $begingroup$
        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
        Mar 31 at 6:50


















      • $begingroup$
        I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
        $endgroup$
        – Anonymous
        Mar 31 at 6:47






      • 1




        $begingroup$
        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
        Mar 31 at 6:50
















      $begingroup$
      I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
      $endgroup$
      – Anonymous
      Mar 31 at 6:47




      $begingroup$
      I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
      $endgroup$
      – Anonymous
      Mar 31 at 6:47




      1




      1




      $begingroup$
      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
      Mar 31 at 6:50




      $begingroup$
      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
      Mar 31 at 6:50











      3












      $begingroup$

      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|improve this answer









      $endgroup$


















        3












        $begingroup$

        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|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          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|improve this answer









          $endgroup$



          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|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 31 at 6:50









          TravisTravis

          64.1k769151




          64.1k769151























              1












              $begingroup$

              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









              $endgroup$


















                1












                $begingroup$

                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









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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 Mar 31 at 6:37









                  TravisTravis

                  64.1k769151




                  64.1k769151






























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