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












2












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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







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  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    17 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.






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.







$endgroup$








  • 1




    $begingroup$
    How about using the floor function?
    $endgroup$
    – John. P
    17 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.






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.







$endgroup$




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




sequences-and-series






share|cite|improve this question







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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







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asked 21 mins ago









AnonymousAnonymous

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  • 1




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
















  • 1




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










1




1




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






$begingroup$
How about using the floor function?
$endgroup$
– John. P
17 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
    4 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
    1 min ago



















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












    $begingroup$

    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









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      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
        4 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
        1 min 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
        4 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
        1 min 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 17 mins ago









      FlowersFlowers

      638410




      638410












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


















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
















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











      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 15 mins ago









          TravisTravis

          63.8k769151




          63.8k769151























              0












              $begingroup$

              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












                $begingroup$

                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












                  0








                  0





                  $begingroup$

                  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 1 min ago









                  TravisTravis

                  63.8k769151




                  63.8k769151






















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