How to find the nth term in the following sequence: 1,1,2,2,4,4,8,8,16,16 The Next CEO of...

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



The Next CEO of Stack OverflowHow to interpret the OEIS function for the “even fractal sequence” A103391 (1, 2, 2, 3, 2, 4, 3, 5, …)What will be nth term of the following sequence?How to find the nth term of this sequence?Number of possible ordered sequencesFind nth term of sequenceHow can i find the decimal values with a list of integers?Given a sequence find nth termFind nth term for below sequenceProve $lim_{ntoinfty}U_{n} = 1$ given $0 lt U_{n} - {1over U_{n}}lt {1over n}$ and $U_n>0$How to find the nth term in quadratic sequence?












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$begingroup$


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
    16 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
    16 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







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






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asked 20 mins ago









AnonymousAnonymous

111




111




New contributor




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New contributor





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Check out our Code of Conduct.








  • 1




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
















  • 1




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










1




1




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






$begingroup$
How about using the floor function?
$endgroup$
– John. P
16 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
    3 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
    46 secs 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









    $endgroup$














      Your Answer





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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
        3 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
        46 secs 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
        3 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
        46 secs 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 16 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
        3 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
        46 secs ago


















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
















      $begingroup$
      I'm not sure this works for everything, for example the 6th term should be 4 but 2^(3) = 8
      $endgroup$
      – Anonymous
      3 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
      3 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
      46 secs 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
      46 secs 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 14 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 44 secs ago









                  TravisTravis

                  63.8k769151




                  63.8k769151






















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