What number comes next in this sequence?












1












$begingroup$


$4, 15, 13, 7, 22, -1, 31, -9, 40, -17, 49$.



What comes next? The answer is $-25$, but why?






sequences-and-series pattern-recognition







share|cite|improve this question









$endgroup$








  • 16




    $begingroup$
    My answer is 42. The reason is that I like the number 42. And there is no one who can prove me wrong. That being said, if I were to try to read the mind of whoever made this problem, I would look at every other term.
    $endgroup$
    – Arthur
    Nov 24 '18 at 21:52








  • 3




    $begingroup$
    Any finite sequence of integers can be continued any way you like. Sometimes there are patterns that suggest that one continuation is more natural than another. I see no such pattern here. If you [edit' the question to tell us where the sequence comes from we may be able to hlep. Otherwise the question will probably be closed.
    $endgroup$
    – Ethan Bolker
    Nov 24 '18 at 21:53






  • 1




    $begingroup$
    wolframalpha.com/input/…
    $endgroup$
    – AccidentalFourierTransform
    Nov 25 '18 at 0:11
















1












$begingroup$


$4, 15, 13, 7, 22, -1, 31, -9, 40, -17, 49$.



What comes next? The answer is $-25$, but why?






sequences-and-series pattern-recognition







share|cite|improve this question









$endgroup$








  • 16




    $begingroup$
    My answer is 42. The reason is that I like the number 42. And there is no one who can prove me wrong. That being said, if I were to try to read the mind of whoever made this problem, I would look at every other term.
    $endgroup$
    – Arthur
    Nov 24 '18 at 21:52








  • 3




    $begingroup$
    Any finite sequence of integers can be continued any way you like. Sometimes there are patterns that suggest that one continuation is more natural than another. I see no such pattern here. If you [edit' the question to tell us where the sequence comes from we may be able to hlep. Otherwise the question will probably be closed.
    $endgroup$
    – Ethan Bolker
    Nov 24 '18 at 21:53






  • 1




    $begingroup$
    wolframalpha.com/input/…
    $endgroup$
    – AccidentalFourierTransform
    Nov 25 '18 at 0:11














1












1








1


1



$begingroup$


$4, 15, 13, 7, 22, -1, 31, -9, 40, -17, 49$.



What comes next? The answer is $-25$, but why?






sequences-and-series pattern-recognition







share|cite|improve this question









$endgroup$




$4, 15, 13, 7, 22, -1, 31, -9, 40, -17, 49$.



What comes next? The answer is $-25$, but why?






sequences-and-series pattern-recognition




sequences-and-series pattern-recognition






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 '18 at 21:49









stackofhay42stackofhay42

1696




1696








  • 16




    $begingroup$
    My answer is 42. The reason is that I like the number 42. And there is no one who can prove me wrong. That being said, if I were to try to read the mind of whoever made this problem, I would look at every other term.
    $endgroup$
    – Arthur
    Nov 24 '18 at 21:52








  • 3




    $begingroup$
    Any finite sequence of integers can be continued any way you like. Sometimes there are patterns that suggest that one continuation is more natural than another. I see no such pattern here. If you [edit' the question to tell us where the sequence comes from we may be able to hlep. Otherwise the question will probably be closed.
    $endgroup$
    – Ethan Bolker
    Nov 24 '18 at 21:53






  • 1




    $begingroup$
    wolframalpha.com/input/…
    $endgroup$
    – AccidentalFourierTransform
    Nov 25 '18 at 0:11














  • 16




    $begingroup$
    My answer is 42. The reason is that I like the number 42. And there is no one who can prove me wrong. That being said, if I were to try to read the mind of whoever made this problem, I would look at every other term.
    $endgroup$
    – Arthur
    Nov 24 '18 at 21:52








  • 3




    $begingroup$
    Any finite sequence of integers can be continued any way you like. Sometimes there are patterns that suggest that one continuation is more natural than another. I see no such pattern here. If you [edit' the question to tell us where the sequence comes from we may be able to hlep. Otherwise the question will probably be closed.
    $endgroup$
    – Ethan Bolker
    Nov 24 '18 at 21:53






  • 1




    $begingroup$
    wolframalpha.com/input/…
    $endgroup$
    – AccidentalFourierTransform
    Nov 25 '18 at 0:11








16




16




$begingroup$
My answer is 42. The reason is that I like the number 42. And there is no one who can prove me wrong. That being said, if I were to try to read the mind of whoever made this problem, I would look at every other term.
$endgroup$
– Arthur
Nov 24 '18 at 21:52






$begingroup$
My answer is 42. The reason is that I like the number 42. And there is no one who can prove me wrong. That being said, if I were to try to read the mind of whoever made this problem, I would look at every other term.
$endgroup$
– Arthur
Nov 24 '18 at 21:52






3




3




$begingroup$
Any finite sequence of integers can be continued any way you like. Sometimes there are patterns that suggest that one continuation is more natural than another. I see no such pattern here. If you [edit' the question to tell us where the sequence comes from we may be able to hlep. Otherwise the question will probably be closed.
$endgroup$
– Ethan Bolker
Nov 24 '18 at 21:53




$begingroup$
Any finite sequence of integers can be continued any way you like. Sometimes there are patterns that suggest that one continuation is more natural than another. I see no such pattern here. If you [edit' the question to tell us where the sequence comes from we may be able to hlep. Otherwise the question will probably be closed.
$endgroup$
– Ethan Bolker
Nov 24 '18 at 21:53




1




1




$begingroup$
wolframalpha.com/input/…
$endgroup$
– AccidentalFourierTransform
Nov 25 '18 at 0:11




$begingroup$
wolframalpha.com/input/…
$endgroup$
– AccidentalFourierTransform
Nov 25 '18 at 0:11










3 Answers
3






active

oldest

votes


















15












$begingroup$

Break up the sequence into the even ordered terms and odd ordered.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Wow! I was looking for some very complicated patterns, and it is actually so easy.
    $endgroup$
    – Mark
    Nov 24 '18 at 21:55



















4












$begingroup$


  • Firs observation, first term and second term add up to 19, third and fourth add to 20, fifth and sixth add to 21 and so on..


According to that, the the next number is $49+x = 24 implies x = -25$




  • Second observation, second and third terms add to 28, the fourth and fifth add to 29 and so on...


Therefore, you can generate the next number using these two observations anywhere in the sequence.



I know this is not the best way to predict the next number. However, it is not a bad try.



:)






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    As a general rule, the simplest kind of sequences of numbers are linear recurrence sequences. It is a matter of finding the recurrence relation. Using the first $10$ terms of the sequence, and linear algebra, the generating function appears to be



    $$ A(x) := frac{-23 x^4 + 5 x^3 + 15 x^2 + 4 x}{(x^2 - 1)^2} = 4 x + 15x^2 + 13x^3 +dots $$



    with the $11$th term $49$ being consistent with the generating function.



    The $12$th term is then $-25$ as you stated. The polynomial numerator and denominator coefficients can be found using linear algebra in the general case. In your case, by looking at every other term, you can find that they are both arithmetic progressions with constant differences $9$ and $-8$.






    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









      15












      $begingroup$

      Break up the sequence into the even ordered terms and odd ordered.






      share|cite|improve this answer









      $endgroup$









      • 2




        $begingroup$
        Wow! I was looking for some very complicated patterns, and it is actually so easy.
        $endgroup$
        – Mark
        Nov 24 '18 at 21:55
















      15












      $begingroup$

      Break up the sequence into the even ordered terms and odd ordered.






      share|cite|improve this answer









      $endgroup$









      • 2




        $begingroup$
        Wow! I was looking for some very complicated patterns, and it is actually so easy.
        $endgroup$
        – Mark
        Nov 24 '18 at 21:55














      15












      15








      15





      $begingroup$

      Break up the sequence into the even ordered terms and odd ordered.






      share|cite|improve this answer









      $endgroup$



      Break up the sequence into the even ordered terms and odd ordered.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Nov 24 '18 at 21:53









      TurlocTheRedTurlocTheRed

      838311




      838311








      • 2




        $begingroup$
        Wow! I was looking for some very complicated patterns, and it is actually so easy.
        $endgroup$
        – Mark
        Nov 24 '18 at 21:55














      • 2




        $begingroup$
        Wow! I was looking for some very complicated patterns, and it is actually so easy.
        $endgroup$
        – Mark
        Nov 24 '18 at 21:55








      2




      2




      $begingroup$
      Wow! I was looking for some very complicated patterns, and it is actually so easy.
      $endgroup$
      – Mark
      Nov 24 '18 at 21:55




      $begingroup$
      Wow! I was looking for some very complicated patterns, and it is actually so easy.
      $endgroup$
      – Mark
      Nov 24 '18 at 21:55











      4












      $begingroup$


      • Firs observation, first term and second term add up to 19, third and fourth add to 20, fifth and sixth add to 21 and so on..


      According to that, the the next number is $49+x = 24 implies x = -25$




      • Second observation, second and third terms add to 28, the fourth and fifth add to 29 and so on...


      Therefore, you can generate the next number using these two observations anywhere in the sequence.



      I know this is not the best way to predict the next number. However, it is not a bad try.



      :)






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$


        • Firs observation, first term and second term add up to 19, third and fourth add to 20, fifth and sixth add to 21 and so on..


        According to that, the the next number is $49+x = 24 implies x = -25$




        • Second observation, second and third terms add to 28, the fourth and fifth add to 29 and so on...


        Therefore, you can generate the next number using these two observations anywhere in the sequence.



        I know this is not the best way to predict the next number. However, it is not a bad try.



        :)






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$


          • Firs observation, first term and second term add up to 19, third and fourth add to 20, fifth and sixth add to 21 and so on..


          According to that, the the next number is $49+x = 24 implies x = -25$




          • Second observation, second and third terms add to 28, the fourth and fifth add to 29 and so on...


          Therefore, you can generate the next number using these two observations anywhere in the sequence.



          I know this is not the best way to predict the next number. However, it is not a bad try.



          :)






          share|cite|improve this answer











          $endgroup$




          • Firs observation, first term and second term add up to 19, third and fourth add to 20, fifth and sixth add to 21 and so on..


          According to that, the the next number is $49+x = 24 implies x = -25$




          • Second observation, second and third terms add to 28, the fourth and fifth add to 29 and so on...


          Therefore, you can generate the next number using these two observations anywhere in the sequence.



          I know this is not the best way to predict the next number. However, it is not a bad try.



          :)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 '18 at 22:21

























          answered Nov 24 '18 at 21:59









          Maged SaeedMaged Saeed

          8471417




          8471417























              1












              $begingroup$

              As a general rule, the simplest kind of sequences of numbers are linear recurrence sequences. It is a matter of finding the recurrence relation. Using the first $10$ terms of the sequence, and linear algebra, the generating function appears to be



              $$ A(x) := frac{-23 x^4 + 5 x^3 + 15 x^2 + 4 x}{(x^2 - 1)^2} = 4 x + 15x^2 + 13x^3 +dots $$



              with the $11$th term $49$ being consistent with the generating function.



              The $12$th term is then $-25$ as you stated. The polynomial numerator and denominator coefficients can be found using linear algebra in the general case. In your case, by looking at every other term, you can find that they are both arithmetic progressions with constant differences $9$ and $-8$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                As a general rule, the simplest kind of sequences of numbers are linear recurrence sequences. It is a matter of finding the recurrence relation. Using the first $10$ terms of the sequence, and linear algebra, the generating function appears to be



                $$ A(x) := frac{-23 x^4 + 5 x^3 + 15 x^2 + 4 x}{(x^2 - 1)^2} = 4 x + 15x^2 + 13x^3 +dots $$



                with the $11$th term $49$ being consistent with the generating function.



                The $12$th term is then $-25$ as you stated. The polynomial numerator and denominator coefficients can be found using linear algebra in the general case. In your case, by looking at every other term, you can find that they are both arithmetic progressions with constant differences $9$ and $-8$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  As a general rule, the simplest kind of sequences of numbers are linear recurrence sequences. It is a matter of finding the recurrence relation. Using the first $10$ terms of the sequence, and linear algebra, the generating function appears to be



                  $$ A(x) := frac{-23 x^4 + 5 x^3 + 15 x^2 + 4 x}{(x^2 - 1)^2} = 4 x + 15x^2 + 13x^3 +dots $$



                  with the $11$th term $49$ being consistent with the generating function.



                  The $12$th term is then $-25$ as you stated. The polynomial numerator and denominator coefficients can be found using linear algebra in the general case. In your case, by looking at every other term, you can find that they are both arithmetic progressions with constant differences $9$ and $-8$.






                  share|cite|improve this answer









                  $endgroup$



                  As a general rule, the simplest kind of sequences of numbers are linear recurrence sequences. It is a matter of finding the recurrence relation. Using the first $10$ terms of the sequence, and linear algebra, the generating function appears to be



                  $$ A(x) := frac{-23 x^4 + 5 x^3 + 15 x^2 + 4 x}{(x^2 - 1)^2} = 4 x + 15x^2 + 13x^3 +dots $$



                  with the $11$th term $49$ being consistent with the generating function.



                  The $12$th term is then $-25$ as you stated. The polynomial numerator and denominator coefficients can be found using linear algebra in the general case. In your case, by looking at every other term, you can find that they are both arithmetic progressions with constant differences $9$ and $-8$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 '18 at 3:21









                  SomosSomos

                  13.1k11034




                  13.1k11034






























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