Prove that function on naturals defined recursively is idempotent on odd numbers












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Consider the function $f$ on natural numbers defined by the following recursion:




  • $f(1)=1$

  • $f(3)=3$

  • $f(2n)=f(n)$

  • $f(4n+1)=2f(2n+1)-f(n)$

  • $f(4n+3)=3f(2n+1)-2f(n)$


Numerical evidence shows that for odd $k$ we have $f(f(k))=k$, but I have no clue on how to prove it. Any ideas?










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    2












    $begingroup$


    Consider the function $f$ on natural numbers defined by the following recursion:




    • $f(1)=1$

    • $f(3)=3$

    • $f(2n)=f(n)$

    • $f(4n+1)=2f(2n+1)-f(n)$

    • $f(4n+3)=3f(2n+1)-2f(n)$


    Numerical evidence shows that for odd $k$ we have $f(f(k))=k$, but I have no clue on how to prove it. Any ideas?










    share|cite|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      Consider the function $f$ on natural numbers defined by the following recursion:




      • $f(1)=1$

      • $f(3)=3$

      • $f(2n)=f(n)$

      • $f(4n+1)=2f(2n+1)-f(n)$

      • $f(4n+3)=3f(2n+1)-2f(n)$


      Numerical evidence shows that for odd $k$ we have $f(f(k))=k$, but I have no clue on how to prove it. Any ideas?










      share|cite|improve this question









      $endgroup$




      Consider the function $f$ on natural numbers defined by the following recursion:




      • $f(1)=1$

      • $f(3)=3$

      • $f(2n)=f(n)$

      • $f(4n+1)=2f(2n+1)-f(n)$

      • $f(4n+3)=3f(2n+1)-2f(n)$


      Numerical evidence shows that for odd $k$ we have $f(f(k))=k$, but I have no clue on how to prove it. Any ideas?







      recurrence-relations recursion






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      asked Dec 19 '18 at 18:37









      A. BellmuntA. Bellmunt

      895515




      895515






















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          $begingroup$

          Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112






          share|cite|improve this answer











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          • $begingroup$
            Brilliant! Thanks a lot.
            $endgroup$
            – A. Bellmunt
            Dec 19 '18 at 18:54











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          active

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          $begingroup$

          Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Brilliant! Thanks a lot.
            $endgroup$
            – A. Bellmunt
            Dec 19 '18 at 18:54
















          1












          $begingroup$

          Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Brilliant! Thanks a lot.
            $endgroup$
            – A. Bellmunt
            Dec 19 '18 at 18:54














          1












          1








          1





          $begingroup$

          Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112






          share|cite|improve this answer











          $endgroup$



          Hint: Consider a binary expansion of natural numbers. Take a detailed solution here https://artofproblemsolving.com/community/c6h60400p365112







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 19 '18 at 18:51

























          answered Dec 19 '18 at 18:41









          greedoidgreedoid

          42.7k1153105




          42.7k1153105












          • $begingroup$
            Brilliant! Thanks a lot.
            $endgroup$
            – A. Bellmunt
            Dec 19 '18 at 18:54


















          • $begingroup$
            Brilliant! Thanks a lot.
            $endgroup$
            – A. Bellmunt
            Dec 19 '18 at 18:54
















          $begingroup$
          Brilliant! Thanks a lot.
          $endgroup$
          – A. Bellmunt
          Dec 19 '18 at 18:54




          $begingroup$
          Brilliant! Thanks a lot.
          $endgroup$
          – A. Bellmunt
          Dec 19 '18 at 18:54


















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