Inequality involving number of binary Lyndon words of length $n$ and $n+1$












5












$begingroup$



Let $f(n)$ be the number of binary Lyndon words of length $n$. This sequence is given by OEIS entry A001037. Is it true that $2f(n) ge f(n+1)$ for all positive $n$?




I have found a general formula to calculate $f(n)$:
$$
f(n) = frac{1}{n} sum_{d mid n} mu(n/d) 2^d.
$$

Symmetry also allows us to rewrite it another way:
$$
f(n) = frac{1}{n} sum_{d mid n} mu(d) 2^{n/d}.
$$

Here $mu(x)$ is the Möbius function, and $n$ is a positive integer (the case $n=0$ can be handled separately). It is possible to substitute these identities into the inequality above and perform some simplifications, but it does not seem to go anywhere. Any help is appreciated.



EDIT: My original formula was incorrect, I forgot the factor of $1/n$ in front of the summation.










share|cite|improve this question











$endgroup$

















    5












    $begingroup$



    Let $f(n)$ be the number of binary Lyndon words of length $n$. This sequence is given by OEIS entry A001037. Is it true that $2f(n) ge f(n+1)$ for all positive $n$?




    I have found a general formula to calculate $f(n)$:
    $$
    f(n) = frac{1}{n} sum_{d mid n} mu(n/d) 2^d.
    $$

    Symmetry also allows us to rewrite it another way:
    $$
    f(n) = frac{1}{n} sum_{d mid n} mu(d) 2^{n/d}.
    $$

    Here $mu(x)$ is the Möbius function, and $n$ is a positive integer (the case $n=0$ can be handled separately). It is possible to substitute these identities into the inequality above and perform some simplifications, but it does not seem to go anywhere. Any help is appreciated.



    EDIT: My original formula was incorrect, I forgot the factor of $1/n$ in front of the summation.










    share|cite|improve this question











    $endgroup$















      5












      5








      5


      2



      $begingroup$



      Let $f(n)$ be the number of binary Lyndon words of length $n$. This sequence is given by OEIS entry A001037. Is it true that $2f(n) ge f(n+1)$ for all positive $n$?




      I have found a general formula to calculate $f(n)$:
      $$
      f(n) = frac{1}{n} sum_{d mid n} mu(n/d) 2^d.
      $$

      Symmetry also allows us to rewrite it another way:
      $$
      f(n) = frac{1}{n} sum_{d mid n} mu(d) 2^{n/d}.
      $$

      Here $mu(x)$ is the Möbius function, and $n$ is a positive integer (the case $n=0$ can be handled separately). It is possible to substitute these identities into the inequality above and perform some simplifications, but it does not seem to go anywhere. Any help is appreciated.



      EDIT: My original formula was incorrect, I forgot the factor of $1/n$ in front of the summation.










      share|cite|improve this question











      $endgroup$





      Let $f(n)$ be the number of binary Lyndon words of length $n$. This sequence is given by OEIS entry A001037. Is it true that $2f(n) ge f(n+1)$ for all positive $n$?




      I have found a general formula to calculate $f(n)$:
      $$
      f(n) = frac{1}{n} sum_{d mid n} mu(n/d) 2^d.
      $$

      Symmetry also allows us to rewrite it another way:
      $$
      f(n) = frac{1}{n} sum_{d mid n} mu(d) 2^{n/d}.
      $$

      Here $mu(x)$ is the Möbius function, and $n$ is a positive integer (the case $n=0$ can be handled separately). It is possible to substitute these identities into the inequality above and perform some simplifications, but it does not seem to go anywhere. Any help is appreciated.



      EDIT: My original formula was incorrect, I forgot the factor of $1/n$ in front of the summation.







      sequences-and-series combinatorics necklace-and-bracelets






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      edited Dec 4 '18 at 3:09









      Saad

      19.7k92352




      19.7k92352










      asked Jul 10 '15 at 3:46









      RyanRyan

      596628




      596628






















          2 Answers
          2






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          1





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          $begingroup$

          $$f(n)=frac{1}{n} sum_{d | n} mu(d) 2^{n/d} le frac{ 2^n + d(n)cdot 2^{n/2} }n = frac{ 2^n + O(sqrt{n}cdot 2^{n/2}) }n,$$



          where $d(n)=O(sqrt n)$ denotes the umber of divisors of $n$.
          We have $$2f(n)!!-!!f(n+1)!ge! frac{ 2^{n+1} + O(sqrt{n}cdot 2^{n/2}) }n !-!frac{ 2^{n+1} + O(sqrt{n+1}cdot 2^{(n+1)/2}) }{n+1}!ge! frac{ 2^{n+1}}{n(n+1)}! - ! O(2^{n/2}),$$



          which tends to infinity, thus it is enough to check the first few terms, which can be done in OEIS (though I cheated a bit, because I didn't look up the constant in the $O$ notation, but it's definitely not that big).






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Where does that first identity come from? The limiting behavior makes sense to me but I'm not sure how to derive it.
            $endgroup$
            – Ryan
            Dec 3 '18 at 18:17










          • $begingroup$
            @Ryan I've added $d(n)$, I hope that explains all.
            $endgroup$
            – domotorp
            Dec 3 '18 at 20:23



















          2












          $begingroup$

          Step 1: For $n in mathbb{N}_+$,$$
          frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} leqslant f(n) leqslant frac{2^n}{n} + frac{2^{frac{n}{6}}}{6}. tag{1}
          $$



          Proof: For the upper bound,$$
          n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d leqslant 2^n + sum_{substack{d mid n, d < n\μ(n / d) = 1}} 2^d. tag{2}
          $$

          If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = 1$, then $dfrac{n}{d}$ has at least two distinct prime divisors, which implies $dfrac{n}{d} geqslant 6$, i.e. $d leqslant dfrac{n}{6}$. Thus $(2) leqslant 2^n + dfrac{n}{6} 2^{frac{n}{6}} Rightarrow f(n) leqslant dfrac{2^n}{n} + dfrac{2^{frac{n}{6}}}{6}$.



          For the lower bound,$$
          n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d geqslant 2^n - sum_{substack{d mid n, d < n\μ(n / d) = -1}} 2^d. tag{3}
          $$

          If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = -1$, then $dfrac{n}{d}$ has at least one prime divisor, which implies $dfrac{n}{d} geqslant 2$, i.e. $d leqslant dfrac{n}{2}$. Thus $(3) geqslant 2^n - dfrac{n}{2} 2^{frac{n}{2}} Rightarrow f(n) geqslant dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2}$. Therefore, (1) holds.



          Step 2: $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



          Proof: By (1. 1), it suffices to prove that $2 left( dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2} right) geqslant dfrac{2^{n + 1}}{n + 1} + dfrac{2^{frac{n + 1}{6}}}{6}$. Because$$
          2 left( frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} right) geqslant frac{2^{n + 1}}{n + 1} + frac{2^{frac{n + 1}{6}}}{6} Longleftrightarrow frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6},
          $$

          and $2^{frac{n + 1}{4}} geqslant n + 1$ for $n geqslant 15$, then$$
          frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n + 1}{2}} = 2^{frac{n}{2}} + (sqrt{2} - 1) 2^{frac{n}{2}} geqslant 2^{frac{n}{2}} + frac{2^{frac{n}{2}}}{6} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6}.
          $$

          Therefore, $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



          Step 3: $2f(n) geqslant f(n + 1)$ holds for $0 leqslant n leqslant 14$.



          Proof: Looking up in A001037 verifies this.






          share|cite|improve this answer









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            2 Answers
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            2 Answers
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            1





            +250







            $begingroup$

            $$f(n)=frac{1}{n} sum_{d | n} mu(d) 2^{n/d} le frac{ 2^n + d(n)cdot 2^{n/2} }n = frac{ 2^n + O(sqrt{n}cdot 2^{n/2}) }n,$$



            where $d(n)=O(sqrt n)$ denotes the umber of divisors of $n$.
            We have $$2f(n)!!-!!f(n+1)!ge! frac{ 2^{n+1} + O(sqrt{n}cdot 2^{n/2}) }n !-!frac{ 2^{n+1} + O(sqrt{n+1}cdot 2^{(n+1)/2}) }{n+1}!ge! frac{ 2^{n+1}}{n(n+1)}! - ! O(2^{n/2}),$$



            which tends to infinity, thus it is enough to check the first few terms, which can be done in OEIS (though I cheated a bit, because I didn't look up the constant in the $O$ notation, but it's definitely not that big).






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Where does that first identity come from? The limiting behavior makes sense to me but I'm not sure how to derive it.
              $endgroup$
              – Ryan
              Dec 3 '18 at 18:17










            • $begingroup$
              @Ryan I've added $d(n)$, I hope that explains all.
              $endgroup$
              – domotorp
              Dec 3 '18 at 20:23
















            1





            +250







            $begingroup$

            $$f(n)=frac{1}{n} sum_{d | n} mu(d) 2^{n/d} le frac{ 2^n + d(n)cdot 2^{n/2} }n = frac{ 2^n + O(sqrt{n}cdot 2^{n/2}) }n,$$



            where $d(n)=O(sqrt n)$ denotes the umber of divisors of $n$.
            We have $$2f(n)!!-!!f(n+1)!ge! frac{ 2^{n+1} + O(sqrt{n}cdot 2^{n/2}) }n !-!frac{ 2^{n+1} + O(sqrt{n+1}cdot 2^{(n+1)/2}) }{n+1}!ge! frac{ 2^{n+1}}{n(n+1)}! - ! O(2^{n/2}),$$



            which tends to infinity, thus it is enough to check the first few terms, which can be done in OEIS (though I cheated a bit, because I didn't look up the constant in the $O$ notation, but it's definitely not that big).






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Where does that first identity come from? The limiting behavior makes sense to me but I'm not sure how to derive it.
              $endgroup$
              – Ryan
              Dec 3 '18 at 18:17










            • $begingroup$
              @Ryan I've added $d(n)$, I hope that explains all.
              $endgroup$
              – domotorp
              Dec 3 '18 at 20:23














            1





            +250







            1





            +250



            1




            +250



            $begingroup$

            $$f(n)=frac{1}{n} sum_{d | n} mu(d) 2^{n/d} le frac{ 2^n + d(n)cdot 2^{n/2} }n = frac{ 2^n + O(sqrt{n}cdot 2^{n/2}) }n,$$



            where $d(n)=O(sqrt n)$ denotes the umber of divisors of $n$.
            We have $$2f(n)!!-!!f(n+1)!ge! frac{ 2^{n+1} + O(sqrt{n}cdot 2^{n/2}) }n !-!frac{ 2^{n+1} + O(sqrt{n+1}cdot 2^{(n+1)/2}) }{n+1}!ge! frac{ 2^{n+1}}{n(n+1)}! - ! O(2^{n/2}),$$



            which tends to infinity, thus it is enough to check the first few terms, which can be done in OEIS (though I cheated a bit, because I didn't look up the constant in the $O$ notation, but it's definitely not that big).






            share|cite|improve this answer











            $endgroup$



            $$f(n)=frac{1}{n} sum_{d | n} mu(d) 2^{n/d} le frac{ 2^n + d(n)cdot 2^{n/2} }n = frac{ 2^n + O(sqrt{n}cdot 2^{n/2}) }n,$$



            where $d(n)=O(sqrt n)$ denotes the umber of divisors of $n$.
            We have $$2f(n)!!-!!f(n+1)!ge! frac{ 2^{n+1} + O(sqrt{n}cdot 2^{n/2}) }n !-!frac{ 2^{n+1} + O(sqrt{n+1}cdot 2^{(n+1)/2}) }{n+1}!ge! frac{ 2^{n+1}}{n(n+1)}! - ! O(2^{n/2}),$$



            which tends to infinity, thus it is enough to check the first few terms, which can be done in OEIS (though I cheated a bit, because I didn't look up the constant in the $O$ notation, but it's definitely not that big).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 3 '18 at 20:23

























            answered Dec 3 '18 at 12:16









            domotorpdomotorp

            910515




            910515












            • $begingroup$
              Where does that first identity come from? The limiting behavior makes sense to me but I'm not sure how to derive it.
              $endgroup$
              – Ryan
              Dec 3 '18 at 18:17










            • $begingroup$
              @Ryan I've added $d(n)$, I hope that explains all.
              $endgroup$
              – domotorp
              Dec 3 '18 at 20:23


















            • $begingroup$
              Where does that first identity come from? The limiting behavior makes sense to me but I'm not sure how to derive it.
              $endgroup$
              – Ryan
              Dec 3 '18 at 18:17










            • $begingroup$
              @Ryan I've added $d(n)$, I hope that explains all.
              $endgroup$
              – domotorp
              Dec 3 '18 at 20:23
















            $begingroup$
            Where does that first identity come from? The limiting behavior makes sense to me but I'm not sure how to derive it.
            $endgroup$
            – Ryan
            Dec 3 '18 at 18:17




            $begingroup$
            Where does that first identity come from? The limiting behavior makes sense to me but I'm not sure how to derive it.
            $endgroup$
            – Ryan
            Dec 3 '18 at 18:17












            $begingroup$
            @Ryan I've added $d(n)$, I hope that explains all.
            $endgroup$
            – domotorp
            Dec 3 '18 at 20:23




            $begingroup$
            @Ryan I've added $d(n)$, I hope that explains all.
            $endgroup$
            – domotorp
            Dec 3 '18 at 20:23











            2












            $begingroup$

            Step 1: For $n in mathbb{N}_+$,$$
            frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} leqslant f(n) leqslant frac{2^n}{n} + frac{2^{frac{n}{6}}}{6}. tag{1}
            $$



            Proof: For the upper bound,$$
            n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d leqslant 2^n + sum_{substack{d mid n, d < n\μ(n / d) = 1}} 2^d. tag{2}
            $$

            If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = 1$, then $dfrac{n}{d}$ has at least two distinct prime divisors, which implies $dfrac{n}{d} geqslant 6$, i.e. $d leqslant dfrac{n}{6}$. Thus $(2) leqslant 2^n + dfrac{n}{6} 2^{frac{n}{6}} Rightarrow f(n) leqslant dfrac{2^n}{n} + dfrac{2^{frac{n}{6}}}{6}$.



            For the lower bound,$$
            n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d geqslant 2^n - sum_{substack{d mid n, d < n\μ(n / d) = -1}} 2^d. tag{3}
            $$

            If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = -1$, then $dfrac{n}{d}$ has at least one prime divisor, which implies $dfrac{n}{d} geqslant 2$, i.e. $d leqslant dfrac{n}{2}$. Thus $(3) geqslant 2^n - dfrac{n}{2} 2^{frac{n}{2}} Rightarrow f(n) geqslant dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2}$. Therefore, (1) holds.



            Step 2: $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



            Proof: By (1. 1), it suffices to prove that $2 left( dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2} right) geqslant dfrac{2^{n + 1}}{n + 1} + dfrac{2^{frac{n + 1}{6}}}{6}$. Because$$
            2 left( frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} right) geqslant frac{2^{n + 1}}{n + 1} + frac{2^{frac{n + 1}{6}}}{6} Longleftrightarrow frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6},
            $$

            and $2^{frac{n + 1}{4}} geqslant n + 1$ for $n geqslant 15$, then$$
            frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n + 1}{2}} = 2^{frac{n}{2}} + (sqrt{2} - 1) 2^{frac{n}{2}} geqslant 2^{frac{n}{2}} + frac{2^{frac{n}{2}}}{6} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6}.
            $$

            Therefore, $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



            Step 3: $2f(n) geqslant f(n + 1)$ holds for $0 leqslant n leqslant 14$.



            Proof: Looking up in A001037 verifies this.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Step 1: For $n in mathbb{N}_+$,$$
              frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} leqslant f(n) leqslant frac{2^n}{n} + frac{2^{frac{n}{6}}}{6}. tag{1}
              $$



              Proof: For the upper bound,$$
              n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d leqslant 2^n + sum_{substack{d mid n, d < n\μ(n / d) = 1}} 2^d. tag{2}
              $$

              If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = 1$, then $dfrac{n}{d}$ has at least two distinct prime divisors, which implies $dfrac{n}{d} geqslant 6$, i.e. $d leqslant dfrac{n}{6}$. Thus $(2) leqslant 2^n + dfrac{n}{6} 2^{frac{n}{6}} Rightarrow f(n) leqslant dfrac{2^n}{n} + dfrac{2^{frac{n}{6}}}{6}$.



              For the lower bound,$$
              n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d geqslant 2^n - sum_{substack{d mid n, d < n\μ(n / d) = -1}} 2^d. tag{3}
              $$

              If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = -1$, then $dfrac{n}{d}$ has at least one prime divisor, which implies $dfrac{n}{d} geqslant 2$, i.e. $d leqslant dfrac{n}{2}$. Thus $(3) geqslant 2^n - dfrac{n}{2} 2^{frac{n}{2}} Rightarrow f(n) geqslant dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2}$. Therefore, (1) holds.



              Step 2: $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



              Proof: By (1. 1), it suffices to prove that $2 left( dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2} right) geqslant dfrac{2^{n + 1}}{n + 1} + dfrac{2^{frac{n + 1}{6}}}{6}$. Because$$
              2 left( frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} right) geqslant frac{2^{n + 1}}{n + 1} + frac{2^{frac{n + 1}{6}}}{6} Longleftrightarrow frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6},
              $$

              and $2^{frac{n + 1}{4}} geqslant n + 1$ for $n geqslant 15$, then$$
              frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n + 1}{2}} = 2^{frac{n}{2}} + (sqrt{2} - 1) 2^{frac{n}{2}} geqslant 2^{frac{n}{2}} + frac{2^{frac{n}{2}}}{6} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6}.
              $$

              Therefore, $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



              Step 3: $2f(n) geqslant f(n + 1)$ holds for $0 leqslant n leqslant 14$.



              Proof: Looking up in A001037 verifies this.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Step 1: For $n in mathbb{N}_+$,$$
                frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} leqslant f(n) leqslant frac{2^n}{n} + frac{2^{frac{n}{6}}}{6}. tag{1}
                $$



                Proof: For the upper bound,$$
                n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d leqslant 2^n + sum_{substack{d mid n, d < n\μ(n / d) = 1}} 2^d. tag{2}
                $$

                If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = 1$, then $dfrac{n}{d}$ has at least two distinct prime divisors, which implies $dfrac{n}{d} geqslant 6$, i.e. $d leqslant dfrac{n}{6}$. Thus $(2) leqslant 2^n + dfrac{n}{6} 2^{frac{n}{6}} Rightarrow f(n) leqslant dfrac{2^n}{n} + dfrac{2^{frac{n}{6}}}{6}$.



                For the lower bound,$$
                n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d geqslant 2^n - sum_{substack{d mid n, d < n\μ(n / d) = -1}} 2^d. tag{3}
                $$

                If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = -1$, then $dfrac{n}{d}$ has at least one prime divisor, which implies $dfrac{n}{d} geqslant 2$, i.e. $d leqslant dfrac{n}{2}$. Thus $(3) geqslant 2^n - dfrac{n}{2} 2^{frac{n}{2}} Rightarrow f(n) geqslant dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2}$. Therefore, (1) holds.



                Step 2: $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



                Proof: By (1. 1), it suffices to prove that $2 left( dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2} right) geqslant dfrac{2^{n + 1}}{n + 1} + dfrac{2^{frac{n + 1}{6}}}{6}$. Because$$
                2 left( frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} right) geqslant frac{2^{n + 1}}{n + 1} + frac{2^{frac{n + 1}{6}}}{6} Longleftrightarrow frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6},
                $$

                and $2^{frac{n + 1}{4}} geqslant n + 1$ for $n geqslant 15$, then$$
                frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n + 1}{2}} = 2^{frac{n}{2}} + (sqrt{2} - 1) 2^{frac{n}{2}} geqslant 2^{frac{n}{2}} + frac{2^{frac{n}{2}}}{6} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6}.
                $$

                Therefore, $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



                Step 3: $2f(n) geqslant f(n + 1)$ holds for $0 leqslant n leqslant 14$.



                Proof: Looking up in A001037 verifies this.






                share|cite|improve this answer









                $endgroup$



                Step 1: For $n in mathbb{N}_+$,$$
                frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} leqslant f(n) leqslant frac{2^n}{n} + frac{2^{frac{n}{6}}}{6}. tag{1}
                $$



                Proof: For the upper bound,$$
                n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d leqslant 2^n + sum_{substack{d mid n, d < n\μ(n / d) = 1}} 2^d. tag{2}
                $$

                If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = 1$, then $dfrac{n}{d}$ has at least two distinct prime divisors, which implies $dfrac{n}{d} geqslant 6$, i.e. $d leqslant dfrac{n}{6}$. Thus $(2) leqslant 2^n + dfrac{n}{6} 2^{frac{n}{6}} Rightarrow f(n) leqslant dfrac{2^n}{n} + dfrac{2^{frac{n}{6}}}{6}$.



                For the lower bound,$$
                n f(n) = sum_{d mid n} μleft( frac{n}{d} right) 2^d geqslant 2^n - sum_{substack{d mid n, d < n\μ(n / d) = -1}} 2^d. tag{3}
                $$

                If $d mid n$, $d < n$ and $μleft( dfrac{n}{d} right) = -1$, then $dfrac{n}{d}$ has at least one prime divisor, which implies $dfrac{n}{d} geqslant 2$, i.e. $d leqslant dfrac{n}{2}$. Thus $(3) geqslant 2^n - dfrac{n}{2} 2^{frac{n}{2}} Rightarrow f(n) geqslant dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2}$. Therefore, (1) holds.



                Step 2: $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



                Proof: By (1. 1), it suffices to prove that $2 left( dfrac{2^n}{n} - dfrac{2^{frac{n}{2}}}{2} right) geqslant dfrac{2^{n + 1}}{n + 1} + dfrac{2^{frac{n + 1}{6}}}{6}$. Because$$
                2 left( frac{2^n}{n} - frac{2^{frac{n}{2}}}{2} right) geqslant frac{2^{n + 1}}{n + 1} + frac{2^{frac{n + 1}{6}}}{6} Longleftrightarrow frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6},
                $$

                and $2^{frac{n + 1}{4}} geqslant n + 1$ for $n geqslant 15$, then$$
                frac{2^{n + 1}}{n(n + 1)} geqslant 2^{frac{n + 1}{2}} = 2^{frac{n}{2}} + (sqrt{2} - 1) 2^{frac{n}{2}} geqslant 2^{frac{n}{2}} + frac{2^{frac{n}{2}}}{6} geqslant 2^{frac{n}{2}} + frac{2^{frac{n + 1}{6}}}{6}.
                $$

                Therefore, $2f(n) geqslant f(n + 1)$ holds for $n geqslant 15$.



                Step 3: $2f(n) geqslant f(n + 1)$ holds for $0 leqslant n leqslant 14$.



                Proof: Looking up in A001037 verifies this.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 4 '18 at 3:06









                SaadSaad

                19.7k92352




                19.7k92352






























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