Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$












0












$begingroup$



Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$.




My solution:
$N = 2^n$



$C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$



$C_N = 3*(5/2)*...*(2+1/lgN)$



How can I simplify further?






recurrence-relations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
    $endgroup$
    – Did
    Jan 11 at 14:05












  • $begingroup$
    Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
    $endgroup$
    – Did
    Jan 11 at 14:15










  • $begingroup$
    ((Previous comment addresses a now deleted comment by the OP.))
    $endgroup$
    – Did
    Jan 11 at 14:16










  • $begingroup$
    Thank you, after a second look I realized your point of view, it s really simple!
    $endgroup$
    – Hmmman
    Jan 11 at 14:17










  • $begingroup$
    Yes. Write an answer yourself then?
    $endgroup$
    – Did
    Jan 11 at 14:18
















0












$begingroup$



Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$.




My solution:
$N = 2^n$



$C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$



$C_N = 3*(5/2)*...*(2+1/lgN)$



How can I simplify further?






recurrence-relations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
    $endgroup$
    – Did
    Jan 11 at 14:05












  • $begingroup$
    Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
    $endgroup$
    – Did
    Jan 11 at 14:15










  • $begingroup$
    ((Previous comment addresses a now deleted comment by the OP.))
    $endgroup$
    – Did
    Jan 11 at 14:16










  • $begingroup$
    Thank you, after a second look I realized your point of view, it s really simple!
    $endgroup$
    – Hmmman
    Jan 11 at 14:17










  • $begingroup$
    Yes. Write an answer yourself then?
    $endgroup$
    – Did
    Jan 11 at 14:18














0












0








0





$begingroup$



Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$.




My solution:
$N = 2^n$



$C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$



$C_N = 3*(5/2)*...*(2+1/lgN)$



How can I simplify further?






recurrence-relations







share|cite|improve this question











$endgroup$





Solve $C_N = (2 + 1/log_2N)C_{N/2}$ for $N ge2$ and $C_1 = 1$.




My solution:
$N = 2^n$



$C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$



$C_N = 3*(5/2)*...*(2+1/lgN)$



How can I simplify further?






recurrence-relations




recurrence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 14:15









Did

247k23223460




247k23223460










asked Jan 11 at 14:02









HmmmanHmmman

134




134












  • $begingroup$
    Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
    $endgroup$
    – Did
    Jan 11 at 14:05












  • $begingroup$
    Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
    $endgroup$
    – Did
    Jan 11 at 14:15










  • $begingroup$
    ((Previous comment addresses a now deleted comment by the OP.))
    $endgroup$
    – Did
    Jan 11 at 14:16










  • $begingroup$
    Thank you, after a second look I realized your point of view, it s really simple!
    $endgroup$
    – Hmmman
    Jan 11 at 14:17










  • $begingroup$
    Yes. Write an answer yourself then?
    $endgroup$
    – Did
    Jan 11 at 14:18


















  • $begingroup$
    Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
    $endgroup$
    – Did
    Jan 11 at 14:05












  • $begingroup$
    Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
    $endgroup$
    – Did
    Jan 11 at 14:15










  • $begingroup$
    ((Previous comment addresses a now deleted comment by the OP.))
    $endgroup$
    – Did
    Jan 11 at 14:16










  • $begingroup$
    Thank you, after a second look I realized your point of view, it s really simple!
    $endgroup$
    – Hmmman
    Jan 11 at 14:17










  • $begingroup$
    Yes. Write an answer yourself then?
    $endgroup$
    – Did
    Jan 11 at 14:18
















$begingroup$
Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
$endgroup$
– Did
Jan 11 at 14:05






$begingroup$
Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=frac{(2n+1)!}{2^n(n!)^2}$$
$endgroup$
– Did
Jan 11 at 14:05














$begingroup$
Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
$endgroup$
– Did
Jan 11 at 14:15




$begingroup$
Simply rewrote your formula $C_{2^n} = 3*(5/2)*(7/3)*(9/4)*...*(2+1/n)$ as a single ratio and multiplied by $2cdot4cdotcdotscdot(2n-2)(2n)=2^ncdot n!$ both terms of the ratio.
$endgroup$
– Did
Jan 11 at 14:15












$begingroup$
((Previous comment addresses a now deleted comment by the OP.))
$endgroup$
– Did
Jan 11 at 14:16




$begingroup$
((Previous comment addresses a now deleted comment by the OP.))
$endgroup$
– Did
Jan 11 at 14:16












$begingroup$
Thank you, after a second look I realized your point of view, it s really simple!
$endgroup$
– Hmmman
Jan 11 at 14:17




$begingroup$
Thank you, after a second look I realized your point of view, it s really simple!
$endgroup$
– Hmmman
Jan 11 at 14:17












$begingroup$
Yes. Write an answer yourself then?
$endgroup$
– Did
Jan 11 at 14:18




$begingroup$
Yes. Write an answer yourself then?
$endgroup$
– Did
Jan 11 at 14:18










1 Answer
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$begingroup$

$C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
$(2 + 1/n) = (2n + 1)/n$

or
$C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$

or
$C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$






share|cite|improve this answer









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    1 Answer
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    1 Answer
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    $begingroup$

    $C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
    $(2 + 1/n) = (2n + 1)/n$

    or
    $C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$

    or
    $C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
      $(2 + 1/n) = (2n + 1)/n$

      or
      $C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$

      or
      $C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
        $(2 + 1/n) = (2n + 1)/n$

        or
        $C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$

        or
        $C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$






        share|cite|improve this answer









        $endgroup$



        $C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$
        $(2 + 1/n) = (2n + 1)/n$

        or
        $C_{2^n} = (3 * 5 * 7 * ... * (2n + 1)) / (1 * 2 * 3 ... * n)$

        or
        $C_{2^n} = (2n + 1)! / ((2^n)* n! * n!)$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 6:50









        HmmmanHmmman

        134




        134






























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