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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?

Did
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Hmmman
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  • Not sure what you are trying to achieve exactly here (since you do not say), but anyway, $$C_{2^n}=\frac{(2n+1)!}{2^n(n!)^2}$$ – Did Jan 11 '19 at 14:05
  • Simply rewrote your formula $C_{2^n} = 3(5/2)(7/3)(9/4)...*(2+1/n)$ as a single ratio and multiplied by $2\cdot4\cdot\cdots\cdot(2n-2)(2n)=2^n\cdot n!$ both terms of the ratio. – Did Jan 11 '19 at 14:15
  • ((Previous comment addresses a now deleted comment by the OP.)) – Did Jan 11 '19 at 14:16
  • Thank you, after a second look I realized your point of view, it s really simple! – Hmmman Jan 11 '19 at 14:17
  • Yes. Write an answer yourself then? – Did Jan 11 '19 at 14:18
  • Hmm, sorry to disturb you once again, but: – Hmmman Jan 11 '19 at 14:28
  • $C_{2^n} = 3 * (5/2) * (7/3) * ... * (2 + 1/n)$ &\n&

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

     – Hmmman Jan 11 '19 at 14:28
  • it feels like I was too optimistic with my understanding, from where came: $1 / (2^n*(n!))$ – Hmmman Jan 11 '19 at 14:30
  • No, $3\cdot5\cdot\cdots\cdot(2n+1)$ is not $(2n+1)!$, only $3\cdot5\cdot\cdots\cdot(2n+1)\cdot(2\cdot4\cdot\cdots\cdot(2n))$ is. And this is why the factor $2\cdot4\cdot\cdots\cdot(2n)=2^n\cdot n!$ apears in the denominator. – Did Jan 11 '19 at 16:08
  • Thanks once again! – Hmmman Jan 12 '19 at 06:47

1 Answers1

1

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

Hmmman
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