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I was considering an alternate simplification of $\binom {2 n} {n} $ by pairing the components of one of the denominator factorials with the even terms in the numerator and pairing the other denominator factorial with the product of odd terms.

$$ \frac{(2n)(2n-2)(2n-4)\dots(2n-1)(2n-3)(2n-5)\dots}{n (n-1)(n-2)\dots n (n-1)(n-2)\dots} = 2^n \frac{1~3~5~7 \dots}{2~4~6~8 \dots} $$.

I was trying to figure out how many terms in the resulting ratio of product terms of were required to cancel a power of 2 in the leading $2^n$ term. The ratio of the products of odd and even terms can be considered to be products of separate ratio terms $\left(\frac{1}{2}\right) \left(\frac{3}{4}\right) \left(\frac{5}{6}\right)\dots$. If this product of ratios is represented as $t_0 t_1 t_2 t_3 \dots$, with $t_0=1$, $t_1=\frac{1}{2}$, $t_2=\frac{3}{4}$ etc., then the number of required fraction terms for which the next consecutive partial product undergoes a transition from above $\frac{1}{2}$ to below or equal to $\frac{1}{2}$ appears to be exactly $4^m$ where $m$ is the number of earlier partial products, partial products being ones that cancel powers of 2 approximately by virtue having a value not greater than $\frac{1}{2}$: By direct calculation, the transition of the next partial product $P_m$ from $P_m>\frac{1}{2}$ to $P_m\le\frac{1}{2}$ always seems to be occurring exactly for partial products of length $4^m$ even when this transition from above $\frac{1}{2}$ to at most $\frac{1}{2}$ occurs at small valued decimal places of $P_m$.
$$ P_0 = \prod_{j=0}^{4^0} t_j;~~~ P_1=\prod_{j=4^0+1}^{4^0+4^1} t_j; ~~~ P_2=\prod_{j=4^0+4^1+1}^{4^0+4^1+4^2} t_j;~~~etc.$$ e.g. the transitions are at $t_1$, $t_{(4^0-1)}$ to $t_{(4^0)} $ (if $t_0=1$), at $t_5$, $t_{(4^0 + 4^1-1)}$ to $t_{(4^0+4^1)}$, at $t_{21}$, $t_{(4^0 + 4^1 + 4^2-1)}$ to $t_{(4^0+4^1+4^2)}$, with the considered partial products starting from one index after the transition point (e.g. $t_{4^0 + 4^1 + 4^2 + 1}$). This continues at least up to the term for which the calculated subscript included $4^{13}$ for which the transition of the partial product values looked like: 0.5000000084 to 0.4999999972;

I do not know if this is already known and follows from some math that is unknown to me or, if not, how to consider proving the generality of this. For my purposes. $4^{13}$ is big enough for all of the uses I am considering.

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  • $\begingroup$ Tried a similar thing with trinomial $\frac{(3n)!}{n!~n!~n!} = 3^n \left(\frac{1~4~7~10\dots}{3~6~9~12\dots}\right)\left(\frac{2~5~8~11\dots}{3~6~9~12\dots}\right)$. Here each fraction from the modulo$_3=1$ ratio was paired with the corresponding fraction from the modulo$_3=2$ ratio to count as "1 term" in the partial product. (Transition at $\frac{1}{3}$). Again there was a pattern, but this time with the gap between transitions being $3^m$. The same did not work for a pentanomial, but instead the gaps were not exact and the ratio of adjacent partial product lengths approached $\sqrt{5}$. $\endgroup$
    – jmf7
    Nov 24, 2014 at 1:46
  • $\begingroup$ Tried it for the N-nomials from 2 through 10 now. The pattern that emerges is that the limiting partial product lengths are exact for 2 & 3 and approach a fractional power of N for $N>3$. There is a pattern in the values of the powers $N^p$: $N=2$, limit $2^2$; $N=3$, $3^1$; $N=4$, $4^\frac{2}{3}$; $N=5$, $5^\frac{1}{2}$; $N=6$, $6^\frac{2}{5}$, etc. through, $N=10$, ${10}^\frac{2}{9}$. In general $N^p$ with $p= 2 \times \frac{1}{2} \times\frac{2}{3} \times\frac{3}{4} \times\frac{4}{5}\dots$ $\endgroup$
    – jmf7
    Nov 24, 2014 at 6:48
  • $\begingroup$ I guess this is equivalent to the limiting power $p = \frac{2}{N-1}$ $\endgroup$
    – jmf7
    Nov 24, 2014 at 7:04

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