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Nate
Nate
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Let's look at the ratio of two adjacent $q$-binomials as we move away from the center, for simplicity I'll do the even case.

$\binom{2n}{n-a}_q / \binom{2n}{n-a-1}_q = \frac{[n+a+1]_q}{[n-a]_q} > q^{2a+1}$

In particular these are decaying faster than geometrically. With some possible exceptions for say $q =2$ or $3$ and$q=3$ with $n$ small, it looks like the middle binomial coefficient will be bigger than the sum of all the others.

Let's look at the ratio of two adjacent $q$-binomials as we move away from the center, for simplicity I'll do the even case.

$\binom{2n}{n-a}_q / \binom{2n}{n-a-1}_q = \frac{[n+a+1]_q}{[n-a]_q} > q^{2a+1}$

In particular these are decaying faster than geometrically. With some possible exceptions for say $q =2$ or $3$ and $n$ small, it looks like the middle binomial coefficient will be bigger than the sum of all the others.

Let's look at the ratio of two adjacent $q$-binomials as we move away from the center, for simplicity I'll do the even case.

$\binom{2n}{n-a}_q / \binom{2n}{n-a-1}_q = \frac{[n+a+1]_q}{[n-a]_q} > q^{2a+1}$

In particular these are decaying faster than geometrically. With some possible exceptions for say $q =2$ or $q=3$ with $n$ small, it looks like the middle binomial coefficient will be bigger than the sum of all the others.

Source Link
Nate
Nate
  • 2.2k
  • 14
  • 16

Let's look at the ratio of two adjacent $q$-binomials as we move away from the center, for simplicity I'll do the even case.

$\binom{2n}{n-a}_q / \binom{2n}{n-a-1}_q = \frac{[n+a+1]_q}{[n-a]_q} > q^{2a+1}$

In particular these are decaying faster than geometrically. With some possible exceptions for say $q =2$ or $3$ and $n$ small, it looks like the middle binomial coefficient will be bigger than the sum of all the others.