How can we find the gcd for elements (binomial coefficient)?

Question

$\gcd\left(\binom{2n}1,\binom{2n}3,\binom{2n}5,\ldots,\binom{2n}{2n-1}\right)$

i want to know what is specialty of such a series.I am not able to generalize the problem solution.Is there any rule for such gcd computation.

HINT: try using the sum of binomial coefficients.. The GCD must divide it.. — user94529, Sep 25 '13 at 16:41
not only GCD but there will be lot of numbers deviding this :) — Sagar Nishad, Sep 25 '13 at 16:44
@Ross why u deleted ur answer ..? please suggest some appropriate logic? — Sagar Nishad, Sep 25 '13 at 16:46
But the peculiarity is that the sum of these coefficient can be proved to be a pure power of 2 — user94529, Sep 25 '13 at 16:47
like suppose total is divisible by 24.. but the gcd is 8(3*8)... then how can we prove .. please explain with an example? — kumar, Sep 25 '13 at 16:50

user94529 · Answer 1 · 2013-09-25T17:59:55.053

4

Use - $$(1+x)^{2n}=\binom{2n}0x^0+\binom{2n}1x^1+\binom{2n}2x^2+\binom{2n}3x^3+\binom{2n}4x^4+...+\binom{2n}{2n}x^{2n}$$ First substituting $ x=-1$ $$0=\binom{2n}0-\binom{2n}1+\binom{2n}2...-\binom{2n}{2n-1}+\binom{2n}{2n}$$$$\implies \binom{2n}0+\binom{2n}2+\binom{2n}4+...+\binom{2n}{2n}=\binom{2n}1+\binom{2n}3+\binom{2n}5+...+\binom{2n}{2n-1} $$ But substituting $x=1$, we have$$\binom{2n}0+\binom{2n}1+\binom{2n}2+\binom{2n}3+\binom{2n}4+...+\binom{2n}{2n}=2^{2n}$$ $$\implies\binom{2n}1+\binom{2n}3+\binom{2n}5+...+\binom{2n}{2n-1}=2^{2n-1}$$ Hence the gcd is a power of 2.. Now it is sufficient only study the powers of 2 present in the different terms. This will require some analysis and one way to begin would be assuming$n=\sum2^{k} i.e.$ writing the number in binary form.

edited Sep 25 '13 at 17:59

answered Sep 25 '13 at 16:54

user94529

181

i think u are right but how can we find the general solution for such a problem? – kumar Sep 25 '13 at 17:03
it seems correct ... but can anyone provide proof for correctness ..or contradicts.. @user94529 you are sure for its correctness – Sagar Nishad Sep 25 '13 at 17:21
So how can we compute gcd ,still not answered :)?? – Sagar Nishad Sep 25 '13 at 17:42
2

im still trying the same.. – user94529 Sep 25 '13 at 17:43
it is sure it will be the smallest multiple (of 2^k) with some coefficient .. we need to get that coefficient that has least number of 2 as factor...r8? – Sagar Nishad Sep 25 '13 at 17:51
Yes u are correct but this problem seems to become extremely complex.. – user94529 Sep 25 '13 at 17:56
i think it doesnt hs any absolute formula ... perhaps it can be converted in any other form ....but i am sure it has solution .. my frnd has done this problem but he is not helping :( – Sagar Nishad Sep 25 '13 at 18:00
I will think over it and post the same if i can think of anything else.. – user94529 Sep 25 '13 at 18:05
let us continue this discussion in chat – user94529 Sep 25 '13 at 18:06
@user94529 i am not getting what you want to say with binary conversion?? – Sagar Nishad Sep 25 '13 at 18:18

score 2 · Answer 2 · answered Oct 22 '13 at 13:08

Following on user94529's answer, since the sum $\sum_k \binom{2n}{2k+1} = 2^{2n-1}$ is a power of two, we will be done if we look only at the greatest power of two dividng them all. By experimenting a bit, it seems that $\binom{2n}{1} = 2n$ has the smallest power of 2 of all.

This is indeed the case, looking at $\binom{2n}{2k+1} = \frac{2n}{2k+1} \times \binom{2n-1}{2k}$ since $2k+1$ is odd and $\binom{2n-1}{2k}$ is an integer, we get that the greatest power of 2 dividing $\binom{2n}{2k+1}$ is greater or equal to that of $2n$.

Hence, if $n=2^j \times m$ where $m$ is odd, then the answer is

$$ \gcd\left( \binom{2n}{1},\ldots,\binom{2n}{2n-1} \right) = 2^{j+1} $$

score 0 · Answer 3 · edited Apr 13 '17 at 12:20

As user94529 showed, the gcd is a power of 2.

Looking just at the first term, the gcd must divide $2n$.

In particular, if $n$ is odd, the gcd is 2 or 1.

I now use the answer to this question: Odd Binomial Coefficients?

The answer by Arturo Magidin says "This follows easily from Kummer's Theorem, that the highest power of a prime $p$ that divides $\binom{n}{m}$ is equal to the number of "carries" when adding $n-m$ and $m$ in base $p$. In particular, $\binom{n}{m}$ is odd if and only if there are no carries when adding in base $2$."

Both $2k-1$ and $2n-(2k-1)$ are odd, so there is a carry when they are added in base 2. Therefore, all of $\binom{2n}{2k-1}$ are even.

Therefore, if $n$ is odd, the gcd is 2.

Perhaps by looking at the power of 2 than divides $n$ we could get a result for all $n$.

How can we find the gcd for elements (binomial coefficient)?

3 Answers3