Mixing
Parity of some binomial coefficients
2
$begingroup$
Consider for fixed integer $n>1$ the binomial coefficients $binom{n-1+2^j}{n+1-2^j}$ for $j>0.$
It seems that precisely one of these numbers is odd.
Is there a simple proof of this fact?
binomial-coefficients parity
asked Feb 1 at 16:45
Johann CiglerJohann Cigler
44329
$endgroup$
add a comment |
2
$begingroup$
Consider for fixed integer $n>1$ the binomial coefficients $binom{n-1+2^j}{n+1-2^j}$ for $j>0.$
It seems that precisely one of these numbers is odd.
Is there a simple proof of this fact?
binomial-coefficients parity
asked Feb 1 at 16:45
Johann CiglerJohann Cigler
44329
$endgroup$
1
$begingroup$
Hmm, it depends on what you mean by "simple". I see an inductive argument. Recall that $dbinom{2a}{2b} equiv dbinom{2a+1}{2b+1} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. Thus, $dbinom{n-1+2^j}{n+1-2^j} equiv dbinom{m+2^{j-1}}{m+1-2^{j-1}} mod 2$, where $m = leftlfloor dfrac{n-1}{2} rightrfloor$. Thus, it remains to prove that for each nonnegative integer $m$, there is exactly one nonnegative integer $i$ such that $dbinom{m+2^i}{m+1-2^i}$ is odd. This can be proven by strong induction on $m$, where ...
$endgroup$
– darij grinberg
Feb 2 at 3:04
1
$begingroup$
... the induction step uses the facts that $dbinom{2a}{2b+1} equiv 0 mod 2$ and $dbinom{2a+1}{2b} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. There is probably a slicker way to put this -- maybe using generating functions?
$endgroup$
– darij grinberg
Feb 2 at 3:05
1
$begingroup$
Ah, I see how to compute this unique $i$ (non-inductively): Write $m$ in binary as $m = m_k 2^k + m_{k-1} 2^{k-1} + cdots + m_0 2^0$ with all $m_h$ belonging to $left{0,1right}$. Then, $i$ is the smallest $h$ such that $m_h = 0$. This can be easily checked using Lucas's theorem (though the inductive proof still has the advantage of being easier to write up).
$endgroup$
– darij grinberg
Feb 2 at 3:07
$begingroup$
Thank you. Nice proof.
$endgroup$
– Johann Cigler
Feb 2 at 7:25
add a comment |
2
2
2
0
$begingroup$
Consider for fixed integer $n>1$ the binomial coefficients $binom{n-1+2^j}{n+1-2^j}$ for $j>0.$
It seems that precisely one of these numbers is odd.
Is there a simple proof of this fact?
binomial-coefficients parity
asked Feb 1 at 16:45
Johann CiglerJohann Cigler
44329
$endgroup$
Consider for fixed integer $n>1$ the binomial coefficients $binom{n-1+2^j}{n+1-2^j}$ for $j>0.$
It seems that precisely one of these numbers is odd.
Is there a simple proof of this fact?
binomial-coefficients parity
binomial-coefficients parity
asked Feb 1 at 16:45
Johann CiglerJohann Cigler
44329
asked Feb 1 at 16:45
Johann CiglerJohann Cigler
44329
asked Feb 1 at 16:45
Johann CiglerJohann Cigler
44329
asked Feb 1 at 16:45
Johann CiglerJohann Cigler
44329
asked Feb 1 at 16:45
Johann CiglerJohann Cigler
44329
44329
1
$begingroup$
Hmm, it depends on what you mean by "simple". I see an inductive argument. Recall that $dbinom{2a}{2b} equiv dbinom{2a+1}{2b+1} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. Thus, $dbinom{n-1+2^j}{n+1-2^j} equiv dbinom{m+2^{j-1}}{m+1-2^{j-1}} mod 2$, where $m = leftlfloor dfrac{n-1}{2} rightrfloor$. Thus, it remains to prove that for each nonnegative integer $m$, there is exactly one nonnegative integer $i$ such that $dbinom{m+2^i}{m+1-2^i}$ is odd. This can be proven by strong induction on $m$, where ...
$endgroup$
– darij grinberg
Feb 2 at 3:04
1
$begingroup$
... the induction step uses the facts that $dbinom{2a}{2b+1} equiv 0 mod 2$ and $dbinom{2a+1}{2b} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. There is probably a slicker way to put this -- maybe using generating functions?
$endgroup$
– darij grinberg
Feb 2 at 3:05
1
$begingroup$
Ah, I see how to compute this unique $i$ (non-inductively): Write $m$ in binary as $m = m_k 2^k + m_{k-1} 2^{k-1} + cdots + m_0 2^0$ with all $m_h$ belonging to $left{0,1right}$. Then, $i$ is the smallest $h$ such that $m_h = 0$. This can be easily checked using Lucas's theorem (though the inductive proof still has the advantage of being easier to write up).
$endgroup$
– darij grinberg
Feb 2 at 3:07
$begingroup$
Thank you. Nice proof.
$endgroup$
– Johann Cigler
Feb 2 at 7:25
add a comment |
1
$begingroup$
Hmm, it depends on what you mean by "simple". I see an inductive argument. Recall that $dbinom{2a}{2b} equiv dbinom{2a+1}{2b+1} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. Thus, $dbinom{n-1+2^j}{n+1-2^j} equiv dbinom{m+2^{j-1}}{m+1-2^{j-1}} mod 2$, where $m = leftlfloor dfrac{n-1}{2} rightrfloor$. Thus, it remains to prove that for each nonnegative integer $m$, there is exactly one nonnegative integer $i$ such that $dbinom{m+2^i}{m+1-2^i}$ is odd. This can be proven by strong induction on $m$, where ...
$endgroup$
– darij grinberg
Feb 2 at 3:04
1
$begingroup$
... the induction step uses the facts that $dbinom{2a}{2b+1} equiv 0 mod 2$ and $dbinom{2a+1}{2b} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. There is probably a slicker way to put this -- maybe using generating functions?
$endgroup$
– darij grinberg
Feb 2 at 3:05
1
$begingroup$
Ah, I see how to compute this unique $i$ (non-inductively): Write $m$ in binary as $m = m_k 2^k + m_{k-1} 2^{k-1} + cdots + m_0 2^0$ with all $m_h$ belonging to $left{0,1right}$. Then, $i$ is the smallest $h$ such that $m_h = 0$. This can be easily checked using Lucas's theorem (though the inductive proof still has the advantage of being easier to write up).
$endgroup$
– darij grinberg
Feb 2 at 3:07
$begingroup$
Thank you. Nice proof.
$endgroup$
– Johann Cigler
Feb 2 at 7:25
1
1
$begingroup$
Hmm, it depends on what you mean by "simple". I see an inductive argument. Recall that $dbinom{2a}{2b} equiv dbinom{2a+1}{2b+1} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. Thus, $dbinom{n-1+2^j}{n+1-2^j} equiv dbinom{m+2^{j-1}}{m+1-2^{j-1}} mod 2$, where $m = leftlfloor dfrac{n-1}{2} rightrfloor$. Thus, it remains to prove that for each nonnegative integer $m$, there is exactly one nonnegative integer $i$ such that $dbinom{m+2^i}{m+1-2^i}$ is odd. This can be proven by strong induction on $m$, where ...
$endgroup$
– darij grinberg
Feb 2 at 3:04
$begingroup$
Hmm, it depends on what you mean by "simple". I see an inductive argument. Recall that $dbinom{2a}{2b} equiv dbinom{2a+1}{2b+1} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. Thus, $dbinom{n-1+2^j}{n+1-2^j} equiv dbinom{m+2^{j-1}}{m+1-2^{j-1}} mod 2$, where $m = leftlfloor dfrac{n-1}{2} rightrfloor$. Thus, it remains to prove that for each nonnegative integer $m$, there is exactly one nonnegative integer $i$ such that $dbinom{m+2^i}{m+1-2^i}$ is odd. This can be proven by strong induction on $m$, where ...
$endgroup$
– darij grinberg
Feb 2 at 3:04
1
1
$begingroup$
... the induction step uses the facts that $dbinom{2a}{2b+1} equiv 0 mod 2$ and $dbinom{2a+1}{2b} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. There is probably a slicker way to put this -- maybe using generating functions?
$endgroup$
– darij grinberg
Feb 2 at 3:05
$begingroup$
... the induction step uses the facts that $dbinom{2a}{2b+1} equiv 0 mod 2$ and $dbinom{2a+1}{2b} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. There is probably a slicker way to put this -- maybe using generating functions?
$endgroup$
– darij grinberg
Feb 2 at 3:05
1
1
$begingroup$
Ah, I see how to compute this unique $i$ (non-inductively): Write $m$ in binary as $m = m_k 2^k + m_{k-1} 2^{k-1} + cdots + m_0 2^0$ with all $m_h$ belonging to $left{0,1right}$. Then, $i$ is the smallest $h$ such that $m_h = 0$. This can be easily checked using Lucas's theorem (though the inductive proof still has the advantage of being easier to write up).
$endgroup$
– darij grinberg
Feb 2 at 3:07
$begingroup$
Ah, I see how to compute this unique $i$ (non-inductively): Write $m$ in binary as $m = m_k 2^k + m_{k-1} 2^{k-1} + cdots + m_0 2^0$ with all $m_h$ belonging to $left{0,1right}$. Then, $i$ is the smallest $h$ such that $m_h = 0$. This can be easily checked using Lucas's theorem (though the inductive proof still has the advantage of being easier to write up).
$endgroup$
– darij grinberg
Feb 2 at 3:07
$begingroup$
Thank you. Nice proof.
$endgroup$
– Johann Cigler
Feb 2 at 7:25
$begingroup$
Thank you. Nice proof.
$endgroup$
– Johann Cigler
Feb 2 at 7:25
add a comment |
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1
$begingroup$
Hmm, it depends on what you mean by "simple". I see an inductive argument. Recall that $dbinom{2a}{2b} equiv dbinom{2a+1}{2b+1} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. Thus, $dbinom{n-1+2^j}{n+1-2^j} equiv dbinom{m+2^{j-1}}{m+1-2^{j-1}} mod 2$, where $m = leftlfloor dfrac{n-1}{2} rightrfloor$. Thus, it remains to prove that for each nonnegative integer $m$, there is exactly one nonnegative integer $i$ such that $dbinom{m+2^i}{m+1-2^i}$ is odd. This can be proven by strong induction on $m$, where ...
$endgroup$
– darij grinberg
Feb 2 at 3:04
1
$begingroup$
... the induction step uses the facts that $dbinom{2a}{2b+1} equiv 0 mod 2$ and $dbinom{2a+1}{2b} equiv dbinom{a}{b} mod 2$ for all nonnegative integers $a$ and $b$. There is probably a slicker way to put this -- maybe using generating functions?
$endgroup$
– darij grinberg
Feb 2 at 3:05
1
$begingroup$
Ah, I see how to compute this unique $i$ (non-inductively): Write $m$ in binary as $m = m_k 2^k + m_{k-1} 2^{k-1} + cdots + m_0 2^0$ with all $m_h$ belonging to $left{0,1right}$. Then, $i$ is the smallest $h$ such that $m_h = 0$. This can be easily checked using Lucas's theorem (though the inductive proof still has the advantage of being easier to write up).
$endgroup$
– darij grinberg
Feb 2 at 3:07
$begingroup$
Thank you. Nice proof.
$endgroup$
– Johann Cigler
Feb 2 at 7:25