Result on partitions with distinct odd parts The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)integer partitionsInteger Partitions FormulasQuestion on combinatorics, partitions.Prove : $p$(n│even number of ODD parts)=$p$(n│distinct parts ,number of ODD parts is even )Counting integer partitions of n into exactly k distinct parts size at most MProve that the number of partitions of $2010$ into $10$ parts is equal to the number of partitions of $2055$ into $10$ distinct parts.Partition identity with generating functionshow can we prove the number of partitions of $n$ into odd parts equals the number of partitions of $n$ into distinct parts using Ferrers graphPartitions of $n$ with exactly 3 partsGenerating function for number of partitions with only distinct even parts
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Result on partitions with distinct odd parts
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)integer partitionsInteger Partitions FormulasQuestion on combinatorics, partitions.Prove : $p$(n│even number of ODD parts)=$p$(n│distinct parts ,number of ODD parts is even )Counting integer partitions of n into exactly k distinct parts size at most MProve that the number of partitions of $2010$ into $10$ parts is equal to the number of partitions of $2055$ into $10$ distinct parts.Partition identity with generating functionshow can we prove the number of partitions of $n$ into odd parts equals the number of partitions of $n$ into distinct parts using Ferrers graphPartitions of $n$ with exactly 3 partsGenerating function for number of partitions with only distinct even parts
$begingroup$
Let $pdo(n)$ be the number of partitions of n into distinct odd parts. Then $p(n)$ is odd if and only if $pdo(n)$ is odd.
I am well aware that a proof of this is available here but I want to do it algebraically using generating functions.
So $pdo(n)$ $=$ $displaystyle prod_i=odd (1+x^i)$
I know that I can do this by showing that $pdo(n)$ $equiv$ $p(n)$ $mod 2$ but I am not sure how I can do that. Any hint or help would be appreciated.
abstract-algebra number-theory generating-functions integer-partitions
asked Apr 8 at 12:53
JakeJake
32
New contributor
Jake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
Let $pdo(n)$ be the number of partitions of n into distinct odd parts. Then $p(n)$ is odd if and only if $pdo(n)$ is odd.
I am well aware that a proof of this is available here but I want to do it algebraically using generating functions.
So $pdo(n)$ $=$ $displaystyle prod_i=odd (1+x^i)$
I know that I can do this by showing that $pdo(n)$ $equiv$ $p(n)$ $mod 2$ but I am not sure how I can do that. Any hint or help would be appreciated.
abstract-algebra number-theory generating-functions integer-partitions
asked Apr 8 at 12:53
JakeJake
32
New contributor
Jake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $pdo(n)$ be the number of partitions of n into distinct odd parts. Then $p(n)$ is odd if and only if $pdo(n)$ is odd.
I am well aware that a proof of this is available here but I want to do it algebraically using generating functions.
So $pdo(n)$ $=$ $displaystyle prod_i=odd (1+x^i)$
I know that I can do this by showing that $pdo(n)$ $equiv$ $p(n)$ $mod 2$ but I am not sure how I can do that. Any hint or help would be appreciated.
abstract-algebra number-theory generating-functions integer-partitions
asked Apr 8 at 12:53
JakeJake
32
New contributor
Jake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $pdo(n)$ be the number of partitions of n into distinct odd parts. Then $p(n)$ is odd if and only if $pdo(n)$ is odd.
I am well aware that a proof of this is available here but I want to do it algebraically using generating functions.
So $pdo(n)$ $=$ $displaystyle prod_i=odd (1+x^i)$
I know that I can do this by showing that $pdo(n)$ $equiv$ $p(n)$ $mod 2$ but I am not sure how I can do that. Any hint or help would be appreciated.
abstract-algebra number-theory generating-functions integer-partitions
abstract-algebra number-theory generating-functions integer-partitions
asked Apr 8 at 12:53
JakeJake
32
New contributor
Jake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 12:53
JakeJake
32
New contributor
Jake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 12:53
JakeJake
32
New contributor
Jake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 12:53
JakeJake
32
asked Apr 8 at 12:53
JakeJake
32
32
New contributor
Jake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jake is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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$begingroup$
You have g.f. $$prod_i=0^infty (1 + x^2i+1)$$ for partitions into distinct odd parts. The g.f. for general partitions is $$frac1prod_i=1^infty(1 - x^i)$$
So $pdo(n) equiv p(n) pmod 2$ iff $$prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i)$$ has only even coefficients.
Now,
$$
begineqnarray
prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i) &=&
fracprod_i=1^infty (1 + x^i)prod_i=1^infty (1 + x^2i) - frac1prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 + x^i)prod_i=1^infty(1 - x^i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 - x^2i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)(1 - x^i) \
endeqnarray
$$
Is that enough to say something about the parity of the coefficients?
answered Apr 9 at 21:51
Peter TaylorPeter Taylor
9,18212343
$endgroup$
add a comment |
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1 Answer
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$begingroup$
You have g.f. $$prod_i=0^infty (1 + x^2i+1)$$ for partitions into distinct odd parts. The g.f. for general partitions is $$frac1prod_i=1^infty(1 - x^i)$$
So $pdo(n) equiv p(n) pmod 2$ iff $$prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i)$$ has only even coefficients.
Now,
$$
begineqnarray
prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i) &=&
fracprod_i=1^infty (1 + x^i)prod_i=1^infty (1 + x^2i) - frac1prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 + x^i)prod_i=1^infty(1 - x^i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 - x^2i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)(1 - x^i) \
endeqnarray
$$
Is that enough to say something about the parity of the coefficients?
answered Apr 9 at 21:51
Peter TaylorPeter Taylor
9,18212343
$endgroup$
add a comment |
$begingroup$
You have g.f. $$prod_i=0^infty (1 + x^2i+1)$$ for partitions into distinct odd parts. The g.f. for general partitions is $$frac1prod_i=1^infty(1 - x^i)$$
So $pdo(n) equiv p(n) pmod 2$ iff $$prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i)$$ has only even coefficients.
Now,
$$
begineqnarray
prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i) &=&
fracprod_i=1^infty (1 + x^i)prod_i=1^infty (1 + x^2i) - frac1prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 + x^i)prod_i=1^infty(1 - x^i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 - x^2i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)(1 - x^i) \
endeqnarray
$$
Is that enough to say something about the parity of the coefficients?
answered Apr 9 at 21:51
Peter TaylorPeter Taylor
9,18212343
$endgroup$
add a comment |
$begingroup$
You have g.f. $$prod_i=0^infty (1 + x^2i+1)$$ for partitions into distinct odd parts. The g.f. for general partitions is $$frac1prod_i=1^infty(1 - x^i)$$
So $pdo(n) equiv p(n) pmod 2$ iff $$prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i)$$ has only even coefficients.
Now,
$$
begineqnarray
prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i) &=&
fracprod_i=1^infty (1 + x^i)prod_i=1^infty (1 + x^2i) - frac1prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 + x^i)prod_i=1^infty(1 - x^i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 - x^2i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)(1 - x^i) \
endeqnarray
$$
Is that enough to say something about the parity of the coefficients?
answered Apr 9 at 21:51
Peter TaylorPeter Taylor
9,18212343
$endgroup$
You have g.f. $$prod_i=0^infty (1 + x^2i+1)$$ for partitions into distinct odd parts. The g.f. for general partitions is $$frac1prod_i=1^infty(1 - x^i)$$
So $pdo(n) equiv p(n) pmod 2$ iff $$prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i)$$ has only even coefficients.
Now,
$$
begineqnarray
prod_i=0^infty (1 + x^2i+1) - frac1prod_i=1^infty(1 - x^i) &=&
fracprod_i=1^infty (1 + x^i)prod_i=1^infty (1 + x^2i) - frac1prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 + x^i)prod_i=1^infty(1 - x^i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)prod_i=1^infty(1 - x^i) \
&=& fracprod_i=1^infty (1 - x^2i) - prod_i=1^infty (1 + x^2i)prod_i=1^infty (1 + x^2i)(1 - x^i) \
endeqnarray
$$
Is that enough to say something about the parity of the coefficients?
answered Apr 9 at 21:51
Peter TaylorPeter Taylor
9,18212343
answered Apr 9 at 21:51
Peter TaylorPeter Taylor
9,18212343
answered Apr 9 at 21:51
Peter TaylorPeter Taylor
9,18212343
answered Apr 9 at 21:51
Peter TaylorPeter Taylor
9,18212343
9,18212343
add a comment |
add a comment |
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Jake is a new contributor. Be nice, and check out our Code of Conduct.
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