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Generating function for 3 -core partitions: Part II
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Let $lambda$ be an integer partition: $lambda=(lambda_1geqlambda_2geqdotsgeq0)$. Further, let $h_u$ denote the hook-length of the cell $u$.
We call $lambda$ a $t$-core partition if none of its hooks $h_u$ equals $t$. Define $c_t(n)$ to be the number of partitions of $n$ that are $t$-core partitions. It's well-known that
$$sum_ngeq0c_t(n),q^n=prod_k=1^inftyfrac(1-q^tk)^t1-q^k.$$
For example, $sum_n=0^inftyc_2(n),q^n=sum_k=0^inftyq^binomk+12$.
Now, consider only those partitions of $n$ with odd parts and let $O_t(n)$ be the number of such partitions that are $t$-cores.
QUESTION. Is this true?
$$sum_ngeq1O_3(n),q^n=(1+q)sum_kgeq1q^k^2.$$
co.combinatorics soft-question partitions elementary-proofs
asked 1 hour ago
T. Amdeberhan
16.5k228122
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2
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Let $lambda$ be an integer partition: $lambda=(lambda_1geqlambda_2geqdotsgeq0)$. Further, let $h_u$ denote the hook-length of the cell $u$.
We call $lambda$ a $t$-core partition if none of its hooks $h_u$ equals $t$. Define $c_t(n)$ to be the number of partitions of $n$ that are $t$-core partitions. It's well-known that
$$sum_ngeq0c_t(n),q^n=prod_k=1^inftyfrac(1-q^tk)^t1-q^k.$$
For example, $sum_n=0^inftyc_2(n),q^n=sum_k=0^inftyq^binomk+12$.
Now, consider only those partitions of $n$ with odd parts and let $O_t(n)$ be the number of such partitions that are $t$-cores.
QUESTION. Is this true?
$$sum_ngeq1O_3(n),q^n=(1+q)sum_kgeq1q^k^2.$$
co.combinatorics soft-question partitions elementary-proofs
asked 1 hour ago
T. Amdeberhan
16.5k228122
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $lambda$ be an integer partition: $lambda=(lambda_1geqlambda_2geqdotsgeq0)$. Further, let $h_u$ denote the hook-length of the cell $u$.
We call $lambda$ a $t$-core partition if none of its hooks $h_u$ equals $t$. Define $c_t(n)$ to be the number of partitions of $n$ that are $t$-core partitions. It's well-known that
$$sum_ngeq0c_t(n),q^n=prod_k=1^inftyfrac(1-q^tk)^t1-q^k.$$
For example, $sum_n=0^inftyc_2(n),q^n=sum_k=0^inftyq^binomk+12$.
Now, consider only those partitions of $n$ with odd parts and let $O_t(n)$ be the number of such partitions that are $t$-cores.
QUESTION. Is this true?
$$sum_ngeq1O_3(n),q^n=(1+q)sum_kgeq1q^k^2.$$
co.combinatorics soft-question partitions elementary-proofs
asked 1 hour ago
T. Amdeberhan
16.5k228122
Let $lambda$ be an integer partition: $lambda=(lambda_1geqlambda_2geqdotsgeq0)$. Further, let $h_u$ denote the hook-length of the cell $u$.
We call $lambda$ a $t$-core partition if none of its hooks $h_u$ equals $t$. Define $c_t(n)$ to be the number of partitions of $n$ that are $t$-core partitions. It's well-known that
$$sum_ngeq0c_t(n),q^n=prod_k=1^inftyfrac(1-q^tk)^t1-q^k.$$
For example, $sum_n=0^inftyc_2(n),q^n=sum_k=0^inftyq^binomk+12$.
Now, consider only those partitions of $n$ with odd parts and let $O_t(n)$ be the number of such partitions that are $t$-cores.
QUESTION. Is this true?
$$sum_ngeq1O_3(n),q^n=(1+q)sum_kgeq1q^k^2.$$
co.combinatorics soft-question partitions elementary-proofs
co.combinatorics soft-question partitions elementary-proofs
asked 1 hour ago
T. Amdeberhan
16.5k228122
asked 1 hour ago
T. Amdeberhan
16.5k228122
asked 1 hour ago
T. Amdeberhan
16.5k228122
asked 1 hour ago
T. Amdeberhan
16.5k228122
asked 1 hour ago
T. Amdeberhan
16.5k228122
16.5k228122
add a comment |Â
add a comment |Â
1 Answer
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Suppose the partition with $k$ parts $lambda=lambda_1geq lambda_2geqdotsgeqlambda_kgeq lambda_k+1=0$ is a partition with only odd parts. Then we have $lambda_i-lambda_i+1$ is even for all $ileq k-1$ and odd for $i=k$. Using the same characterization I mentioned in the previous answer we see that the sequence
$lambda_1-lambda_2,lambda_2-lambda_3,dots,lambda_k-lambda_k+1$ for such a $3$-core has to be equal to $2,2,dots,2,1$ or $2,2,dots,2,0,1$. The first case is a partition with size $k^2$ and the second has size $1+k^2$ which implies the generating function
$$sum_ngeq1O_3(n),q^n=sum_kgeq1(q^k^2+q^k^2+1).$$
answered 1 hour ago
Gjergji Zaimi
60.1k4154296
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Suppose the partition with $k$ parts $lambda=lambda_1geq lambda_2geqdotsgeqlambda_kgeq lambda_k+1=0$ is a partition with only odd parts. Then we have $lambda_i-lambda_i+1$ is even for all $ileq k-1$ and odd for $i=k$. Using the same characterization I mentioned in the previous answer we see that the sequence
$lambda_1-lambda_2,lambda_2-lambda_3,dots,lambda_k-lambda_k+1$ for such a $3$-core has to be equal to $2,2,dots,2,1$ or $2,2,dots,2,0,1$. The first case is a partition with size $k^2$ and the second has size $1+k^2$ which implies the generating function
$$sum_ngeq1O_3(n),q^n=sum_kgeq1(q^k^2+q^k^2+1).$$
answered 1 hour ago
Gjergji Zaimi
60.1k4154296
add a comment |Â
up vote
4
down vote
Suppose the partition with $k$ parts $lambda=lambda_1geq lambda_2geqdotsgeqlambda_kgeq lambda_k+1=0$ is a partition with only odd parts. Then we have $lambda_i-lambda_i+1$ is even for all $ileq k-1$ and odd for $i=k$. Using the same characterization I mentioned in the previous answer we see that the sequence
$lambda_1-lambda_2,lambda_2-lambda_3,dots,lambda_k-lambda_k+1$ for such a $3$-core has to be equal to $2,2,dots,2,1$ or $2,2,dots,2,0,1$. The first case is a partition with size $k^2$ and the second has size $1+k^2$ which implies the generating function
$$sum_ngeq1O_3(n),q^n=sum_kgeq1(q^k^2+q^k^2+1).$$
answered 1 hour ago
Gjergji Zaimi
60.1k4154296
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Suppose the partition with $k$ parts $lambda=lambda_1geq lambda_2geqdotsgeqlambda_kgeq lambda_k+1=0$ is a partition with only odd parts. Then we have $lambda_i-lambda_i+1$ is even for all $ileq k-1$ and odd for $i=k$. Using the same characterization I mentioned in the previous answer we see that the sequence
$lambda_1-lambda_2,lambda_2-lambda_3,dots,lambda_k-lambda_k+1$ for such a $3$-core has to be equal to $2,2,dots,2,1$ or $2,2,dots,2,0,1$. The first case is a partition with size $k^2$ and the second has size $1+k^2$ which implies the generating function
$$sum_ngeq1O_3(n),q^n=sum_kgeq1(q^k^2+q^k^2+1).$$
answered 1 hour ago
Gjergji Zaimi
60.1k4154296
Suppose the partition with $k$ parts $lambda=lambda_1geq lambda_2geqdotsgeqlambda_kgeq lambda_k+1=0$ is a partition with only odd parts. Then we have $lambda_i-lambda_i+1$ is even for all $ileq k-1$ and odd for $i=k$. Using the same characterization I mentioned in the previous answer we see that the sequence
$lambda_1-lambda_2,lambda_2-lambda_3,dots,lambda_k-lambda_k+1$ for such a $3$-core has to be equal to $2,2,dots,2,1$ or $2,2,dots,2,0,1$. The first case is a partition with size $k^2$ and the second has size $1+k^2$ which implies the generating function
$$sum_ngeq1O_3(n),q^n=sum_kgeq1(q^k^2+q^k^2+1).$$
answered 1 hour ago
Gjergji Zaimi
60.1k4154296
answered 1 hour ago
Gjergji Zaimi
60.1k4154296
answered 1 hour ago
Gjergji Zaimi
60.1k4154296
answered 1 hour ago
Gjergji Zaimi
60.1k4154296
60.1k4154296
add a comment |Â
add a comment |Â
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