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# Extension of Andrews’ identity

In document Hook type tableaux and Partition identities (Stránka 111-156)

Theorem (Andrews (2019))

LetOd(n)denote the number of partitions of n in which the odd parts are distinct and each positive odd integer smaller than the largest odd part must appear as a part. Then

podeu(n) =Od(n),

where podeu(n)denotes the number of partitions of n in which each even part is less than each odd part and odd parts are distinct.

Ex: The 6 partitions enumerated byOd(9) are 8 + 1, 6 + 2 + 1, 5 + 3 + 1, 4 + 4 + 1, 4 + 2 + 2 + 1, 2 + 2 + 2 + 2 + 1 and those enumerated by peuod(9) are 9, 7 + 2, 5 + 4, 5 + 3 + 1, 5 + 2 + 2, 3 + 2 + 2 + 2.

### Extension of Andrews’ identity

Theorem (Andrews (2019))

LetOd(n)denote the number of partitions of n in which the odd parts are distinct and each positive odd integer smaller than the largest odd part must appear as a part. Then

podeu(n) =Od(n),

where podeu(n)denotes the number of partitions of n in which each even part is less than each odd part and odd parts are distinct.

Ex: The 6 partitions enumerated byOd(9) are 8 + 1, 6 + 2 + 1, 5 + 3 + 1, 4 + 4 + 1, 4 + 2 + 2 + 1, 2 + 2 + 2 + 2 + 1 and those enumerated by peuod(9) are 9, 7 + 2, 5 + 4, 5 + 3 + 1, 5 + 2 + 2, 3 + 2 + 2 + 2.

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### Extension of Andrews’ identity

Definition 1:

Peuou(n) :=

(

λ`n: (1) all the odd parts ofλare unrestricted, (2) each even part ofλis less than each odd part ofλ

) ,

peuou(n) := #{λ`n:λ∈Peuou(n)}. For example,peuou(9) = 12

(9,7+2,7+1+1,5+4,5+3+1,5+2+2,5+1+1+1+1,3+3+3,3+3+ 1+1+1,3+2+2+2,3+1+1+1+1+1+1,1+1+1+1+1+1+1+1+1).

### Extension of Andrews’ identity

Definition 1:

Peuou(n) :=

(

λ`n: (1) all the odd parts ofλare unrestricted, (2) each even part ofλis less than each odd part ofλ

) ,

peuou(n) := #{λ`n:λ∈Peuou(n)}.

For example,peuou(9) = 12

(9,7+2,7+1+1,5+4,5+3+1,5+2+2,5+1+1+1+1,3+3+3,3+3+ 1+1+1,3+2+2+2,3+1+1+1+1+1+1,1+1+1+1+1+1+1+1+1).

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### Extension of Andrews’ identity

Definition 1:

Peuou(n) :=

(

λ`n: (1) all the odd parts ofλare unrestricted, (2) each even part ofλis less than each odd part ofλ

) ,

peuou(n) := #{λ`n:λ∈Peuou(n)}.

For example,peuou(9) = 12

(9,7+2,7+1+1,5+4,5+3+1,5+2+2,5+1+1+1+1,3+3+3,3+3+

1+1+1,3+2+2+2,3+1+1+1+1+1+1,1+1+1+1+1+1+1+1+1).

### Extension of Andrews’ identity

Definition 2: For λ`nsuch that an odd integer must appear as a part ofλ,

OMax(λ) := greatest odd part of λ, EMax(λ) :=

greatest even part of λ, if even parts occur inλ,

0, otherwise

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### Extension of Andrews’ identity

Definition 2: For λ`nsuch that an odd integer must appear as a part ofλ,

OMax(λ) := greatest odd part of λ,

EMax(λ) :=

greatest even part of λ, if even parts occur inλ,

0, otherwise

### Extension of Andrews’ identity

Definition 2: For λ`nsuch that an odd integer must appear as a part ofλ,

OMax(λ) := greatest odd part of λ, EMax(λ) :=

greatest even part of λ, if even parts occur inλ,

0, otherwise

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### Extension of Andrews’ identity

Definition 2: For λ`nsuch that an odd integer must appear as a part ofλ,

OMax(λ) := greatest odd part of λ, EMax(λ) :=

greatest even part of λ, if even parts occur inλ,

0, otherwise

### Extension of Andrews’ identity

Definition 2: For λ`nsuch that an odd integer must appear as a part ofλ,

OMax(λ) := greatest odd part of λ, EMax(λ) :=

greatest even part of λ, if even parts occur inλ,

0, otherwise

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### Extension of Andrews’ identity

Definition 2: For λ`nsuch that an odd integer must appear as a part ofλ,

OMax(λ) := greatest odd part of λ, EMax(λ) :=

greatest even part of λ, if even parts occur inλ,

0, otherwise

### Extension of Andrews’ identity

Definition 4:

OEMaxDiff(λ) = min{OEMaxDiff (λ0) :λ0 ∈Ou(n)}.

Ou(n) :={λ∈Ou(n) : OEMaxDiff(λ)}. For example,ou(9) = 12

(8 + 1,6 + 2 + 1,5 + 3 + 1,4 + 4 + 1,4 + 3 + 1 + 1,4 + 2 + 2 + 1,3 + 2 + 1 + 1 + 1 + 1,2 + 2 + 2 + 2 + 1,3 + 3 + 1 + 1 + 1,3 + 1 + 1 + 1 + 1 + 1 + 1,2 + 2 + 1 + 1 + 1 + 1 + 1,1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1). According to our definition, the partition λ= (6,1,1,1)∈/Ou(9) but the partition (4,3,1,1)∈Ou(9).

Theorem (B., Dastidar (2019)) ou(n) =peuou(n)

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### Extension of Andrews’ identity

Definition 4:

OEMaxDiff(λ) = min{OEMaxDiff (λ0) :λ0 ∈Ou(n)}.

Ou(n) :={λ∈Ou(n) : OEMaxDiff(λ)}.

For example,ou(9) = 12

(8 + 1,6 + 2 + 1,5 + 3 + 1,4 + 4 + 1,4 + 3 + 1 + 1,4 + 2 + 2 + 1,3 + 2 + 1 + 1 + 1 + 1,2 + 2 + 2 + 2 + 1,3 + 3 + 1 + 1 + 1,3 + 1 + 1 + 1 + 1 + 1 + 1,2 + 2 + 1 + 1 + 1 + 1 + 1,1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1). According to our definition, the partition λ= (6,1,1,1)∈/Ou(9) but the partition (4,3,1,1)∈Ou(9).

Theorem (B., Dastidar (2019)) ou(n) =peuou(n)

### Extension of Andrews’ identity

Definition 4:

OEMaxDiff(λ) = min{OEMaxDiff (λ0) :λ0 ∈Ou(n)}.

Ou(n) :={λ∈Ou(n) : OEMaxDiff(λ)}.

For example,ou(9) = 12

(8 + 1,6 + 2 + 1,5 + 3 + 1,4 + 4 + 1,4 + 3 + 1 + 1,4 + 2 + 2 + 1,3 + 2 + 1 + 1 + 1 + 1,2 + 2 + 2 + 2 + 1,3 + 3 + 1 + 1 + 1,3 + 1 + 1 + 1 + 1 + 1 + 1,2 + 2 + 1 + 1 + 1 + 1 + 1,1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1).

According to our definition, the partition λ= (6,1,1,1)∈/Ou(9) but the partition (4,3,1,1)∈Ou(9).

Theorem (B., Dastidar (2019)) ou(n) =peuou(n)

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### Extension of Andrews’ identity

Definition 4:

OEMaxDiff(λ) = min{OEMaxDiff (λ0) :λ0 ∈Ou(n)}.

Ou(n) :={λ∈Ou(n) : OEMaxDiff(λ)}.

For example,ou(9) = 12

(8 + 1,6 + 2 + 1,5 + 3 + 1,4 + 4 + 1,4 + 3 + 1 + 1,4 + 2 + 2 + 1,3 + 2 + 1 + 1 + 1 + 1,2 + 2 + 2 + 2 + 1,3 + 3 + 1 + 1 + 1,3 + 1 + 1 + 1 + 1 + 1 + 1,2 + 2 + 1 + 1 + 1 + 1 + 1,1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1).

According to our definition, the partition λ= (6,1,1,1)∈/Ou(9) but the partition (4,3,1,1)∈Ou(9).

Theorem (B., Dastidar (2019)) ou(n) =peuou(n)

### Extension of Andrews’ identity

Definition 4:

OEMaxDiff(λ) = min{OEMaxDiff (λ0) :λ0 ∈Ou(n)}.

Ou(n) :={λ∈Ou(n) : OEMaxDiff(λ)}.

For example,ou(9) = 12

(8 + 1,6 + 2 + 1,5 + 3 + 1,4 + 4 + 1,4 + 3 + 1 + 1,4 + 2 + 2 + 1,3 + 2 + 1 + 1 + 1 + 1,2 + 2 + 2 + 2 + 1,3 + 3 + 1 + 1 + 1,3 + 1 + 1 + 1 + 1 + 1 + 1,2 + 2 + 1 + 1 + 1 + 1 + 1,1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1).

According to our definition, the partition λ= (6,1,1,1)∈/Ou(9) but the partition (4,3,1,1)∈Ou(9).

Theorem (B., Dastidar (2019)) ou(n) =peuou(n)

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### Extension of Andrews’ identity

Proof Sketch: (Ou(n)−→Peuou(n))

First, consider the Young diagramYλ for the partition λ= (λ1, λ2, . . . , λ`)∈Ou(n).

We separateλintoλ0 = (λo1, λo2, . . . , λor) where 1≤oi≤`and λ00= (λe1, λe2, . . . , λet) where 1≤oj≤l according to the odd and even parts, respectively with corresponding Young diagrams Yλ0 and Yλ00.

Next, we joinYλ0 andYλ00 by successively adjoining their rows with respect to the ordering of the parts inλ0, λ00, respectively, starting with the largest one and end with the smallest one with restricting Young diagram, say,Yλ000.

Now, we consider the following three cases:

(1) If the number of odd parts is equal to the number of even parts in a partition λ∈Ou(n), then Yλ000 is withλ000∈Peuou(n) as for λ0 = (λo1, . . . , λor) andλ00= (λe1, . . . , λer), the resulting partition λ000 = (λo1e1, . . . , λorer).

### Extension of Andrews’ identity

Proof Sketch: (Ou(n)−→Peuou(n))

First, consider the Young diagramYλ for the partition λ= (λ1, λ2, . . . , λ`)∈Ou(n).

We separateλintoλ0 = (λo1, λo2, . . . , λor) where 1≤oi≤`and λ00= (λe1, λe2, . . . , λet) where 1≤oj≤l according to the odd and even parts, respectively with corresponding Young diagrams Yλ0 and Yλ00.

Next, we joinYλ0 andYλ00 by successively adjoining their rows with respect to the ordering of the parts inλ0, λ00, respectively, starting with the largest one and end with the smallest one with restricting Young diagram, say,Yλ000.

Now, we consider the following three cases:

(1) If the number of odd parts is equal to the number of even parts in a partition λ∈Ou(n), then Yλ000 is withλ000∈Peuou(n) as for λ0 = (λo1, . . . , λor) andλ00= (λe1, . . . , λer), the resulting partition λ000 = (λo1e1, . . . , λorer).

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### Extension of Andrews’ identity

Proof Sketch: (Ou(n)−→Peuou(n))

First, consider the Young diagramYλ for the partition λ= (λ1, λ2, . . . , λ`)∈Ou(n).

We separateλintoλ0 = (λo1, λo2, . . . , λor) where 1≤oi≤`and λ00= (λe1, λe2, . . . , λet) where 1≤oj≤l according to the odd and even parts, respectively with corresponding Young diagrams Yλ0 and Yλ00.

Next, we joinYλ0 andYλ00 by successively adjoining their rows with respect to the ordering of the parts inλ0, λ00, respectively, starting with the largest one and end with the smallest one with restricting Young diagram, say,Yλ000.

Now, we consider the following three cases:

(1) If the number of odd parts is equal to the number of even parts in a partition λ∈Ou(n), then Yλ000 is withλ000∈Peuou(n) as for λ0 = (λo1, . . . , λor) andλ00= (λe1, . . . , λer), the resulting partition λ000 = (λo1e1, . . . , λorer).

### Extension of Andrews’ identity

Proof Sketch: (Ou(n)−→Peuou(n))

First, consider the Young diagramYλ for the partition λ= (λ1, λ2, . . . , λ`)∈Ou(n).

We separateλintoλ0 = (λo1, λo2, . . . , λor) where 1≤oi≤`and λ00= (λe1, λe2, . . . , λet) where 1≤oj≤l according to the odd and even parts, respectively with corresponding Young diagrams Yλ0 and Yλ00.

Next, we joinYλ0 andYλ00 by successively adjoining their rows with respect to the ordering of the parts inλ0, λ00, respectively, starting with the largest one and end with the smallest one with restricting Young diagram, say,Yλ000.

Now, we consider the following three cases:

(1) If the number of odd parts is equal to the number of even parts in a partition λ∈Ou(n), then Yλ000 is withλ000∈Peuou(n) as for λ0 = (λo1, . . . , λor) andλ00= (λe1, . . . , λer), the resulting partition λ000 = (λo1e1, . . . , λorer).

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### Extension of Andrews’ identity

Proof Sketch: (Ou(n)−→Peuou(n))

First, consider the Young diagramYλ for the partition λ= (λ1, λ2, . . . , λ`)∈Ou(n).

We separateλintoλ0 = (λo1, λo2, . . . , λor) where 1≤oi≤`and λ00= (λe1, λe2, . . . , λet) where 1≤oj≤l according to the odd and even parts, respectively with corresponding Young diagrams Yλ0 and Yλ00.

Next, we joinYλ0 andYλ00 by successively adjoining their rows with respect to the ordering of the parts inλ0, λ00, respectively, starting with the largest one and end with the smallest one with restricting Young diagram, say,Yλ000.

Now, we consider the following three cases:

(1) If the number of odd parts is equal to the number of even parts in a partition λ∈Ou(n), then Yλ000 is withλ000∈Peuou(n) as for λ0 = (λo1, . . . , λor) andλ00= (λe1, . . . , λer), the resulting partition λ000 = (λo1e1, . . . , λorer).

### Extension of Andrews’ identity

The remaining two cases are:

(2) If number of odd parts is greater than the number of even parts in a partition λ∈Ou(n) and let the difference bet. Then a similar argument shows that thet rows in Yλ0 remain left after adjoining of rows of Yλ0 andYλ00. Therefore, in the resultingYλ000 with

λ000 ∈Peuou(n),t rows will be positioned in the same order as in Yλ0. (3) Last, if the number of even parts is greater than the number of odd parts in a partition λ∈Ou(n) and let the difference beu. Similarly, we see that urows inYλ00 remain left after adjoining the rows of Yλ0 andYλ00 and hereu rows will be inserted intoYλ0 so that the resulting Yλ000 withλ000∈Peuou(n) does not violate the structure of the Young diagram.

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### Extension of Andrews’ identity

The remaining two cases are:

(2) If number of odd parts is greater than the number of even parts in a partition λ∈Ou(n) and let the difference bet. Then a similar argument shows that thet rows in Yλ0 remain left after adjoining of rows of Yλ0 andYλ00. Therefore, in the resultingYλ000 with

λ000 ∈Peuou(n),t rows will be positioned in the same order as in Yλ0.

(3) Last, if the number of even parts is greater than the number of odd parts in a partition λ∈Ou(n) and let the difference beu. Similarly, we see that urows inYλ00 remain left after adjoining the rows of Yλ0 andYλ00 and hereu rows will be inserted intoYλ0 so that the resulting Yλ000 withλ000∈Peuou(n) does not violate the structure of the Young diagram.

### Extension of Andrews’ identity

The remaining two cases are:

(2) If number of odd parts is greater than the number of even parts in a partition λ∈Ou(n) and let the difference bet. Then a similar argument shows that thet rows in Yλ0 remain left after adjoining of rows of Yλ0 andYλ00. Therefore, in the resultingYλ000 with

λ000 ∈Peuou(n),t rows will be positioned in the same order as in Yλ0. (3) Last, if the number of even parts is greater than the number of odd parts in a partition λ∈Ou(n) and let the difference beu.

Similarly, we see that urows inYλ00 remain left after adjoining the rows of Yλ0 andYλ00 and hereu rows will be inserted intoYλ0 so that the resulting Yλ000 withλ000 ∈Peuou(n) does not violate the structure of the Young diagram.

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### Extension of Andrews’ identity

For example, givenYλwith the partition λ= (5,4,3,2,1,1)∈Ou(16):

Step 1: SeparatingYλ into the odd and even parts; i.e., into Yλ0 with λ0 = (5,3,1,1) andYλ00 withλ00 = (4,2) yields;

### Extension of Andrews’ identity

For example, givenYλwith the partition λ= (5,4,3,2,1,1)∈Ou(16):

Step 1: SeparatingYλ into the odd and even parts; i.e., into Yλ0 with λ0 = (5,3,1,1) andYλ00 withλ00 = (4,2) yields;

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### Extension of Andrews’ identity

For example, givenYλwith the partition λ= (5,4,3,2,1,1)∈Ou(16):

Step 1: SeparatingYλ into the odd and even parts; i.e., into Yλ0 with λ0 = (5,3,1,1) andYλ00 withλ00 = (4,2) yields;

### Extension of Andrews’ identity

For example, givenYλwith the partition λ= (5,4,3,2,1,1)∈Ou(16):

Step 1: SeparatingYλ into the odd and even parts; i.e., into Yλ0 with λ0 = (5,3,1,1) andYλ00 withλ00 = (4,2) yields;

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### Extension of Andrews’ identity

Step 2: Adjoining the rows ofYλ0 andYλ00 givesYλ000 with the partition λ000 = (9,5,1,1)∈Peuou(16);

### Extension of Andrews’ identity

Step 2: Adjoining the rows ofYλ0 andYλ00 givesYλ000 with the partition λ000 = (9,5,1,1)∈Peuou(16);

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### Extension of Andrews’ identity

(1) All odd parts of µare distinct; i.e., there arei distinct odd values withµoi < µoi−1<· · ·< µo1. For allj (1≤j ≤i), we extract 2j−1 boxes from thejth row ofYµ0 and attach 2j−1 boxes toYµ0 without violating the structure of the Young diagram Yµ0. Explicitly, we break an odd part µot of the partitionµ0 into

ot−(2v−1),2v−1) where the part µot corresponds to the number of boxes in thevth row ofYµ0. The Young diagramYµ000 obtained from Yµ0 by the above construction and adjoiningYµ00 with it to get the unique resulting Young diagram, sayYπ with π∈Ou(n).

### Extension of Andrews’ identity

(1) All odd parts of µare distinct; i.e., there arei distinct odd values withµoi < µoi−1<· · ·< µo1. For allj (1≤j ≤i), we extract 2j−1 boxes from thejth row ofYµ0 and attach 2j−1 boxes toYµ0 without violating the structure of the Young diagram Yµ0. Explicitly, we break an odd part µot of the partitionµ0 into

ot−(2v−1),2v−1) where the partµot corresponds to the number of boxes in thevth row ofYµ0. The Young diagramYµ000 obtained from Yµ0 by the above construction and adjoiningYµ00 with it to get the unique resulting Young diagram, sayYπ with π∈Ou(n).

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### Extension of Andrews’ identity

(1) All odd parts of µare distinct; i.e., there arei distinct odd values withµoi < µoi−1<· · ·< µo1. For all j (1≤j ≤i), we extract 2j−1 boxes from thejth row ofYµ0 and attach 2j−1 boxes toYµ0 without violating the structure of the Young diagram Yµ0. Explicitly, we break an odd part µot of the partitionµ0 into

ot−(2v−1),2v−1) where the partµot corresponds to the number of boxes in thevth row ofYµ0. The Young diagramYµ000 obtained from Yµ0 by the above construction and adjoiningYµ00 with it to get the unique resulting Young diagram, sayYπ with π∈Ou(n).

### Extension of Andrews’ identity

For example,Yµ withµ= (9,7,4,2)∈Peuou(22) breaks intoYµ0 with µ0= (9,7) andYµ00 withµ00= (4,2);

Step 1:

= +

Step 2: Following the above construction,Yµ0 resultsYµ000 with µ000= (1,3,6,6);

=

Step 3: Then the resulting diagramYπ with

π= (6,6,4,3,2,1)∈Ou(22) is the unique pre-image of µ;

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### Extension of Andrews’ identity

For example,Yµ withµ= (9,7,4,2)∈Peuou(22) breaks intoYµ0 with µ0= (9,7) andYµ00 withµ00= (4,2);

Step 1:

= +

Step 2: Following the above construction,Yµ0 resultsYµ000 with µ000= (1,3,6,6);

=

Step 3: Then the resulting diagramYπ with

π= (6,6,4,3,2,1)∈Ou(22) is the unique pre-image of µ;

### Extension of Andrews’ identity

For example,Yµ withµ= (9,7,4,2)∈Peuou(22) breaks intoYµ0 with µ0= (9,7) andYµ00 withµ00= (4,2);

Step 1:

= +

Step 2: Following the above construction,Yµ0 resultsYµ000 with µ000= (1,3,6,6);

=

Step 3: Then the resulting diagramYπ with

π= (6,6,4,3,2,1)∈Ou(22) is the unique pre-image of µ;

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### Extension of Andrews’ identity

For example,Yµ withµ= (9,7,4,2)∈Peuou(22) breaks intoYµ0 with µ0= (9,7) andYµ00 withµ00= (4,2);

Step 1:

= +

Step 2: Following the above construction,Yµ0 resultsYµ000 with µ000= (1,3,6,6);

=

Step 3: Then the resulting diagramYπ with

π= (6,6,4,3,2,1)∈Ou(22) is the unique pre-image of µ;

### Extension of Andrews’ identity

For example,Yµ withµ= (9,7,4,2)∈Peuou(22) breaks intoYµ0 with µ0= (9,7) andYµ00 withµ00= (4,2);

Step 1:

= +

Step 2: Following the above construction,Yµ0 resultsYµ000 with µ000= (1,3,6,6);

=

Step 3: Then the resulting diagramYπ with

π= (6,6,4,3,2,1)∈Ou(22) is the unique pre-image of µ;

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### Extension of Andrews’ identity

For example,Yµ withµ= (9,7,4,2)∈Peuou(22) breaks intoYµ0 with µ0= (9,7) andYµ00 withµ00= (4,2);

Step 1:

= +

Step 2: Following the above construction,Yµ0 resultsYµ000 with µ000= (1,3,6,6);

=

Step 3: Then the resulting diagramYπ with

π= (6,6,4,3,2,1)∈Ou(22) is the unique pre-image of µ;

### Extension of Andrews’ identity

Proof Sketch: (Peuou(n)−→Ou(n))

The remaining case:

(2) Odd parts of µrepeats; i.e.,µ0 = (µo1, . . . , µoi) with

µoi < µoi−1 <· · ·< µo1 with the assumption thatµo1, . . . , µoi occurs with multiplicityk1,k2, . . . ,ki, respectively. Now, for all 1≤t ≤i, we break thekt tuple (µot, . . . , µot) into

((µot−(2v−1),2v−1), . . . ,(µot−(2v−1),2v−1)), where the partµot corresponds to the number of boxes in thevth row ofYµ0. Similar argument shows that the resulting partition, say π∈Ou(n).

For example, the pre-image ofµ= (7,7,5,1,1,1)∈Peuou(22) is π= (5,5,3,2,2,2,1,1,1)∈Ou(22);

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### Extension of Andrews’ identity

Proof Sketch: (Peuou(n)−→Ou(n)) The remaining case:

(2) Odd parts of µrepeats; i.e.,µ0 = (µo1, . . . , µoi) with

µoi < µoi−1 <· · ·< µo1 with the assumption thatµo1, . . . , µoi occurs with multiplicityk1,k2, . . . ,ki, respectively. Now, for all 1≤t ≤i, we break thekt tuple (µot, . . . , µot) into

((µot−(2v−1),2v−1), . . . ,(µot−(2v−1),2v−1)), where the partµot corresponds to the number of boxes in thevth row ofYµ0. Similar argument shows that the resulting partition, say π∈Ou(n). For example, the pre-image ofµ= (7,7,5,1,1,1)∈Peuou(22) is π= (5,5,3,2,2,2,1,1,1)∈Ou(22);

### Extension of Andrews’ identity

Proof Sketch: (Peuou(n)−→Ou(n)) The remaining case:

(2) Odd parts of µrepeats; i.e.,µ0 = (µo1, . . . , µoi) with

µoi < µoi−1 <· · ·< µo1 with the assumption thatµo1, . . . , µoi occurs with multiplicityk1,k2, . . . ,ki, respectively. Now, for all 1≤t ≤i, we break thekt tuple (µot, . . . , µot) into

((µot−(2v−1),2v−1), . . . ,(µot−(2v−1),2v−1)), where the partµot corresponds to the number of boxes in thevth row ofYµ0. Similar argument shows that the resulting partition, say π∈Ou(n).

For example, the pre-image ofµ= (7,7,5,1,1,1)∈Peuou(22) is π= (5,5,3,2,2,2,1,1,1)∈Ou(22);

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### Extension of Andrews’ identity

Proof Sketch: (Peuou(n)−→Ou(n)) The remaining case:

(2) Odd parts of µrepeats; i.e.,µ0 = (µo1, . . . , µoi) with

µoi < µoi−1 <· · ·< µo1 with the assumption thatµo1, . . . , µoi occurs with multiplicityk1,k2, . . . ,ki, respectively. Now, for all 1≤t ≤i, we break thekt tuple (µot, . . . , µot) into

((µot−(2v−1),2v−1), . . . ,(µot−(2v−1),2v−1)), where the partµot corresponds to the number of boxes in thevth row ofYµ0. Similar argument shows that the resulting partition, say π∈Ou(n).

For example, the pre-image ofµ= (7,7,5,1,1,1)∈Peuou(22) is π= (5,5,3,2,2,2,1,1,1)∈Ou(22);

### Extension of Andrews’ identity

Proof Sketch: (Peuou(n)−→Ou(n)) The remaining case:

(2) Odd parts of µrepeats; i.e.,µ0 = (µo1, . . . , µoi) with

µoi < µoi−1 <· · ·< µo1 with the assumption thatµo1, . . . , µoi occurs with multiplicityk1,k2, . . . ,ki, respectively. Now, for all 1≤t ≤i, we break thekt tuple (µot, . . . , µot) into

((µot−(2v−1),2v−1), . . . ,(µot−(2v−1),2v−1)), where the partµot corresponds to the number of boxes in thevth row ofYµ0. Similar argument shows that the resulting partition, say π∈Ou(n).

For example, the pre-image ofµ= (7,7,5,1,1,1)∈Peuou(22) is π= (5,5,3,2,2,2,1,1,1)∈Ou(22);

Koustav Banerjee BSP and Partition identities 43 / 45

### References

G.E. Andrews, The Theory of Partitions, Addison-Wesley Pub. Co., NY, 300 pp. (1976). Reissued, Cambridge University Press, New York, 1998.

C. Bessenrodt, On hooks of Young diagrams, Annals of Combinatorics2(1998), 103-110.

R. Bacher and L. Manivel,Hooks and powers of parts in partitions, S´eminaire Lotharingien de Combinatoire47(2002).

H.C. Chan,Ramanujan’s cubic continued fraction and an analogue of his most beautiful identity, International Journal of Number Theory06 (2010), 673–680.

M.G. Dastidar and S. Sengupta,Generalization of a few results in integer Partitions, Notes in Number theory and Discrete

Mathematics 19(2013), 69-76.

G.-N. Han, Some conjectures and open problems on partition hook lengths, Experimental Mathematics18(2009), 97-106.

R. Honsberger,Mathematical Gems III, Washington, DC: Math.

Thank you!

Koustav Banerjee BSP and Partition identities 45 / 45

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