The Jacobi-Stirling Numbers Eric S. Egge (joint work with G. - PowerPoint PPT Presentation

The Jacobi-Stirling Numbers Eric S. Egge (joint work with G. Andrews, L. Littlejohn, and W. Gawronski) Carleton College March 18, 2012 Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 1 / 12

The Differential Operator xD D = d y = y ( x ) dx Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 2 / 12

The Differential Operator xD D = d y = y ( x ) dx xD [ y ] = 1 xy ′ ( xD ) 2 [ y ] = 1 xy ′ + 1 x 2 y (2) Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 2 / 12

The Differential Operator xD D = d y = y ( x ) dx xD [ y ] = 1 xy ′ ( xD ) 2 [ y ] = 1 xy ′ + 1 x 2 y (2) ( xD ) 3 [ y ] = 1 xy ′ + 3 x 2 y (2) + 1 x 3 y (3) Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 2 / 12

The Differential Operator xD D = d y = y ( x ) dx xD [ y ] = 1 xy ′ ( xD ) 2 [ y ] = 1 xy ′ + 1 x 2 y (2) ( xD ) 3 [ y ] = 1 xy ′ + 3 x 2 y (2) + 1 x 3 y (3) ( xD ) 4 [ y ] = 1 xy ′ + 7 x 2 y (2) + 6 x 3 y (3) + 1 x 4 y (4) Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 2 / 12

The Differential Operator xD D = d y = y ( x ) dx xD [ y ] = 1 xy ′ ( xD ) 2 [ y ] = 1 xy ′ + 1 x 2 y (2) ( xD ) 3 [ y ] = 1 xy ′ + 3 x 2 y (2) + 1 x 3 y (3) ( xD ) 4 [ y ] = 1 xy ′ + 7 x 2 y (2) + 6 x 3 y (3) + 1 x 4 y (4) n � n � � n � � n − 1 � � n − 1 � ( xD ) n [ y ] = � x j y ( j ) = + j n + 1 − j j j − 1 j j =1 Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 2 / 12

The Jacobi-Stirling Numbers ℓ α,β [ y ] = − (1 − x 2 ) y ′′ + ( α − β + ( α + β + 2) x ) y ′ Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 3 / 12

The Jacobi-Stirling Numbers ℓ α,β [ y ] = − (1 − x 2 ) y ′′ + ( α − β + ( α + β + 2) x ) y ′ Theorem (Everitt, Kwon, Littlejohn, Wellman, Yoon) n 1 � n � (1 − x ) α + j (1 + x ) β + j y ( j ) � ( j ) ( − 1) j � � ℓ n α,β [ y ] = w α,β ( x ) j α,β j =1 Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 3 / 12

The Jacobi-Stirling Numbers ℓ α,β [ y ] = − (1 − x 2 ) y ′′ + ( α − β + ( α + β + 2) x ) y ′ Theorem (Everitt, Kwon, Littlejohn, Wellman, Yoon) n 1 � n � (1 − x ) α + j (1 + x ) β + j y ( j ) � ( j ) ( − 1) j � � ℓ n α,β [ y ] = w α,β ( x ) j α,β j =1 � n � � n − 1 � � n − 1 � = + j ( j + α + β + 1) j j − 1 j α,β α,β α,β Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 3 / 12

� n � What Does α,β Count? j 2 γ − 1 = α + β + 1 [ n ] 2 := { 1 1 , 1 2 , 2 1 , 2 2 , . . . , n 1 , n 2 } Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 4 / 12

� n � What Does α,β Count? j 2 γ − 1 = α + β + 1 [ n ] 2 := { 1 1 , 1 2 , 2 1 , 2 2 , . . . , n 1 , n 2 } Theorem (AEGL) � n � For any positive integer γ , the Jacobi-Stirling number γ counts set j partitions of [ n ] 2 into j + γ blocks such that 1 There are γ distinguishable zero blocks, any of which may be empty. 2 There are j indistinguishable nonzero blocks, all nonempty. 3 The union of the zero blocks does not contain both copies of any number. 4 Each nonzero block contains both copies of its smallest element does not contain both copies of any other number. These are Jacobi-Stirling set partitions. Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 4 / 12

A Generating Function Interpretation z = α + β Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 5 / 12

A Generating Function Interpretation z = α + β S ( n , j ) := Jacobi-Stirling set partitions of [ n ] 2 into 1 zero block and j nonzero blocks. Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 5 / 12

A Generating Function Interpretation z = α + β S ( n , j ) := Jacobi-Stirling set partitions of [ n ] 2 into 1 zero block and j nonzero blocks. Theorem (Gelineau, Zeng) � n � z is the generating function in z for S ( n , j ) The Jacobi-Stirling number j with respect to the number of numbers with subscript 1 in the zero block. Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 5 / 12

A Generating Function Interpretation z = α + β S ( n , j ) := Jacobi-Stirling set partitions of [ n ] 2 into 1 zero block and j nonzero blocks. Theorem (Gelineau, Zeng) � n � z is the generating function in z for S ( n , j ) The Jacobi-Stirling number j with respect to the number of numbers with subscript 1 in the zero block. Corollary � n � n � � The leading coefficient in z is the Stirling number . j j Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 5 / 12

Jacobi-Stirling Numbers of the First Kind j − 1 n � n � x n = � � ( x − k ( k + α + β + 1)) j α,β j =0 k =0 Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 6 / 12

Jacobi-Stirling Numbers of the First Kind j − 1 n � n � x n = � � ( x − k ( k + α + β + 1)) j α,β j =0 k =0 n − 1 n � n � � � ( − 1) n + j x j ( x − k ( k + α + β + 1)) = j α,β k =0 j =0 Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 6 / 12

Jacobi-Stirling Numbers of the First Kind j − 1 n � n � x n = � � ( x − k ( k + α + β + 1)) j α,β j =0 k =0 n − 1 n � n � � � ( − 1) n + j x j ( x − k ( k + α + β + 1)) = j α,β k =0 j =0 � n � � n − 1 � � n − 1 � = + ( n − 1)( n + α + β ) j − 1 j j α,β α,β α,β Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 6 / 12

Jacobi-Stirling Numbers of the First Kind j − 1 n � n � x n = � � ( x − k ( k + α + β + 1)) j α,β j =0 k =0 n − 1 n � n � � � ( − 1) n + j x j ( x − k ( k + α + β + 1)) = j α,β k =0 j =0 � n � � n − 1 � � n − 1 � = + ( n − 1)( n + α + β ) j − 1 j j α,β α,β α,β Theorem (AEGL) � − j � � n � = ( − 1) n + j − n j 1 − γ γ Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 6 / 12

Balanced Jacobi-Stirling Permutations Theorem (AEGL) � n � For any positive integer γ , the Jacobi-Stirling number of the first kind j γ is the number of ordered pairs ( π 1 , π 2 ) of permutations with π 1 ∈ S n + γ and π 2 ∈ S n + γ − 1 such that 1 π 1 has γ + j cycles and π 2 has γ + j − 1 cycles. 2 The cycle maxima of π 1 which are less than n + γ are exactly the cycle maxima of π 2 . 3 For each non cycle maximum k, at least one of π 1 ( k ) and π 2 ( k ) is less than or equal to n. Such ordered pairs are balanced Jacobi-Stirling permutations. Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 7 / 12

Unbalanced Jacobi-Stirling Permutations Theorem (AEGL) If 2 γ is a positive integer, then the Jacobi-Stirling number of the first kind � n � γ is the number of ordered pairs ( π 1 , π 2 ) of permutations with j π 1 ∈ S n + γ and π 2 ∈ S n such that 1 π 1 has j + 2 γ − 1 cycles and π 2 has j cycles. 2 The cycle maxima of π 1 which are less than n + 1 are exactly the cycle maxima of π 2 . 3 For each non cycle maximum k, at least one of π 1 ( k ) and π 2 ( k ) is less than or equal to n. Such ordered pairs are unbalanced Jacobi-Stirling permutations. Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 8 / 12

More Generating Functions Σ( n , j ) := all ( σ, τ ) such that σ is a permutation of { 0 , 1 , . . . , n } , τ is a permutation of { 1 , 2 , . . . , n } , and both have j cycles. 1 and 0 are in the same cycle in σ . Among their nonzero entries, σ and τ have the same cycle minima. Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 9 / 12

More Generating Functions Σ( n , j ) := all ( σ, τ ) such that σ is a permutation of { 0 , 1 , . . . , n } , τ is a permutation of { 1 , 2 , . . . , n } , and both have j cycles. 1 and 0 are in the same cycle in σ . Among their nonzero entries, σ and τ have the same cycle minima. Theorem (Gelineau, Zeng) � n � z is the generating function in z for Σ( n , j ) with respect to the number j of nonzero left-to-right minima in the cycle containing 0 in σ , written as a word beginning with σ (0) . Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 9 / 12

Where are the Symmetric Functions? h n − j ( x 1 , . . . , x j ) = h n − j ( x 1 , . . . , x j − 1 ) + x j h n − j − 1 ( x 1 , . . . , x j ) Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 10 / 12

Where are the Symmetric Functions? h n − j ( x 1 , . . . , x j ) = h n − j ( x 1 , . . . , x j − 1 ) + x j h n − j − 1 ( x 1 , . . . , x j ) � n � = h n − j (1(1 + z ) , 2(2 + z ) , . . . , j ( j + z )) j z Eric S. Egge (Carleton College) The Jacobi-Stirling Numbers March 18, 2012 10 / 12