Integer quotients of factorials and algebraic generating functions - PowerPoint PPT Presentation

Integer quotients of factorials and algebraic generating functions Ira M. Gessel Department of Mathematics Brandeis University MIT Combinatorics Seminar September 30, 2011

Integer quotients of factorials There are many formulas for integer quotients of factorials: ( 2 n )! n ! ( n + 1 )! ,

Integer quotients of factorials There are many formulas for integer quotients of factorials: ( 2 n )! n ! ( 6 n )! n ! ( n + 1 )! , ( 2 n )! 2 ( 3 n )! ,

Integer quotients of factorials There are many formulas for integer quotients of factorials: ( 2 n )! n ! ( 6 n )! n ! ( 30 n )! n ! ( n + 1 )! , ( 2 n )! 2 ( 3 n )! , ( 6 n )! ( 10 n )! ( 15 n )!

Integer quotients of factorials There are many formulas for integer quotients of factorials: ( 2 n )! n ! ( 6 n )! n ! ( 30 n )! n ! ( n + 1 )! , ( 2 n )! 2 ( 3 n )! , ( 6 n )! ( 10 n )! ( 15 n )! ( m + n )! , m ! n !

Integer quotients of factorials There are many formulas for integer quotients of factorials: ( 2 n )! n ! ( 6 n )! n ! ( 30 n )! n ! ( n + 1 )! , ( 2 n )! 2 ( 3 n )! , ( 6 n )! ( 10 n )! ( 15 n )! ( m + n )! ( 2 m )! ( 2 n )! , m ! n ! ( m + n )! , m ! n !

Integer quotients of factorials There are many formulas for integer quotients of factorials: ( 2 n )! n ! ( 6 n )! n ! ( 30 n )! n ! ( n + 1 )! , ( 2 n )! 2 ( 3 n )! , ( 6 n )! ( 10 n )! ( 15 n )! ( m + n )! ( 2 m )! ( 2 n )! ( 4 m )! ( 4 n )! , m ! n ! ( m + n )! , m ! n ! m ! n ! ( 2 m + n )! ( m + 2 n )!

Integer quotients of factorials There are many formulas for integer quotients of factorials: ( 2 n )! n ! ( 6 n )! n ! ( 30 n )! n ! ( n + 1 )! , ( 2 n )! 2 ( 3 n )! , ( 6 n )! ( 10 n )! ( 15 n )! ( m + n )! ( 2 m )! ( 2 n )! ( 4 m )! ( 4 n )! , m ! n ! ( m + n )! , m ! n ! m ! n ! ( 2 m + n )! ( m + 2 n )! ( k + 2 l )! ( k + 2 m )! ( k + 2 n )! ( k + l + m + n )! k ! l ! m ! n ! ( k + l + m )! ( k + l + n )! ( k + m + n )!

Some known facts ◮ It is straightforward to check that a factorial quotient is integral (Landau’s theorem).

Some known facts ◮ It is straightforward to check that a factorial quotient is integral (Landau’s theorem). ◮ In all “interesting” cases where � i ( a i m + b i n )! � j ( c j m + d j n )! is integral, � i a i = � j c j and � i b i = � j d j (and similarly for more than two parameters).

The one-parameter case ◮ Suppose that u n = � r i = 1 ( a i n )! / � s j = 1 ( b j n )! is integral. Then n = 0 u n x n is algebraic if and only if s = r + 1. � ∞ (Rodriguez-Villegas, based on work of Beukers and Heckman).

The one-parameter case ◮ Suppose that u n = � r i = 1 ( a i n )! / � s j = 1 ( b j n )! is integral. Then n = 0 u n x n is algebraic if and only if s = r + 1. � ∞ (Rodriguez-Villegas, based on work of Beukers and Heckman). ◮ These algebraic cases are all known. There are three two-parameter formulas and 52 sporadic cases. (Bober, based on work of Beukers and Heckman.) As a corollary to the classification, we have

The one-parameter case ◮ Suppose that u n = � r i = 1 ( a i n )! / � s j = 1 ( b j n )! is integral. Then n = 0 u n x n is algebraic if and only if s = r + 1. � ∞ (Rodriguez-Villegas, based on work of Beukers and Heckman). ◮ These algebraic cases are all known. There are three two-parameter formulas and 52 sporadic cases. (Bober, based on work of Beukers and Heckman.) As a corollary to the classification, we have ◮ If � r i = 1 ( a i n )! / � s j = 1 ( b j n )! is integral and nontrivial (and � i a i = � j b j ) then s ≥ r + 1.

The one-parameter case ◮ Suppose that u n = � r i = 1 ( a i n )! / � s j = 1 ( b j n )! is integral. Then n = 0 u n x n is algebraic if and only if s = r + 1. � ∞ (Rodriguez-Villegas, based on work of Beukers and Heckman). ◮ These algebraic cases are all known. There are three two-parameter formulas and 52 sporadic cases. (Bober, based on work of Beukers and Heckman.) As a corollary to the classification, we have ◮ If � r i = 1 ( a i n )! / � s j = 1 ( b j n )! is integral and nontrivial (and � i a i = � j b j ) then s ≥ r + 1. Is there a simple proof?

The two-parameter formulas The two-parameter formulas are ( m + n )! m ! ( 2 m + 2 n )! ( 2 m )! ( 2 n )! , ( 2 m )! n ! ( m + n )! , m ! n ! m ! n ! ( m + n )!

The two-parameter formulas The two-parameter formulas are ( m + n )! m ! ( 2 m + 2 n )! ( 2 m )! ( 2 n )! , ( 2 m )! n ! ( m + n )! , m ! n ! m ! n ! ( m + n )! They are all essentially binomial coefficients, since � − m − 1 m ! ( 2 m + 2 n )! � ( 2 m )! n ! ( m + n )! = ( − 1 ) n 2 2 n 2 n � m − 1 ( 2 m )! ( 2 n )! � m ! n ! ( m + n )! = ( − 1 ) n 2 2 m + 2 n 2 m + n

They are clearly integers, because they are coefficients of odd powers of ∞ � 2 n � 1 � z n . √ 1 − 4 z = n n = 0

They are clearly integers, because they are coefficients of odd powers of ∞ � 2 n � 1 � z n . √ 1 − 4 z = n n = 0 More precisely, for m a nonnegative integer, ∞ ( 1 − 4 z ) − m − 1 m ! ( 2 m + 2 n )! � ( 2 m )! n ! ( m + n )! z n 2 = n = 0 m − 1 ( − 1 ) n ( m − n )! ( 2 m )! ( 1 − 4 z ) m − 1 � ( 2 m − 2 n )! n ! m ! z n 2 = n = 0 ∞ ( 2 m )! ( 2 n )! + ( − 1 ) m � m ! n ! ( m + n )! z m + n n = 0

The formula � ∞ � 2 m + 1 ∞ � 2 k � m ! ( 2 m + 2 n )! ( 1 − 4 z ) − m − 1 � � z k ( 2 m )! n ! ( m + n )! z n 2 = = k k = 0 n = 0 gives a combinatorial interpretation to the number m ! ( 2 m + 2 n )! ( 2 m )! n ! ( m + n )! .

The formula � ∞ � 2 m + 1 ∞ � 2 k � m ! ( 2 m + 2 n )! ( 1 − 4 z ) − m − 1 � � z k ( 2 m )! n ! ( m + n )! z n 2 = = k k = 0 n = 0 gives a combinatorial interpretation to the number m ! ( 2 m + 2 n )! ( 2 m )! n ! ( m + n )! . It is the number of ( 2 m + 1 ) -tuples of paths, each with as many up steps as down steps, with a total of 2 n steps.

The formula � ∞ � 2 m + 1 ∞ � 2 k � m ! ( 2 m + 2 n )! ( 1 − 4 z ) − m − 1 � � z k ( 2 m )! n ! ( m + n )! z n 2 = = k k = 0 n = 0 gives a combinatorial interpretation to the number m ! ( 2 m + 2 n )! ( 2 m )! n ! ( m + n )! . It is the number of ( 2 m + 1 ) -tuples of paths, each with as many up steps as down steps, with a total of 2 n steps. Note that the special case n = 2 m gives m ! ( 6 m !) ( 2 m )! 2 ( 3 m )! .

However, the formula m − 1 ( − 1 ) n ( m − n )! ( 2 m )! ( 1 − 4 z ) m − 1 � ( 2 m − 2 n )! n ! m ! z n 2 = n = 0 ∞ ( 2 m )! ( 2 n )! � + ( − 1 ) m m ! n ! ( m + n )! z m + n n = 0 does not give a combinatorial interpretation, because of the sign changes.

What this talk is about

What this talk is about There are some multivariable formulas of the form fg α = multiple hypergeometric series where f and g are algebraic functions (either rational or quadratic) with integer coefficients. Then taking α to be an integer (or sometimes a half-integer) gives an integral quotient of factorials.

What is a multiple hypergeometric series? It is a power series whose terms are quotients of rising factorials: ( α ) n = α ( α + 1 ) · · · ( α + n − 1 ) ( − 1 ) n ( α ) − n = . ( 1 − α ) n

What is a multiple hypergeometric series? It is a power series whose terms are quotients of rising factorials: ( α ) n = α ( α + 1 ) · · · ( α + n − 1 ) ( − 1 ) n ( α ) − n = . ( 1 − α ) n If α is a nonnegative integer then ( α + 1 ) n = ( α + n )! α ! and 2 ) n = α ! ( 2 α + 2 n )! 2 2 n ( α + 1 ( 2 α )! ( α + n )! .

A simple example is 1 ( 2 α + 1 ) m + n ( α + 1 ) m + n � x m y n , = � α + 1 m ! n ! ( α + 1 ) m ( α + 1 ) n ( 1 − x − y ) 2 − 4 xy � 2 m , n where ( α ) n = α ( α + 1 ) · · · ( α + n − 1 ) .

A simple example is 1 ( 2 α + 1 ) m + n ( α + 1 ) m + n � x m y n , = � α + 1 m ! n ! ( α + 1 ) m ( α + 1 ) n ( 1 − x − y ) 2 − 4 xy � 2 m , n where ( α ) n = α ( α + 1 ) · · · ( α + n − 1 ) . Taking α to be a nonnegative integer l gives 1 l ! ( 2 l + m + n )! ( l + m + n )! � ( 2 l )! m ! n ! ( l + m )! ( l + n )! x m y n = � l + 1 ( 1 − x − y ) 2 − 4 xy � 2 m , n

A simple example is 1 ( 2 α + 1 ) m + n ( α + 1 ) m + n � x m y n , = � α + 1 m ! n ! ( α + 1 ) m ( α + 1 ) n ( 1 − x − y ) 2 − 4 xy � 2 m , n where ( α ) n = α ( α + 1 ) · · · ( α + n − 1 ) . Taking α to be a nonnegative integer l gives 1 l ! ( 2 l + m + n )! ( l + m + n )! � ( 2 l )! m ! n ! ( l + m )! ( l + n )! x m y n = � l + 1 ( 1 − x − y ) 2 − 4 xy � 2 m , n Thus l ! ( 2 l + m + n )! ( l + m + n )! ( 2 l )! m ! n ! ( l + m )! ( l + n )! is integral, since � 2 1 � m + n � x m y n = � 1 m ( 1 − x − y ) 2 − 4 xy � 2 m , n

We also have a combinatorial interpretation (though not an especially interesting one) of the coefficients. Since their generating function is � 2 � 2 l + 1 �� � i + j x i y j , i i , j � 2 counts pairs of paths ending at the same point. So � i + j and i ( 2 l + m + n )! ( l + m + n )! l ! ( 2 l )! m ! n ! ( l + m )! ( l + n )! counts ( 2 l + 1 ) -tuples of such pairs of paths.