The Coin Exchange Problem of Frobenius Matthias Beck San Francisco - PowerPoint PPT Presentation

The Frobenius problem is well defined Given two positive integers a and b with no common factor, we can write the (positive) integer t as an integral linear combination t = ma + nb . Once we have one such representation of t we can find many more: t = ( m − 42 b ) a + ( n + 42 a ) b . There is a unique representation t = ma + nb for which 0 ≤ m ≤ b − 1 . The ”Coin Exchange Problem” of Frobenius Matthias Beck 4

The Frobenius problem is well defined Given two positive integers a and b with no common factor, we can write the (positive) integer t as an integral linear combination t = ma + nb . Once we have one such representation of t we can find many more: t = ( m − 42 b ) a + ( n + 42 a ) b . There is a unique representation t = ma + nb for which 0 ≤ m ≤ b − 1 . So if t is large enough, e.g., ≥ ab , then we can find a nonnegative integral linear combination of a and b . The ”Coin Exchange Problem” of Frobenius Matthias Beck 4

A well-defined homework Prove that the Frobenius problem is well defined for d > 2 . The ”Coin Exchange Problem” of Frobenius Matthias Beck 5

A closer look for two coins Given two positive integers a and b with no common factor, we say the integer t is representable if we can find nonnegative integers m and n such that t = ma + nb. The ”Coin Exchange Problem” of Frobenius Matthias Beck 6

A closer look for two coins Given two positive integers a and b with no common factor, we say the integer t is representable if we can find nonnegative integers m and n such that t = ma + nb. We have seen already that we can always write t as an integral linear combination t = ma + nb for which 0 ≤ m ≤ b − 1 . The ”Coin Exchange Problem” of Frobenius Matthias Beck 6

A closer look for two coins Given two positive integers a and b with no common factor, we say the integer t is representable if we can find nonnegative integers m and n such that t = ma + nb. We have seen already that we can always write t as an integral linear combination t = ma + nb for which 0 ≤ m ≤ b − 1 . If (and only if) we can find such a representation for which also n ≥ 0 then t is representable. The ”Coin Exchange Problem” of Frobenius Matthias Beck 6

A closer look for two coins Given two positive integers a and b with no common factor, we say the integer t is representable if we can find nonnegative integers m and n such that t = ma + nb. We have seen already that we can always write t as an integral linear combination t = ma + nb for which 0 ≤ m ≤ b − 1 . If (and only if) we can find such a representation for which also n ≥ 0 then t is representable. Hence the largest integer t that is not representable is t = ♣ a + ♥ b . The ”Coin Exchange Problem” of Frobenius Matthias Beck 6

A closer look for two coins Given two positive integers a and b with no common factor, we say the integer t is representable if we can find nonnegative integers m and n such that t = ma + nb. We have seen already that we can always write t as an integral linear combination t = ma + nb for which 0 ≤ m ≤ b − 1 . If (and only if) we can find such a representation for which also n ≥ 0 then t is representable. Hence the largest integer t that is not representable is t = ( b − 1) a + ♥ b . The ”Coin Exchange Problem” of Frobenius Matthias Beck 6

A closer look for two coins Given two positive integers a and b with no common factor, we say the integer t is representable if we can find nonnegative integers m and n such that t = ma + nb. We have seen already that we can always write t as an integral linear combination t = ma + nb for which 0 ≤ m ≤ b − 1 . If (and only if) we can find such a representation for which also n ≥ 0 then t is representable. Hence the largest integer t that is not representable is t = ( b − 1) a + ( − 1) b . The ”Coin Exchange Problem” of Frobenius Matthias Beck 6

A closer look for two coins Given two positive integers a and b with no common factor, we say the integer t is representable if we can find nonnegative integers m and n such that t = ma + nb. We have seen already that we can always write t as an integral linear combination t = ma + nb for which 0 ≤ m ≤ b − 1 . If (and only if) we can find such a representation for which also n ≥ 0 then t is representable. Hence the largest integer t that is not representable is t = ab − a − b , a formula most likely known already to Sylvester in the 1880’s. The ”Coin Exchange Problem” of Frobenius Matthias Beck 6

A homework with several representations Given two positive integers a and b with no common factor, we say the integer t is k -representable if there are exactly k solutions ( m, n ) ∈ Z 2 ≥ 0 to t = ma + nb. The ”Coin Exchange Problem” of Frobenius Matthias Beck 7

A homework with several representations Given two positive integers a and b with no common factor, we say the integer t is k -representable if there are exactly k solutions ( m, n ) ∈ Z 2 ≥ 0 to t = ma + nb. We define g k as the largest k -representable integer. (So g 0 is the Frobenius number.) The ”Coin Exchange Problem” of Frobenius Matthias Beck 7

A homework with several representations Given two positive integers a and b with no common factor, we say the integer t is k -representable if there are exactly k solutions ( m, n ) ∈ Z 2 ≥ 0 to t = ma + nb. We define g k as the largest k -representable integer. (So g 0 is the Frobenius number.) Prove: g k is well defined. ◮ The ”Coin Exchange Problem” of Frobenius Matthias Beck 7

A homework with several representations Given two positive integers a and b with no common factor, we say the integer t is k -representable if there are exactly k solutions ( m, n ) ∈ Z 2 ≥ 0 to t = ma + nb. We define g k as the largest k -representable integer. (So g 0 is the Frobenius number.) Prove: g k is well defined. ◮ g k = ( k + 1) ab − a − b ◮ The ”Coin Exchange Problem” of Frobenius Matthias Beck 7

A homework with several representations Given two positive integers a and b with no common factor, we say the integer t is k -representable if there are exactly k solutions ( m, n ) ∈ Z 2 ≥ 0 to t = ma + nb. We define g k as the largest k -representable integer. (So g 0 is the Frobenius number.) Prove: g k is well defined. ◮ g k = ( k + 1) ab − a − b ◮ Given k ≥ 2 , the smallest k -representable integer is ab ( k − 1) . ◮ The ”Coin Exchange Problem” of Frobenius Matthias Beck 7

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + b ) a + ( n − a ) b The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + 2 b ) a + ( n − 2 a ) b The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + 3 b ) a + ( n − 3 a ) b The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + 4 b ) a + ( n − 4 a ) b The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + 5 b ) a + ( n − 5 a ) b The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + 6 b ) a + ( n − 6 a ) b The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + 7 b ) a + ( n − 7 a ) b The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + kb ) a + ( n − ka ) b until n − ka becomes negative. The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + kb ) a + ( n − ka ) b until n − ka becomes negative. But then t + ab = ( m + kb ) a + ( n − ka ) b + ab = ( m + kb ) a + ( n − ( k − 1) a ) b has one more representation The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

Further consequences ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � Let N ( t ) = # If the positive integer t is representable then we can find m, n such that t = ma + nb and 0 ≤ m ≤ b − 1 , n ≥ 0 . Starting from this representation, we can obtain more... t = ( m + kb ) a + ( n − ka ) b until n − ka becomes negative. But then t + ab = ( m + kb ) a + ( n − ka ) b + ab = ( m + kb ) a + ( n − ( k − 1) a ) b has one more representation, i.e., N ( t + ab ) = N ( t ) + 1 . The ”Coin Exchange Problem” of Frobenius Matthias Beck 8

A geometric interlude ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � N ( t ) = # counts integer points in R 2 ≥ 0 on the line ax + by = t . The ”Coin Exchange Problem” of Frobenius Matthias Beck 9

A geometric interlude ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � N ( t ) = # counts integer points in R 2 ≥ 0 on the line ax + by = t . The ”Coin Exchange Problem” of Frobenius Matthias Beck 9

Shameless plug M. Beck & S. Robins Computing the continuous discretely Integer-point enumeration in polyhedra To appear in Springer Undergraduate Texts in Mathematics Preprint available at math.sfsu.edu/beck MSRI Summer Graduate Program at Banff (August 6–20) The ”Coin Exchange Problem” of Frobenius Matthias Beck 10

Generating functions Given a sequence ( a k ) ∞ k =0 we encode it into the generating function � a k x k F ( x ) = k ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 11

Generating functions Given a sequence ( a k ) ∞ k =0 we encode it into the generating function � a k x k F ( x ) = k ≥ 0 Example: Fibonacci sequence f 0 = 0 , f 1 = 1 , and f k +2 = f k +1 + f k for k ≥ 0 . The ”Coin Exchange Problem” of Frobenius Matthias Beck 11

Generating functions Given a sequence ( a k ) ∞ k =0 we encode it into the generating function � a k x k F ( x ) = k ≥ 0 Example: Fibonacci sequence f 0 = 0 , f 1 = 1 , and f k +2 = f k +1 + f k for k ≥ 0 . f k +1 x k + � � � f k +2 x k f k x k = k ≥ 0 k ≥ 0 k ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 11

Generating functions Given a sequence ( a k ) ∞ k =0 we encode it into the generating function � a k x k F ( x ) = k ≥ 0 Example: Fibonacci sequence f 0 = 0 , f 1 = 1 , and f k +2 = f k +1 + f k for k ≥ 0 . f k +1 x k + � � � f k +2 x k f k x k = k ≥ 0 k ≥ 0 k ≥ 0 1 1 f k x k + � � � f k x k f k x k = x 2 x k ≥ 2 k ≥ 1 k ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 11

Generating functions Given a sequence ( a k ) ∞ k =0 we encode it into the generating function � a k x k F ( x ) = k ≥ 0 Example: Fibonacci sequence f 0 = 0 , f 1 = 1 , and f k +2 = f k +1 + f k for k ≥ 0 . f k +1 x k + � � � f k +2 x k f k x k = k ≥ 0 k ≥ 0 k ≥ 0 1 1 f k x k + � � � f k x k f k x k = x 2 x k ≥ 2 k ≥ 1 k ≥ 0 1 1 x 2 ( F ( x ) − x ) = xF ( x ) + F ( x ) The ”Coin Exchange Problem” of Frobenius Matthias Beck 11

Generating functions Given a sequence ( a k ) ∞ k =0 we encode it into the generating function � a k x k F ( x ) = k ≥ 0 Example: Fibonacci sequence f 0 = 0 , f 1 = 1 , and f k +2 = f k +1 + f k for k ≥ 0 . f k +1 x k + � � � f k +2 x k f k x k = k ≥ 0 k ≥ 0 k ≥ 0 1 1 f k x k + � � � f k x k f k x k = x 2 x k ≥ 2 k ≥ 1 k ≥ 0 1 1 x 2 ( F ( x ) − x ) = xF ( x ) + F ( x ) x F ( x ) = 1 − x − x 2 . The ”Coin Exchange Problem” of Frobenius Matthias Beck 11

A Fibonacci homework x f k x k = � Expand F ( x ) = 1 − x − x 2 into partial fractions... k ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 12

A Fibonacci homework x f k x k = � Expand F ( x ) = 1 − x − x 2 into partial fractions... k ≥ 0 √ √  � k  � k � � f k = 1 1 + 5 1 − 5  . √ −  2 2 5 The ”Coin Exchange Problem” of Frobenius Matthias Beck 12

My favorite generating function 1 x k = � The geometric series 1 − x , suspected to converge for | x | < 1 . k ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 13

My favorite generating function 1 x k = � The geometric series 1 − x , suspected to converge for | x | < 1 . k ≥ 0 � 1 if 7 | k Example: a k = 0 otherwise The ”Coin Exchange Problem” of Frobenius Matthias Beck 13

My favorite generating function 1 x k = � The geometric series 1 − x , suspected to converge for | x | < 1 . k ≥ 0 � 1 if 7 | k Example: a k = 0 otherwise � � a k x k x 7 j = k ≥ 0 j ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 13

My favorite generating function 1 x k = � The geometric series 1 − x , suspected to converge for | x | < 1 . k ≥ 0 � 1 if 7 | k Example: a k = 0 otherwise � � a k x k x 7 j = k ≥ 0 j ≥ 0 1 = 1 − x 7 The ”Coin Exchange Problem” of Frobenius Matthias Beck 13

Frobenius generatingfunctionology Given positive integers a and b , recall that ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � N ( t ) = # . The ”Coin Exchange Problem” of Frobenius Matthias Beck 14

Frobenius generatingfunctionology Given positive integers a and b , recall that ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � N ( t ) = # . Consider the product of the geometric series     1 � � x ma x nb  . (1 − x a ) (1 − x b ) =  m ≥ 0 n ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 14

Frobenius generatingfunctionology Given positive integers a and b , recall that ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � N ( t ) = # . Consider the product of the geometric series     1 � � x ma x nb  . (1 − x a ) (1 − x b ) =  m ≥ 0 n ≥ 0 A typical term looks like x ma + nb for some m, n ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 14

Frobenius generatingfunctionology Given positive integers a and b , recall that ( m, n ) ∈ Z 2 : m, n ≥ 0 , ma + nb = t � � N ( t ) = # . Consider the product of the geometric series     1 � � x ma x nb  . (1 − x a ) (1 − x b ) =  m ≥ 0 n ≥ 0 A typical term looks like x ma + nb for some m, n ≥ 0 , and so 1 � N ( t ) x t (1 − x a ) (1 − x b ) = t ≥ 0 is the generating function associated to the counting function N ( t ) . The ”Coin Exchange Problem” of Frobenius Matthias Beck 14

More Frobenius generatingfunctionology Recall that N ( t + ab ) = N ( t ) + 1 The ”Coin Exchange Problem” of Frobenius Matthias Beck 15

More Frobenius generatingfunctionology Recall that N ( t + ab ) = N ( t ) + 1 � � N ( t + ab ) x t ( N ( t ) + 1) x t = t ≥ 0 t ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 15

More Frobenius generatingfunctionology Recall that N ( t + ab ) = N ( t ) + 1 � � N ( t + ab ) x t ( N ( t ) + 1) x t = t ≥ 0 t ≥ 0 1 N ( t ) x t + � � � N ( t ) x t x t = x ab t ≥ ab t ≥ 0 t ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 15

More Frobenius generatingfunctionology Recall that N ( t + ab ) = N ( t ) + 1 � � N ( t + ab ) x t ( N ( t ) + 1) x t = t ≥ 0 t ≥ 0 1 N ( t ) x t + � � � N ( t ) x t x t = x ab t ≥ ab t ≥ 0 t ≥ 0   ab − 1 1 1 1 N ( t ) x t − � � N ( t ) x t = (1 − x a ) (1 − x b ) +  x ab 1 − x t =0 t ≥ 0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 15

More Frobenius generatingfunctionology Recall that N ( t + ab ) = N ( t ) + 1 � � N ( t + ab ) x t ( N ( t ) + 1) x t = t ≥ 0 t ≥ 0 1 N ( t ) x t + � � � N ( t ) x t x t = x ab t ≥ ab t ≥ 0 t ≥ 0   ab − 1 1 1 1 N ( t ) x t − � � N ( t ) x t = (1 − x a ) (1 − x b ) +  x ab 1 − x t =0 t ≥ 0 � ab − 1 � 1 1 1 1 � N ( t ) x t (1 − x a ) (1 − x b ) − = (1 − x a ) (1 − x b ) + x ab 1 − x t =0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 15

More Frobenius generatingfunctionology Recall that N ( t + ab ) = N ( t ) + 1 � � N ( t + ab ) x t ( N ( t ) + 1) x t = t ≥ 0 t ≥ 0 1 N ( t ) x t + � � � N ( t ) x t x t = x ab t ≥ ab t ≥ 0 t ≥ 0   ab − 1 1 1 1 N ( t ) x t − � � N ( t ) x t = (1 − x a ) (1 − x b ) +  x ab 1 − x t =0 t ≥ 0 � � ab − 1 1 1 1 1 � N ( t ) x t (1 − x a ) (1 − x b ) − = (1 − x a ) (1 − x b ) + x ab 1 − x t =0 ab − 1 N ( t ) x t + x ab 1 − x ab � 1 − x = (1 − x a ) (1 − x b ) t =0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 15

More Frobenius generatingfunctionology ab − 1 N ( t ) x t + x ab 1 − x ab � = 1 − x (1 − x a ) (1 − x b ) t =0 The ”Coin Exchange Problem” of Frobenius Matthias Beck 16

More Frobenius generatingfunctionology ab − 1 N ( t ) x t + x ab 1 − x ab � = 1 − x (1 − x a ) (1 − x b ) t =0 ab − 1 1 − x ab N ( t ) x t + � � x t = (1 − x a ) (1 − x b ) t =0 t ≥ ab The ”Coin Exchange Problem” of Frobenius Matthias Beck 16

More Frobenius generatingfunctionology ab − 1 N ( t ) x t + x ab 1 − x ab � = 1 − x (1 − x a ) (1 − x b ) t =0 ab − 1 1 − x ab N ( t ) x t + � � x t = (1 − x a ) (1 − x b ) t =0 t ≥ ab For 0 ≤ t ≤ ab − 1 , N ( t ) is 0 or 1 , and so 1 − x ab x t = � (1 − x a ) (1 − x b ) t representable The ”Coin Exchange Problem” of Frobenius Matthias Beck 16

More Frobenius generatingfunctionology ab − 1 N ( t ) x t + x ab 1 − x ab � = 1 − x (1 − x a ) (1 − x b ) t =0 ab − 1 1 − x ab N ( t ) x t + � � x t = (1 − x a ) (1 − x b ) t =0 t ≥ ab For 0 ≤ t ≤ ab − 1 , N ( t ) is 0 or 1 , and so 1 − x ab x t = � (1 − x a ) (1 − x b ) t representable and 1 − x ab 1 x t = � 1 − x − (1 − x a ) (1 − x b ) . t not representable The ”Coin Exchange Problem” of Frobenius Matthias Beck 16

A higher-dimensional homework We say that t is representable by the positive integers a 1 , a 2 , . . . , a d if there is a solution ( m 1 , m 2 , . . . , m d ) in nonnegative integers to m 1 a 1 + m 2 a 2 + · · · + m d a d = t The ”Coin Exchange Problem” of Frobenius Matthias Beck 17

A higher-dimensional homework We say that t is representable by the positive integers a 1 , a 2 , . . . , a d if there is a solution ( m 1 , m 2 , . . . , m d ) in nonnegative integers to m 1 a 1 + m 2 a 2 + · · · + m d a d = t Prove p ( x ) x t = � (1 − x a 1 ) (1 − x a 2 ) · · · (1 − x a d ) t representable for some polynomial p . The ”Coin Exchange Problem” of Frobenius Matthias Beck 17

Corollaries 1 − x ab 1 � x t = 1 − x − (1 − x a ) (1 − x b ) t not representable The ”Coin Exchange Problem” of Frobenius Matthias Beck 18

Corollaries 1 − x ab 1 � x t = 1 − x − (1 − x a ) (1 − x b ) t not representable 1 − x a − x b + x a + b − 1 − x − x ab + x ab +1 � � = (1 − x ) (1 − x a ) (1 − x b ) The ”Coin Exchange Problem” of Frobenius Matthias Beck 18

Corollaries 1 − x ab 1 � x t = 1 − x − (1 − x a ) (1 − x b ) t not representable 1 − x a − x b + x a + b − 1 − x − x ab + x ab +1 � � = (1 − x ) (1 − x a ) (1 − x b ) x − x a − x b + x a + b + x ab − x ab +1 = (1 − x ) (1 − x a ) (1 − x b ) The ”Coin Exchange Problem” of Frobenius Matthias Beck 18

Corollaries 1 − x ab 1 � x t = 1 − x − (1 − x a ) (1 − x b ) t not representable 1 − x a − x b + x a + b − 1 − x − x ab + x ab +1 � � = (1 − x ) (1 − x a ) (1 − x b ) x − x a − x b + x a + b + x ab − x ab +1 = (1 − x ) (1 − x a ) (1 − x b ) This polynomial has degree ab + 1 − (1 + a + b ) = ab − a − b . The ”Coin Exchange Problem” of Frobenius Matthias Beck 18

Corollaries 1 − x ab 1 � x t = 1 − x − (1 − x a ) (1 − x b ) t not representable 1 − x a − x b + x a + b − 1 − x − x ab + x ab +1 � � = (1 − x ) (1 − x a ) (1 − x b ) x − x a − x b + x a + b + x ab − x ab +1 = (1 − x ) (1 − x a ) (1 − x b ) This polynomial has degree ab + 1 − (1 + a + b ) = ab − a − b . The number of non-representable positive integers is x − x a − x b + x ab + x a + b − x ab +1 lim (1 − x ) (1 − x a ) (1 − x b ) x → 1 The ”Coin Exchange Problem” of Frobenius Matthias Beck 18

Corollaries 1 − x ab 1 � x t = 1 − x − (1 − x a ) (1 − x b ) t not representable 1 − x a − x b + x a + b − 1 − x − x ab + x ab +1 � � = (1 − x ) (1 − x a ) (1 − x b ) x − x a − x b + x a + b + x ab − x ab +1 = (1 − x ) (1 − x a ) (1 − x b ) This polynomial has degree ab + 1 − (1 + a + b ) = ab − a − b . The number of non-representable positive integers is x − x a − x b + x ab + x a + b − x ab +1 = ( a − 1)( b − 1) lim . (1 − x ) (1 − x a ) (1 − x b ) 2 x → 1 The ”Coin Exchange Problem” of Frobenius Matthias Beck 18

Another homework with several representations Recall: Given two positive integers a and b with no common factor, we say the integer t is k -representable if there are exactly k solutions ( m, n ) ∈ Z 2 ≥ 0 to t = ma + nb. The ”Coin Exchange Problem” of Frobenius Matthias Beck 19

Another homework with several representations Recall: Given two positive integers a and b with no common factor, we say the integer t is k -representable if there are exactly k solutions ( m, n ) ∈ Z 2 ≥ 0 to t = ma + nb. Prove: There are exactly ab − 1 integers that are uniquely representable. ◮ The ”Coin Exchange Problem” of Frobenius Matthias Beck 19

Another homework with several representations Recall: Given two positive integers a and b with no common factor, we say the integer t is k -representable if there are exactly k solutions ( m, n ) ∈ Z 2 ≥ 0 to t = ma + nb. Prove: There are exactly ab − 1 integers that are uniquely representable. ◮ Given k ≥ 2 , there are exactly ab k -representable integers. ◮ The ”Coin Exchange Problem” of Frobenius Matthias Beck 19

Beyond d=2 Given integers a 1 , a 2 , . . . , a d with no common factor, let p ( x ) x t = � F ( x ) = (1 − x a 1 ) (1 − x a 2 ) · · · (1 − x a d ) . t representable The ”Coin Exchange Problem” of Frobenius Matthias Beck 20

Beyond d=2 Given integers a 1 , a 2 , . . . , a d with no common factor, let p ( x ) x t = � F ( x ) = (1 − x a 1 ) (1 − x a 2 ) · · · (1 − x a d ) . t representable (Denham 2003) For d = 3 , the polynomial p ( x ) has either 4 or 6 terms, ◮ given in semi-explicit form. The ”Coin Exchange Problem” of Frobenius Matthias Beck 20

Beyond d=2 Given integers a 1 , a 2 , . . . , a d with no common factor, let p ( x ) x t = � F ( x ) = (1 − x a 1 ) (1 − x a 2 ) · · · (1 − x a d ) . t representable (Denham 2003) For d = 3 , the polynomial p ( x ) has either 4 or 6 terms, ◮ given in semi-explicit form. (Bresinsky 1975) For d ≥ 4 , there is no absolute bound for the number ◮ of terms in p ( x ) . The ”Coin Exchange Problem” of Frobenius Matthias Beck 20

Beyond d=2 Given integers a 1 , a 2 , . . . , a d with no common factor, let p ( x ) x t = � F ( x ) = (1 − x a 1 ) (1 − x a 2 ) · · · (1 − x a d ) . t representable (Denham 2003) For d = 3 , the polynomial p ( x ) has either 4 or 6 terms, ◮ given in semi-explicit form. (Bresinsky 1975) For d ≥ 4 , there is no absolute bound for the number ◮ of terms in p ( x ) . (Barvinok-Woods 2003) For fixed d , the rational generating function ◮ F ( x ) can be written as a “short” sum of rational functions. The ”Coin Exchange Problem” of Frobenius Matthias Beck 20

What else is known Frobenius number and number of non-representable integers in special ◮ cases: arithmetic progressions and variations, extension cases The ”Coin Exchange Problem” of Frobenius Matthias Beck 21

What else is known Frobenius number and number of non-representable integers in special ◮ cases: arithmetic progressions and variations, extension cases Upper and lower bounds for the Frobenius number ◮ The ”Coin Exchange Problem” of Frobenius Matthias Beck 21