An Introduction to the Narayana Numbers

Murray Bremner

Department of Mathematics and StatisticsCollege of Arts and ScienceUniversity of Saskatchewan1

Saskatoon Saskatchewan Canada

Combinatorics SeminarTuesday 27 March 2018

1ldquoAt the heart of Saskatoon on Treaty 6 territory and the homeland of the Metis is theUniversity of Saskatchewan one of Canadarsquos top research universities Driven by prairie spiritwe make life better for the people of Saskatchewan and the worldrdquo (wwwusaskca)

1 28

Abstract

The Narayana numbers N(n k) for 1 le k le n form a triangular array ofpositive integers introduced in 1915-1916 by the combinatorialist P AMacMahon and rediscovered in 1955 by the statistician T V NarayanaThe sum of the entries in the n-th row of the triangle is the n-th Catalannumber which explains why Narayana numbers appear unexpectedly inmany combinatorial problems For example N(n k) is the number ofcomplete binary trees with n internal nodes and kminus1 jumps (transitionsfrom a node to a strictly higher node in the depth-first preorder traversal)My interest in these numbers began with my study of Rotarsquos Problem onthe classification of polynomial identities satisfied by linear operators onassociative algebras (such as the derivation law which is the product rulein differential calculus) An operator identity is a linear combination ofoperator monomials the number T (m n) of monomials with m operatorsand n variables turns out to be N(m+n n) This talk will provide anintroduction to the Narayana numbers and their numerous applications

2 28

Binomial formula for the Narayana numbers

The Narayana numbers N(n k) are a triangular array of positive integers

N(n k) =1

n

(n

k

)(n

kminus1

)(1 le k le n)

These numbers (OEIS A001263) were first studied more than a centuryago by the famous early combinatorialist2 Percy Alexander MacMahon(1854ndash1929) and were then rediscovered in 1955 by the statistician andcombinatorialist Tadepalli Venkata Narayana (1930ndash1987)bull P A MacMahon Combinatory Analysis Volumes I and II

Cambridge University Press 1915 and 1916 (especially Article 495)Reprinted by Chelsea (1960) Dover (2004) Online archiveorg

bull T V Narayana Sur les treillis formes par les partitions drsquoun entieret leurs applications a la theorie des probabilites C R Acad Sci Paris240 (1955) 1188ndash1189

2combinatoricist combinatorist 3 28

Binomial formula for the Narayana numbers

The Narayana numbers N(n k) are a triangular array of positive integers

N(n k) =1

n

(n

k

)(n

kminus1

)(1 le k le n)

These numbers (OEIS A001263) were first studied more than a centuryago by the famous early combinatorialist2 Percy Alexander MacMahon(1854ndash1929) and were then rediscovered in 1955 by the statistician andcombinatorialist Tadepalli Venkata Narayana (1930ndash1987)bull P A MacMahon Combinatory Analysis Volumes I and II

Cambridge University Press 1915 and 1916 (especially Article 495)Reprinted by Chelsea (1960) Dover (2004) Online archiveorg

bull T V Narayana Sur les treillis formes par les partitions drsquoun entieret leurs applications a la theorie des probabilites C R Acad Sci Paris240 (1955) 1188ndash1189

2combinatoricist combinatorist 3 28

Part of a page from MacMahonrsquos Combinatory Analysis

Fortunately notation has improved during the last 100 years4 28

Narayanarsquos 1955 paper from Comptes Rendus

In fact notation improved dramatically in only 40 years

5 28

The Narayana triangle (1 le k le n)

Therersquos an obvious symmetry in each row N(n k) = N(n nminusk+1)

kN(n k) 1 2 3 4 5 6 7 8 9 10 11 12

n 1 1

2 1 1

3 1 3 1

4 1 6 6 1

5 1 10 20 10 1

6 1 15 50 50 15 1

7 1 21 105 175 105 21 1

8 1 28 196 490 490 196 28 1

9 1 36 336 1176 1764 1176 336 36 1

10 1 45 540 2520 5292 5292 2520 540 45 1

11 1 55 825 4950 13860 19404 13860 4950 825 55 1

12 1 66 1210 9075 32670 60984 60984 32670 9075 1210 66 1

6 28

Combinatorial interpretations of the Narayana numbers (1)

Lemma

The sum of the n-th row of the triangle is the n-th Catalan number

nsumk=1

N(n k) =nsum

k=1

1

n

(n

k

)(n

kminus1

)=

1

n+1

(2n

n

)= C (n)

A few applications from enwikipediaorgwikiNarayana_number

bull N(n k) is the number of strings containing n pairs of parentheses whichare correctly matched and contain k distinct nestings where a nesting is asubstring () For example N(4 k) gives 1 6 6 1 (four pairs k nestings)

k = 1 (((())))

k = 2 ()((())) (())(()) (()(())) ((()())) ((())()) ((()))()

k = 3 ()()(()) ()(())() (())()() (()())() ()(()()) (()()())

k = 4 ()()()()

7 28

Narayanarsquos 1955 paper from Comptes Rendus

In fact notation improved dramatically in only 40 years

5 28

The Narayana triangle (1 le k le n)

Therersquos an obvious symmetry in each row N(n k) = N(n nminusk+1)

kN(n k) 1 2 3 4 5 6 7 8 9 10 11 12

n 1 1

2 1 1

3 1 3 1

4 1 6 6 1

5 1 10 20 10 1

6 1 15 50 50 15 1

7 1 21 105 175 105 21 1

8 1 28 196 490 490 196 28 1

9 1 36 336 1176 1764 1176 336 36 1

10 1 45 540 2520 5292 5292 2520 540 45 1

11 1 55 825 4950 13860 19404 13860 4950 825 55 1

12 1 66 1210 9075 32670 60984 60984 32670 9075 1210 66 1

6 28

Combinatorial interpretations of the Narayana numbers (1)

Lemma

The sum of the n-th row of the triangle is the n-th Catalan number

nsumk=1

N(n k) =nsum

k=1

1

n

(n

k

)(n

kminus1

)=

1

n+1

(2n

n

)= C (n)

A few applications from enwikipediaorgwikiNarayana_number

bull N(n k) is the number of strings containing n pairs of parentheses whichare correctly matched and contain k distinct nestings where a nesting is asubstring () For example N(4 k) gives 1 6 6 1 (four pairs k nestings)

k = 1 (((())))

k = 2 ()((())) (())(()) (()(())) ((()())) ((())()) ((()))()

k = 3 ()()(()) ()(())() (())()() (()())() ()(()()) (()()())

k = 4 ()()()()

7 28

The Narayana triangle (1 le k le n)

Therersquos an obvious symmetry in each row N(n k) = N(n nminusk+1)

kN(n k) 1 2 3 4 5 6 7 8 9 10 11 12

n 1 1

2 1 1

3 1 3 1

4 1 6 6 1

5 1 10 20 10 1

6 1 15 50 50 15 1

7 1 21 105 175 105 21 1

8 1 28 196 490 490 196 28 1

9 1 36 336 1176 1764 1176 336 36 1

10 1 45 540 2520 5292 5292 2520 540 45 1

11 1 55 825 4950 13860 19404 13860 4950 825 55 1

12 1 66 1210 9075 32670 60984 60984 32670 9075 1210 66 1

6 28

Combinatorial interpretations of the Narayana numbers (1)

Lemma

The sum of the n-th row of the triangle is the n-th Catalan number

nsumk=1

N(n k) =nsum

k=1

1

n

(n

k

)(n

kminus1

)=

1

n+1

(2n

n

)= C (n)

A few applications from enwikipediaorgwikiNarayana_number

bull N(n k) is the number of strings containing n pairs of parentheses whichare correctly matched and contain k distinct nestings where a nesting is asubstring () For example N(4 k) gives 1 6 6 1 (four pairs k nestings)

k = 1 (((())))

k = 2 ()((())) (())(()) (()(())) ((()())) ((())()) ((()))()

k = 3 ()()(()) ()(())() (())()() (()())() ()(()()) (()()())

k = 4 ()()()()

7 28

Combinatorial interpretations of the Narayana numbers (1)

Lemma

The sum of the n-th row of the triangle is the n-th Catalan number

nsumk=1

N(n k) =nsum

k=1

1

n

(n

k

)(n

kminus1

)=

1

n+1

(2n

n

)= C (n)

A few applications from enwikipediaorgwikiNarayana_number

bull N(n k) is the number of strings containing n pairs of parentheses whichare correctly matched and contain k distinct nestings where a nesting is asubstring () For example N(4 k) gives 1 6 6 1 (four pairs k nestings)

k = 1 (((())))

k = 2 ()((())) (())(()) (()(())) ((()())) ((())()) ((()))()

k = 3 ()()(()) ()(())() (())()() (()())() ()(()()) (()()())

k = 4 ()()()()

7 28

Combinatorial interpretations of the Narayana numbers (2)

bull N(n k) is the number of lattice paths from (0 0) to (2n 0) with k peaksusing steps northeast (1 1) and southeast (1minus1) with y ge 0 For n = 4

enwikipediaorgwikiNarayana_number

8 28

Combinatorial interpretations of the Narayana numbers (3)

bull N(n k) is the number of non-crossing partitions with k blocks of a setwith n elements For n = 4 we get once more the sequence 1 6 6 1

enwikipediaorgwikiNarayana_number

9 28

Lagrange Inversion Formula

This is sufficiently important that I have put it on a slide by itself

Theorem

Assume that y = x φ(y) where φ(y) is a formal power series in x withnonzero constant term φ(0) 6= 0 Let [xn]y denote the coefficient of xn

in the power series y Then for n ge 1 we have

[xn]y =1

n[ynminus1]

(φ(y)

)nbull M Bona (editor)

Handbook of Enumerative CombinatoricsCRC Press Boca Raton3 2015See especially sect26 (by H Prodinger) for the statement and proof

3ldquoRatrsquos Mouthrdquo Why the early Spanish explorers thought it looked like that I have no idea

10 28

Enumeration of complete t-ary trees

Definition

For t ge 2 a complete t-ary tree is constructed recursively from a rootwith t subtrees each of which is either a leaf or a complete t-ary tree

Lemma

The generating function G (x) for the number of t-ary trees with n internal(nonleaf) nodes (including the root) satisfies the equation G = 1 + xG t

If we set G = 1 + y then 1 + y = 1 + x(1 + y)t hence y = x(1 + y)t which lets us apply Lagrange inversion with φ(y) = (1 + y)t Hence

[xn]G (x) = [xn]y =1

n[ynminus1](1 + y)tn =

1

n

(tn

n minus 1

)=

1

nmiddot (tn)

(nminus1)((tminus1)n+1

)

=(tn)

n((tminus1)n+1

)

=1

(tminus1)n+1

(tn

n

)

which is the familiar formula for the t-ary Catalan numbers

11 28

Enumeration of general planar trees (1)

Consider the set of all planar trees with n total vertices and k leaves

Let G (z u) be the bivariate generating function

exponent n of z counts total vertices exponent k of u counts leaves

For convenience let y = G (z u) regarded as a power series in z

The recursive construction of planar trees implies a functional equation

G (z u) = zu +zG (z u)

1minusG (z u) y = z

(u+

y

1minusy

)= zφ(y) φ(y) = u+

y

1minusy

Applying Lagrange inversion we obtain

[uk ][zn]G (z u) = [uk ]1

n[ynminus1]

(u +

y

1minusy

)n

=1

n[ynminus1][uk ]

(u +

y

1minusy

)n

=1

n[ynminus1]

(n

k

)ynminusk

(1minusy)nminusk=

1

n

(n

k

)[ynminus1]

ynminusk

(1minusy)nminusk=

1

n

(n

k

)[ykminus1]

1

(1minusy)nminusk

12 28

Enumeration of general planar trees (2)

Explanation of the last step (thanks to Chris Duffy) We need to show that

[ynminus1]ynminusk

(1minusy)nminusk= [ykminus1]

1

(1minusy)nminusk

Consider these two functions

r(y) =ynminusk

(1minusy)nminusk s(y) =

1

(1minusy)nminusk=rArr r(y)

ynminusk= s(y)

So the coefficient of y t in r(y) equals the coefficient of y tminus(nminusk) in s(y)Setting t = n minus 1 we obtain

[ynminus1]r(y) = [y (nminus1)minus(nminusk)]s(y) = [ykminus1]s(y)

Expanding s(y) as a formal power series gives

[ykminus1]s(y) = [ykminus1]1

(1minusy)nminusk=

((nminuskminus1) + (kminus1)

kminus1

)=

(nminus2

kminus1

)13 28

Enumeration of general planar trees (3)

Putting this all together almost gives the Narayana numbers

[uk ][zn]G (z u) =1

n

(n

k

)(nminus2

kminus1

)Replace n by n+1 in the right side and do some trivial rearranging

1

n+1

(n+1

k

)(nminus1

kminus1

)=

1

n+1middot (n+1)

k(nminusk+1)middot (nminus1)

(kminus1)(nminusk)

=1

nmiddot n

k(nminusk+1)middot n

(kminus1)(nminusk)=

1

nmiddot n

k(nminusk)middot n

(kminus1)(nminusk+1)

=1

n

(n

k

)(n

kminus1

)= N(n k) = the Narayana triangle

This is the number of planar trees with n+1 total vertices and k leaves

14 28

Combinatorial interpretations of the Narayana numbers (4)

The next few pages contain exercises from

bull R P Stanley Enumerative Combinatorics Volume 2With a Foreword by G-C Rota and Appendix 1 by S FominCambridge Studies in Advanced Mathematics 62Cambridge University Press 1999

bull Exercise 636 (page 237)

Let Xn be the set of all sequences x = x1x2 middot middot middot x2n in which n terms are 1and the other n terms are minus1

Assume x1 + middot middot middot+ xi ge 0 for 1 le i le 2n partial sums are nonnegative

Exercise 619(r) (page 222) shows that |Xn| = C (n) the Catalan number

Let Xnk sube Xn contain the sequences with k steps from 1 to minus1 that isk is the number of indices 1 le i le 2nminus1 such that xi = 1 xi+1 = minus1

Then |Xnk | = N(n k) the Narayana number

This is one way to prove thatsumn

k=1 N(n k) = C (n)

15 28

Combinatorial interpretations of the Narayana numbers (5)

Consider the bivariate generating function for the Narayana numbers

F (x t) =infinsumn=1

infinsumk=1

N(n k)xntk

The combinatorial interpretation of N(n k) on the previous slide impliesthat F (x t) satisfies the following quadratic functional equation

xF 2 + (xt + x minus 1)F + xt = 0

This immediately implies a formula for F (x t) as an algebraic function

F (x t) =1minus (t+1)x minus

radic1minus 2(t+1)x + (tminus1)2x2

2x

Exercise (by me) Use Newtonrsquos Binomial Theorem to expand thisalgebraic function as a power series in x and t and then derive thebinomial formula for Narayana numbers (Maybe not a good idea)

16 28

Combinatorial interpretations of the Narayana numbers (6)

Exercise 619 (p 219) gives 66 different combinatorial interpretations ofthe Catalan numbers C (n) Stanley (p 238) ldquoIt is interesting to find foreach of the combinatorial interpretations of C (n) given by Exercise 619 acorresponding decomposition into subsets counted by Narayana numbersrdquoStanley also gives some additional basic references on Narayana numbers

bull T V Narayana A partial order and its applications to probability theorySankhya 21 (1959) 91ndash98

bull G Kreweras P Moszkowski A new enumerative property of theNarayana numbers J Statist Plann Inference 14 (1986) no 1 63ndash67

bull G Kreweras Y Poupard Subdivision des nombres de Narayana suivantdeux parametres supplementaires Euro J Combin 7 2 (1986) 141ndash149

bull R A Sulanke Refinements of the Narayana numbers Bull InstCombin Appl 7 (1993) 60ndash66

bull R A Sulanke A symmetric variation of a distribution of Kreweras andPoupard J Statist Plann Inference 34 (1993) no 2 291ndash303

17 28

Why am I interested in these numbers (Introduction)

Gian-Carlo Rota posed the following problem

Let A be an associative algebra and let f Ararr A be a linear operatorClassify all the polynomial identities that can be satisfied by f

Posing the same problem in another way (to clarify the quantifiers)

Classify all the polynomial identities that can be satisfied by a linearoperator on an associative algebra

The term ldquoclassifyrdquo is deliberately ambiguous mdash different interpretationscould lead to different but equally valid solutions

Examples

If f is an endomorphism of the algebra structure on A then

f (xy) = f (x)f (y) (not homogeneous in f )

If f is a derivation of the algebra structure on A then

f (xy) = f (x)y + xf (y) (homogeneous of degree 1 in f )

18 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Enumeration of complete t-ary trees

Definition

For t ge 2 a complete t-ary tree is constructed recursively from a rootwith t subtrees each of which is either a leaf or a complete t-ary tree

Lemma

The generating function G (x) for the number of t-ary trees with n internal(nonleaf) nodes (including the root) satisfies the equation G = 1 + xG t

If we set G = 1 + y then 1 + y = 1 + x(1 + y)t hence y = x(1 + y)t which lets us apply Lagrange inversion with φ(y) = (1 + y)t Hence

[xn]G (x) = [xn]y =1

n[ynminus1](1 + y)tn =

1

n

(tn

n minus 1

)=

1

nmiddot (tn)

(nminus1)((tminus1)n+1

)

=(tn)

n((tminus1)n+1

)

=1

(tminus1)n+1

(tn

n

)

which is the familiar formula for the t-ary Catalan numbers

11 28

Enumeration of general planar trees (1)

Consider the set of all planar trees with n total vertices and k leaves

Let G (z u) be the bivariate generating function

exponent n of z counts total vertices exponent k of u counts leaves

For convenience let y = G (z u) regarded as a power series in z

The recursive construction of planar trees implies a functional equation

G (z u) = zu +zG (z u)

1minusG (z u) y = z

(u+

y

1minusy

)= zφ(y) φ(y) = u+

y

1minusy

Applying Lagrange inversion we obtain

[uk ][zn]G (z u) = [uk ]1

n[ynminus1]

(u +

y

1minusy

)n

=1

n[ynminus1][uk ]

(u +

y

1minusy

)n

=1

n[ynminus1]

(n

k

)ynminusk

(1minusy)nminusk=

1

n

(n

k

)[ynminus1]

ynminusk

(1minusy)nminusk=

1

n

(n

k

)[ykminus1]

1

(1minusy)nminusk

12 28

Enumeration of general planar trees (2)

Explanation of the last step (thanks to Chris Duffy) We need to show that

[ynminus1]ynminusk

(1minusy)nminusk= [ykminus1]

1

(1minusy)nminusk

Consider these two functions

r(y) =ynminusk

(1minusy)nminusk s(y) =

1

(1minusy)nminusk=rArr r(y)

ynminusk= s(y)

So the coefficient of y t in r(y) equals the coefficient of y tminus(nminusk) in s(y)Setting t = n minus 1 we obtain

[ynminus1]r(y) = [y (nminus1)minus(nminusk)]s(y) = [ykminus1]s(y)

Expanding s(y) as a formal power series gives

[ykminus1]s(y) = [ykminus1]1

(1minusy)nminusk=

((nminuskminus1) + (kminus1)

kminus1

)=

(nminus2

kminus1

)13 28

Enumeration of general planar trees (3)

Putting this all together almost gives the Narayana numbers

[uk ][zn]G (z u) =1

n

(n

k

)(nminus2

kminus1

)Replace n by n+1 in the right side and do some trivial rearranging

1

n+1

(n+1

k

)(nminus1

kminus1

)=

1

n+1middot (n+1)

k(nminusk+1)middot (nminus1)

(kminus1)(nminusk)

=1

nmiddot n

k(nminusk+1)middot n

(kminus1)(nminusk)=

1

nmiddot n

k(nminusk)middot n

(kminus1)(nminusk+1)

=1

n

(n

k

)(n

kminus1

)= N(n k) = the Narayana triangle

This is the number of planar trees with n+1 total vertices and k leaves

14 28

Combinatorial interpretations of the Narayana numbers (4)

The next few pages contain exercises from

bull R P Stanley Enumerative Combinatorics Volume 2With a Foreword by G-C Rota and Appendix 1 by S FominCambridge Studies in Advanced Mathematics 62Cambridge University Press 1999

bull Exercise 636 (page 237)

Let Xn be the set of all sequences x = x1x2 middot middot middot x2n in which n terms are 1and the other n terms are minus1

Assume x1 + middot middot middot+ xi ge 0 for 1 le i le 2n partial sums are nonnegative

Exercise 619(r) (page 222) shows that |Xn| = C (n) the Catalan number

Let Xnk sube Xn contain the sequences with k steps from 1 to minus1 that isk is the number of indices 1 le i le 2nminus1 such that xi = 1 xi+1 = minus1

Then |Xnk | = N(n k) the Narayana number

This is one way to prove thatsumn

k=1 N(n k) = C (n)

15 28

Combinatorial interpretations of the Narayana numbers (5)

Consider the bivariate generating function for the Narayana numbers

F (x t) =infinsumn=1

infinsumk=1

N(n k)xntk

The combinatorial interpretation of N(n k) on the previous slide impliesthat F (x t) satisfies the following quadratic functional equation

xF 2 + (xt + x minus 1)F + xt = 0

This immediately implies a formula for F (x t) as an algebraic function

F (x t) =1minus (t+1)x minus

radic1minus 2(t+1)x + (tminus1)2x2

2x

Exercise (by me) Use Newtonrsquos Binomial Theorem to expand thisalgebraic function as a power series in x and t and then derive thebinomial formula for Narayana numbers (Maybe not a good idea)

16 28

Combinatorial interpretations of the Narayana numbers (6)

Exercise 619 (p 219) gives 66 different combinatorial interpretations ofthe Catalan numbers C (n) Stanley (p 238) ldquoIt is interesting to find foreach of the combinatorial interpretations of C (n) given by Exercise 619 acorresponding decomposition into subsets counted by Narayana numbersrdquoStanley also gives some additional basic references on Narayana numbers

bull T V Narayana A partial order and its applications to probability theorySankhya 21 (1959) 91ndash98

bull G Kreweras P Moszkowski A new enumerative property of theNarayana numbers J Statist Plann Inference 14 (1986) no 1 63ndash67

bull G Kreweras Y Poupard Subdivision des nombres de Narayana suivantdeux parametres supplementaires Euro J Combin 7 2 (1986) 141ndash149

bull R A Sulanke Refinements of the Narayana numbers Bull InstCombin Appl 7 (1993) 60ndash66

bull R A Sulanke A symmetric variation of a distribution of Kreweras andPoupard J Statist Plann Inference 34 (1993) no 2 291ndash303

17 28

Why am I interested in these numbers (Introduction)

Gian-Carlo Rota posed the following problem

Let A be an associative algebra and let f Ararr A be a linear operatorClassify all the polynomial identities that can be satisfied by f

Posing the same problem in another way (to clarify the quantifiers)

Classify all the polynomial identities that can be satisfied by a linearoperator on an associative algebra

The term ldquoclassifyrdquo is deliberately ambiguous mdash different interpretationscould lead to different but equally valid solutions

Examples

If f is an endomorphism of the algebra structure on A then

f (xy) = f (x)f (y) (not homogeneous in f )

If f is a derivation of the algebra structure on A then

f (xy) = f (x)y + xf (y) (homogeneous of degree 1 in f )

18 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Enumeration of general planar trees (1)

Consider the set of all planar trees with n total vertices and k leaves

Let G (z u) be the bivariate generating function

exponent n of z counts total vertices exponent k of u counts leaves

For convenience let y = G (z u) regarded as a power series in z

The recursive construction of planar trees implies a functional equation

G (z u) = zu +zG (z u)

1minusG (z u) y = z

(u+

y

1minusy

)= zφ(y) φ(y) = u+

y

1minusy

Applying Lagrange inversion we obtain

[uk ][zn]G (z u) = [uk ]1

n[ynminus1]

(u +

y

1minusy

)n

=1

n[ynminus1][uk ]

(u +

y

1minusy

)n

=1

n[ynminus1]

(n

k

)ynminusk

(1minusy)nminusk=

1

n

(n

k

)[ynminus1]

ynminusk

(1minusy)nminusk=

1

n

(n

k

)[ykminus1]

1

(1minusy)nminusk

12 28

Enumeration of general planar trees (2)

Explanation of the last step (thanks to Chris Duffy) We need to show that

[ynminus1]ynminusk

(1minusy)nminusk= [ykminus1]

1

(1minusy)nminusk

Consider these two functions

r(y) =ynminusk

(1minusy)nminusk s(y) =

1

(1minusy)nminusk=rArr r(y)

ynminusk= s(y)

So the coefficient of y t in r(y) equals the coefficient of y tminus(nminusk) in s(y)Setting t = n minus 1 we obtain

[ynminus1]r(y) = [y (nminus1)minus(nminusk)]s(y) = [ykminus1]s(y)

Expanding s(y) as a formal power series gives

[ykminus1]s(y) = [ykminus1]1

(1minusy)nminusk=

((nminuskminus1) + (kminus1)

kminus1

)=

(nminus2

kminus1

)13 28

Enumeration of general planar trees (3)

Putting this all together almost gives the Narayana numbers

[uk ][zn]G (z u) =1

n

(n

k

)(nminus2

kminus1

)Replace n by n+1 in the right side and do some trivial rearranging

1

n+1

(n+1

k

)(nminus1

kminus1

)=

1

n+1middot (n+1)

k(nminusk+1)middot (nminus1)

(kminus1)(nminusk)

=1

nmiddot n

k(nminusk+1)middot n

(kminus1)(nminusk)=

1

nmiddot n

k(nminusk)middot n

(kminus1)(nminusk+1)

=1

n

(n

k

)(n

kminus1

)= N(n k) = the Narayana triangle

This is the number of planar trees with n+1 total vertices and k leaves

14 28

Combinatorial interpretations of the Narayana numbers (4)

The next few pages contain exercises from

bull R P Stanley Enumerative Combinatorics Volume 2With a Foreword by G-C Rota and Appendix 1 by S FominCambridge Studies in Advanced Mathematics 62Cambridge University Press 1999

bull Exercise 636 (page 237)

Let Xn be the set of all sequences x = x1x2 middot middot middot x2n in which n terms are 1and the other n terms are minus1

Assume x1 + middot middot middot+ xi ge 0 for 1 le i le 2n partial sums are nonnegative

Exercise 619(r) (page 222) shows that |Xn| = C (n) the Catalan number

Let Xnk sube Xn contain the sequences with k steps from 1 to minus1 that isk is the number of indices 1 le i le 2nminus1 such that xi = 1 xi+1 = minus1

Then |Xnk | = N(n k) the Narayana number

This is one way to prove thatsumn

k=1 N(n k) = C (n)

15 28

Combinatorial interpretations of the Narayana numbers (5)

Consider the bivariate generating function for the Narayana numbers

F (x t) =infinsumn=1

infinsumk=1

N(n k)xntk

The combinatorial interpretation of N(n k) on the previous slide impliesthat F (x t) satisfies the following quadratic functional equation

xF 2 + (xt + x minus 1)F + xt = 0

This immediately implies a formula for F (x t) as an algebraic function

F (x t) =1minus (t+1)x minus

radic1minus 2(t+1)x + (tminus1)2x2

2x

Exercise (by me) Use Newtonrsquos Binomial Theorem to expand thisalgebraic function as a power series in x and t and then derive thebinomial formula for Narayana numbers (Maybe not a good idea)

16 28

Combinatorial interpretations of the Narayana numbers (6)

Exercise 619 (p 219) gives 66 different combinatorial interpretations ofthe Catalan numbers C (n) Stanley (p 238) ldquoIt is interesting to find foreach of the combinatorial interpretations of C (n) given by Exercise 619 acorresponding decomposition into subsets counted by Narayana numbersrdquoStanley also gives some additional basic references on Narayana numbers

bull T V Narayana A partial order and its applications to probability theorySankhya 21 (1959) 91ndash98

bull G Kreweras P Moszkowski A new enumerative property of theNarayana numbers J Statist Plann Inference 14 (1986) no 1 63ndash67

bull G Kreweras Y Poupard Subdivision des nombres de Narayana suivantdeux parametres supplementaires Euro J Combin 7 2 (1986) 141ndash149

bull R A Sulanke Refinements of the Narayana numbers Bull InstCombin Appl 7 (1993) 60ndash66

bull R A Sulanke A symmetric variation of a distribution of Kreweras andPoupard J Statist Plann Inference 34 (1993) no 2 291ndash303

17 28

Why am I interested in these numbers (Introduction)

Gian-Carlo Rota posed the following problem

Let A be an associative algebra and let f Ararr A be a linear operatorClassify all the polynomial identities that can be satisfied by f

Posing the same problem in another way (to clarify the quantifiers)

Classify all the polynomial identities that can be satisfied by a linearoperator on an associative algebra

The term ldquoclassifyrdquo is deliberately ambiguous mdash different interpretationscould lead to different but equally valid solutions

Examples

If f is an endomorphism of the algebra structure on A then

f (xy) = f (x)f (y) (not homogeneous in f )

If f is a derivation of the algebra structure on A then

f (xy) = f (x)y + xf (y) (homogeneous of degree 1 in f )

18 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Enumeration of general planar trees (2)

Explanation of the last step (thanks to Chris Duffy) We need to show that

[ynminus1]ynminusk

(1minusy)nminusk= [ykminus1]

1

(1minusy)nminusk

Consider these two functions

r(y) =ynminusk

(1minusy)nminusk s(y) =

1

(1minusy)nminusk=rArr r(y)

ynminusk= s(y)

So the coefficient of y t in r(y) equals the coefficient of y tminus(nminusk) in s(y)Setting t = n minus 1 we obtain

[ynminus1]r(y) = [y (nminus1)minus(nminusk)]s(y) = [ykminus1]s(y)

Expanding s(y) as a formal power series gives

[ykminus1]s(y) = [ykminus1]1

(1minusy)nminusk=

((nminuskminus1) + (kminus1)

kminus1

)=

(nminus2

kminus1

)13 28

Enumeration of general planar trees (3)

Putting this all together almost gives the Narayana numbers

[uk ][zn]G (z u) =1

n

(n

k

)(nminus2

kminus1

)Replace n by n+1 in the right side and do some trivial rearranging

1

n+1

(n+1

k

)(nminus1

kminus1

)=

1

n+1middot (n+1)

k(nminusk+1)middot (nminus1)

(kminus1)(nminusk)

=1

nmiddot n

k(nminusk+1)middot n

(kminus1)(nminusk)=

1

nmiddot n

k(nminusk)middot n

(kminus1)(nminusk+1)

=1

n

(n

k

)(n

kminus1

)= N(n k) = the Narayana triangle

This is the number of planar trees with n+1 total vertices and k leaves

14 28

Combinatorial interpretations of the Narayana numbers (4)

The next few pages contain exercises from

bull R P Stanley Enumerative Combinatorics Volume 2With a Foreword by G-C Rota and Appendix 1 by S FominCambridge Studies in Advanced Mathematics 62Cambridge University Press 1999

bull Exercise 636 (page 237)

Let Xn be the set of all sequences x = x1x2 middot middot middot x2n in which n terms are 1and the other n terms are minus1

Assume x1 + middot middot middot+ xi ge 0 for 1 le i le 2n partial sums are nonnegative

Exercise 619(r) (page 222) shows that |Xn| = C (n) the Catalan number

Let Xnk sube Xn contain the sequences with k steps from 1 to minus1 that isk is the number of indices 1 le i le 2nminus1 such that xi = 1 xi+1 = minus1

Then |Xnk | = N(n k) the Narayana number

This is one way to prove thatsumn

k=1 N(n k) = C (n)

15 28

Combinatorial interpretations of the Narayana numbers (5)

Consider the bivariate generating function for the Narayana numbers

F (x t) =infinsumn=1

infinsumk=1

N(n k)xntk

The combinatorial interpretation of N(n k) on the previous slide impliesthat F (x t) satisfies the following quadratic functional equation

xF 2 + (xt + x minus 1)F + xt = 0

This immediately implies a formula for F (x t) as an algebraic function

F (x t) =1minus (t+1)x minus

radic1minus 2(t+1)x + (tminus1)2x2

2x

Exercise (by me) Use Newtonrsquos Binomial Theorem to expand thisalgebraic function as a power series in x and t and then derive thebinomial formula for Narayana numbers (Maybe not a good idea)

16 28

Combinatorial interpretations of the Narayana numbers (6)

Exercise 619 (p 219) gives 66 different combinatorial interpretations ofthe Catalan numbers C (n) Stanley (p 238) ldquoIt is interesting to find foreach of the combinatorial interpretations of C (n) given by Exercise 619 acorresponding decomposition into subsets counted by Narayana numbersrdquoStanley also gives some additional basic references on Narayana numbers

bull T V Narayana A partial order and its applications to probability theorySankhya 21 (1959) 91ndash98

bull G Kreweras P Moszkowski A new enumerative property of theNarayana numbers J Statist Plann Inference 14 (1986) no 1 63ndash67

bull G Kreweras Y Poupard Subdivision des nombres de Narayana suivantdeux parametres supplementaires Euro J Combin 7 2 (1986) 141ndash149

bull R A Sulanke Refinements of the Narayana numbers Bull InstCombin Appl 7 (1993) 60ndash66

bull R A Sulanke A symmetric variation of a distribution of Kreweras andPoupard J Statist Plann Inference 34 (1993) no 2 291ndash303

17 28

Why am I interested in these numbers (Introduction)

Gian-Carlo Rota posed the following problem

Let A be an associative algebra and let f Ararr A be a linear operatorClassify all the polynomial identities that can be satisfied by f

Posing the same problem in another way (to clarify the quantifiers)

Classify all the polynomial identities that can be satisfied by a linearoperator on an associative algebra

The term ldquoclassifyrdquo is deliberately ambiguous mdash different interpretationscould lead to different but equally valid solutions

Examples

If f is an endomorphism of the algebra structure on A then

f (xy) = f (x)f (y) (not homogeneous in f )

If f is a derivation of the algebra structure on A then

f (xy) = f (x)y + xf (y) (homogeneous of degree 1 in f )

18 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Enumeration of general planar trees (3)

Putting this all together almost gives the Narayana numbers

[uk ][zn]G (z u) =1

n

(n

k

)(nminus2

kminus1

)Replace n by n+1 in the right side and do some trivial rearranging

1

n+1

(n+1

k

)(nminus1

kminus1

)=

1

n+1middot (n+1)

k(nminusk+1)middot (nminus1)

(kminus1)(nminusk)

=1

nmiddot n

k(nminusk+1)middot n

(kminus1)(nminusk)=

1

nmiddot n

k(nminusk)middot n

(kminus1)(nminusk+1)

=1

n

(n

k

)(n

kminus1

)= N(n k) = the Narayana triangle

This is the number of planar trees with n+1 total vertices and k leaves

14 28

Combinatorial interpretations of the Narayana numbers (4)

The next few pages contain exercises from

bull R P Stanley Enumerative Combinatorics Volume 2With a Foreword by G-C Rota and Appendix 1 by S FominCambridge Studies in Advanced Mathematics 62Cambridge University Press 1999

bull Exercise 636 (page 237)

Let Xn be the set of all sequences x = x1x2 middot middot middot x2n in which n terms are 1and the other n terms are minus1

Assume x1 + middot middot middot+ xi ge 0 for 1 le i le 2n partial sums are nonnegative

Exercise 619(r) (page 222) shows that |Xn| = C (n) the Catalan number

Let Xnk sube Xn contain the sequences with k steps from 1 to minus1 that isk is the number of indices 1 le i le 2nminus1 such that xi = 1 xi+1 = minus1

Then |Xnk | = N(n k) the Narayana number

This is one way to prove thatsumn

k=1 N(n k) = C (n)

15 28

Combinatorial interpretations of the Narayana numbers (5)

Consider the bivariate generating function for the Narayana numbers

F (x t) =infinsumn=1

infinsumk=1

N(n k)xntk

The combinatorial interpretation of N(n k) on the previous slide impliesthat F (x t) satisfies the following quadratic functional equation

xF 2 + (xt + x minus 1)F + xt = 0

This immediately implies a formula for F (x t) as an algebraic function

F (x t) =1minus (t+1)x minus

radic1minus 2(t+1)x + (tminus1)2x2

2x

Exercise (by me) Use Newtonrsquos Binomial Theorem to expand thisalgebraic function as a power series in x and t and then derive thebinomial formula for Narayana numbers (Maybe not a good idea)

16 28

Combinatorial interpretations of the Narayana numbers (6)

Exercise 619 (p 219) gives 66 different combinatorial interpretations ofthe Catalan numbers C (n) Stanley (p 238) ldquoIt is interesting to find foreach of the combinatorial interpretations of C (n) given by Exercise 619 acorresponding decomposition into subsets counted by Narayana numbersrdquoStanley also gives some additional basic references on Narayana numbers

bull T V Narayana A partial order and its applications to probability theorySankhya 21 (1959) 91ndash98

bull G Kreweras P Moszkowski A new enumerative property of theNarayana numbers J Statist Plann Inference 14 (1986) no 1 63ndash67

bull G Kreweras Y Poupard Subdivision des nombres de Narayana suivantdeux parametres supplementaires Euro J Combin 7 2 (1986) 141ndash149

bull R A Sulanke Refinements of the Narayana numbers Bull InstCombin Appl 7 (1993) 60ndash66

bull R A Sulanke A symmetric variation of a distribution of Kreweras andPoupard J Statist Plann Inference 34 (1993) no 2 291ndash303

17 28

Why am I interested in these numbers (Introduction)

Gian-Carlo Rota posed the following problem

Let A be an associative algebra and let f Ararr A be a linear operatorClassify all the polynomial identities that can be satisfied by f

Posing the same problem in another way (to clarify the quantifiers)

Classify all the polynomial identities that can be satisfied by a linearoperator on an associative algebra

The term ldquoclassifyrdquo is deliberately ambiguous mdash different interpretationscould lead to different but equally valid solutions

Examples

If f is an endomorphism of the algebra structure on A then

f (xy) = f (x)f (y) (not homogeneous in f )

If f is a derivation of the algebra structure on A then

f (xy) = f (x)y + xf (y) (homogeneous of degree 1 in f )

18 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Combinatorial interpretations of the Narayana numbers (4)

The next few pages contain exercises from

bull R P Stanley Enumerative Combinatorics Volume 2With a Foreword by G-C Rota and Appendix 1 by S FominCambridge Studies in Advanced Mathematics 62Cambridge University Press 1999

bull Exercise 636 (page 237)

Let Xn be the set of all sequences x = x1x2 middot middot middot x2n in which n terms are 1and the other n terms are minus1

Assume x1 + middot middot middot+ xi ge 0 for 1 le i le 2n partial sums are nonnegative

Exercise 619(r) (page 222) shows that |Xn| = C (n) the Catalan number

Let Xnk sube Xn contain the sequences with k steps from 1 to minus1 that isk is the number of indices 1 le i le 2nminus1 such that xi = 1 xi+1 = minus1

Then |Xnk | = N(n k) the Narayana number

This is one way to prove thatsumn

k=1 N(n k) = C (n)

15 28

Combinatorial interpretations of the Narayana numbers (5)

Consider the bivariate generating function for the Narayana numbers

F (x t) =infinsumn=1

infinsumk=1

N(n k)xntk

The combinatorial interpretation of N(n k) on the previous slide impliesthat F (x t) satisfies the following quadratic functional equation

xF 2 + (xt + x minus 1)F + xt = 0

This immediately implies a formula for F (x t) as an algebraic function

F (x t) =1minus (t+1)x minus

radic1minus 2(t+1)x + (tminus1)2x2

2x

Exercise (by me) Use Newtonrsquos Binomial Theorem to expand thisalgebraic function as a power series in x and t and then derive thebinomial formula for Narayana numbers (Maybe not a good idea)

16 28

Combinatorial interpretations of the Narayana numbers (6)

Exercise 619 (p 219) gives 66 different combinatorial interpretations ofthe Catalan numbers C (n) Stanley (p 238) ldquoIt is interesting to find foreach of the combinatorial interpretations of C (n) given by Exercise 619 acorresponding decomposition into subsets counted by Narayana numbersrdquoStanley also gives some additional basic references on Narayana numbers

bull T V Narayana A partial order and its applications to probability theorySankhya 21 (1959) 91ndash98

bull G Kreweras P Moszkowski A new enumerative property of theNarayana numbers J Statist Plann Inference 14 (1986) no 1 63ndash67

bull G Kreweras Y Poupard Subdivision des nombres de Narayana suivantdeux parametres supplementaires Euro J Combin 7 2 (1986) 141ndash149

bull R A Sulanke Refinements of the Narayana numbers Bull InstCombin Appl 7 (1993) 60ndash66

bull R A Sulanke A symmetric variation of a distribution of Kreweras andPoupard J Statist Plann Inference 34 (1993) no 2 291ndash303

17 28

Why am I interested in these numbers (Introduction)

Gian-Carlo Rota posed the following problem

Let A be an associative algebra and let f Ararr A be a linear operatorClassify all the polynomial identities that can be satisfied by f

Posing the same problem in another way (to clarify the quantifiers)

Classify all the polynomial identities that can be satisfied by a linearoperator on an associative algebra

The term ldquoclassifyrdquo is deliberately ambiguous mdash different interpretationscould lead to different but equally valid solutions

Examples

If f is an endomorphism of the algebra structure on A then

f (xy) = f (x)f (y) (not homogeneous in f )

If f is a derivation of the algebra structure on A then

f (xy) = f (x)y + xf (y) (homogeneous of degree 1 in f )

18 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Combinatorial interpretations of the Narayana numbers (5)

Consider the bivariate generating function for the Narayana numbers

F (x t) =infinsumn=1

infinsumk=1

N(n k)xntk

The combinatorial interpretation of N(n k) on the previous slide impliesthat F (x t) satisfies the following quadratic functional equation

xF 2 + (xt + x minus 1)F + xt = 0

This immediately implies a formula for F (x t) as an algebraic function

F (x t) =1minus (t+1)x minus

radic1minus 2(t+1)x + (tminus1)2x2

2x

Exercise (by me) Use Newtonrsquos Binomial Theorem to expand thisalgebraic function as a power series in x and t and then derive thebinomial formula for Narayana numbers (Maybe not a good idea)

16 28

Combinatorial interpretations of the Narayana numbers (6)

Exercise 619 (p 219) gives 66 different combinatorial interpretations ofthe Catalan numbers C (n) Stanley (p 238) ldquoIt is interesting to find foreach of the combinatorial interpretations of C (n) given by Exercise 619 acorresponding decomposition into subsets counted by Narayana numbersrdquoStanley also gives some additional basic references on Narayana numbers

bull T V Narayana A partial order and its applications to probability theorySankhya 21 (1959) 91ndash98

bull G Kreweras P Moszkowski A new enumerative property of theNarayana numbers J Statist Plann Inference 14 (1986) no 1 63ndash67

bull G Kreweras Y Poupard Subdivision des nombres de Narayana suivantdeux parametres supplementaires Euro J Combin 7 2 (1986) 141ndash149

bull R A Sulanke Refinements of the Narayana numbers Bull InstCombin Appl 7 (1993) 60ndash66

bull R A Sulanke A symmetric variation of a distribution of Kreweras andPoupard J Statist Plann Inference 34 (1993) no 2 291ndash303

17 28

Why am I interested in these numbers (Introduction)

Gian-Carlo Rota posed the following problem

Let A be an associative algebra and let f Ararr A be a linear operatorClassify all the polynomial identities that can be satisfied by f

Posing the same problem in another way (to clarify the quantifiers)

Classify all the polynomial identities that can be satisfied by a linearoperator on an associative algebra

The term ldquoclassifyrdquo is deliberately ambiguous mdash different interpretationscould lead to different but equally valid solutions

Examples

If f is an endomorphism of the algebra structure on A then

f (xy) = f (x)f (y) (not homogeneous in f )

If f is a derivation of the algebra structure on A then

f (xy) = f (x)y + xf (y) (homogeneous of degree 1 in f )

18 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Combinatorial interpretations of the Narayana numbers (6)

Exercise 619 (p 219) gives 66 different combinatorial interpretations ofthe Catalan numbers C (n) Stanley (p 238) ldquoIt is interesting to find foreach of the combinatorial interpretations of C (n) given by Exercise 619 acorresponding decomposition into subsets counted by Narayana numbersrdquoStanley also gives some additional basic references on Narayana numbers

bull T V Narayana A partial order and its applications to probability theorySankhya 21 (1959) 91ndash98

bull G Kreweras P Moszkowski A new enumerative property of theNarayana numbers J Statist Plann Inference 14 (1986) no 1 63ndash67

bull G Kreweras Y Poupard Subdivision des nombres de Narayana suivantdeux parametres supplementaires Euro J Combin 7 2 (1986) 141ndash149

bull R A Sulanke Refinements of the Narayana numbers Bull InstCombin Appl 7 (1993) 60ndash66

bull R A Sulanke A symmetric variation of a distribution of Kreweras andPoupard J Statist Plann Inference 34 (1993) no 2 291ndash303

17 28

Why am I interested in these numbers (Introduction)

Gian-Carlo Rota posed the following problem

Let A be an associative algebra and let f Ararr A be a linear operatorClassify all the polynomial identities that can be satisfied by f

Posing the same problem in another way (to clarify the quantifiers)

Classify all the polynomial identities that can be satisfied by a linearoperator on an associative algebra

The term ldquoclassifyrdquo is deliberately ambiguous mdash different interpretationscould lead to different but equally valid solutions

Examples

If f is an endomorphism of the algebra structure on A then

f (xy) = f (x)f (y) (not homogeneous in f )

If f is a derivation of the algebra structure on A then

f (xy) = f (x)y + xf (y) (homogeneous of degree 1 in f )

18 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Why am I interested in these numbers (Introduction)

Gian-Carlo Rota posed the following problem

Let A be an associative algebra and let f Ararr A be a linear operatorClassify all the polynomial identities that can be satisfied by f

Posing the same problem in another way (to clarify the quantifiers)

Classify all the polynomial identities that can be satisfied by a linearoperator on an associative algebra

The term ldquoclassifyrdquo is deliberately ambiguous mdash different interpretationscould lead to different but equally valid solutions

Examples

If f is an endomorphism of the algebra structure on A then

f (xy) = f (x)f (y) (not homogeneous in f )

If f is a derivation of the algebra structure on A then

f (xy) = f (x)y + xf (y) (homogeneous of degree 1 in f )

18 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Rota-Baxter operators (1)

Suppose f Ararr A is a derivation which is invertible and write g = f minus1Then for every x y isin A we have x = g(a) y = g(b) for some a b isin AThe derivation identity can then be written as follows

f(g(a)g(b)

)= f (g(a))g(b) + g(a)f (g(b))

Since f g is the identity operator the last equation becomes

f(g(a)g(b)

)= ag(b) + g(a)b

Since g f is the identity operator we apply g to both sides and obtain

g(a)g(b) = g(ag(b) + g(a)b

)

We obtain the Rota-Baxter identity usually written in this form

g(a)g(b) = g(ag(b)) + g(g(a)b) (homogeneous of degree 2 in g)

bull G E Baxter An analytic problem whose solution follows froma simple algebraic identity Pacific J Math 10 (1960) 731ndash742

bull G-C Rota Baxter algebras and combinatorial identities I IIBull Amer Math Soc 75 (2) (1969) 325ndash329 330ndash334

19 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Rota-Baxter operators (2)

Yoursquove seen the Rota-Baxter identity before without even realizing itRemember that itrsquos satisfied by inverses of derivations Rewrite it like this

g(g(a)b) = g(a)g(b)minus g(ag(b))

Write g(a) = u so a = f (u) and g(b) = v so b = f (v)

g(uf (v)) = uv minus g(f (u)v)

If we regard u and v as functions of the real variable t where f denotesthe derivative and hence g denotes the anti-derivative then we obtainint

u(t)v prime(t) dt = uv minusint

uprime(t)v(t) dt

This is just the well-known formula for integration by parts

bull K Ebrahimi-Fard F PatrasRota-Baxter algebra the combinatorial structure of integral calculusarXiv13041204[mathRA] (17 pages submitted on 3 April 2013)

20 28

Why am I interested in these numbers (Continued)

I will restrict attention to operator identities which are

bull homogeneous of degree m in the linear operator f for some fixedm ge 1 each monomial contains exactly m occurrences of f bull multilinear of degree n in the n variables x1 x2 xn for some fixedn ge 2 the n variables occur in the same order in each monomial

The derivation identity corresponds to m = 1 and n = 2

f (xy) = f (x)y + xf (y)

There are four related identities corresponding to m = 2 and n = 2

Rota-Baxter identity f (x)f (y) = f (f (x)y) + f (xf (y))Left averaging identity f (x)f (y) = f (f (x)y)Right averaging identity f (x)f (y) = f (xf (y))Nijenhuis identity f (x)f (y) = f (f (x)y) + f (xf (y))minus f 2(xy)

bull L Guo W Y Sit R ZhangDifferential type operators and Grobner-Shirshov basesJ Symbolic Comput 52 (2013) 97ndash123

21 28

Enumeration of operator monomials

Write T (m n) for the number of all monomials with m occurrences of

the linear operator f applied to n variables in the order x1 xn

First assume n = 2 For m = 0 we get only the monomial xy

For m = 1 we get the three monomials in the derivation identity

f (xy) f (x)y xf (y)

For m = 2 we obtain the following six monomials

f 2(xy) f 2(x)y xf 2(y) f (x)f (y) f (f (x)y) f (xf (y))

For m = 3 we obtain the following ten monomials

f 3(xy) f 3(x)y xf 3(y) f 2(x)f (y) f (x)f 2(y)

f 2(f (x)y) f 2(xf (y) f (f 2(x)y) f (xf 2(y) f (f (x)f (y)

We get binomial coefficients (proof) column 2 of the Narayana triangle

T (m n) =

(m+2n

)= N(m+2 2)

22 28

What about three variables (n = 3)

For m = 0 we get only the monomial xyz For m = 1 we get the following six monomials

f (xyz) f (xy)z xf (yz) f (x)yz xf (y)z xyf (z)

For m = 2 we get the following 20 monomials

f (f (xyz)) f (f (xy))z xf (f (yz)) f (f (x))yz xf (f (y))z

xyf (f (z)) f (x)f (y)z f (x)yf (z) xf (y)f (z) f (f (xy)z)

f (xf (yz)) f (f (x)yz) f (xyf (z)) f (xf (y)z) f (xy)f (z)

f (x)f (yz) f (f (x)y)z f (xf (y))z xf (f (y)z) xf (yf (z))

For m = 3 we get 50 monomials For m = 4 we get 105 monomials

This looks like the top of column 3 of the Narayana triangle (proof)

T (m 3) =1

m+3

(m+3

3

)(m+3

2

)= N(m+3 3)

1 6 20 50 105 196 336 540 825 1210 1716 2366 3185 4200 5440

23 28

Conjecture on enumeration of operator monomials

The number T (m n) of monomials with m operators and n variables

seems to appear in row m+n and column n of the Narayana triangle

T (m n) = N(m+n n) =1

m+n

(m+n

n

)(m+n

nminus1

)(1)

=1

m+n

(m+n)

mn

(m+n)

(m+1)(nminus1)(2)

=(m+n)(m+nminus1)

(m+1)mn(nminus1)(3)

=(m+1)n

m+n

((m+n)

(m+1)n

)2

(4)

It would be great to have a bijective proof that T (m n) = N(m+n n)

24 28

An interesting special case n = m+1

Assume the number of variables is one more than the number of operatorsSubstitute n = m+1 into formula (2) on the previous page

T (mm+1) =1

m+m+1middot (m+m+1)

m(m+1)middot (m+m+1)

(m+1)(m+1minus1)

=1

2m+1middot (2m+1)

m(m+1)middot (2m+1)

(m+1)m

=(2m)(2m+1)(m(m+1)

)2

The OEIS has a page for this sequence (oeisorgA000891)

1 3 20 175 1764 19404 226512 2760615 34763300 449141836

Its generating function is essentially the elliptic integral of the second kind

G (x) =1

4x

(1minus E (16x)

π2

)where E (x) =

int 1

0

radic1minus x2t2

1minus t2dt

25 28

Enumeration of operator monomials

Write T (m n) for the number of all monomials with m occurrences of

the linear operator f applied to n variables in the order x1 xn

First assume n = 2 For m = 0 we get only the monomial xy

For m = 1 we get the three monomials in the derivation identity

f (xy) f (x)y xf (y)

For m = 2 we obtain the following six monomials

f 2(xy) f 2(x)y xf 2(y) f (x)f (y) f (f (x)y) f (xf (y))

For m = 3 we obtain the following ten monomials

f 3(xy) f 3(x)y xf 3(y) f 2(x)f (y) f (x)f 2(y)

f 2(f (x)y) f 2(xf (y) f (f 2(x)y) f (xf 2(y) f (f (x)f (y)

We get binomial coefficients (proof) column 2 of the Narayana triangle

T (m n) =

(m+2n

)= N(m+2 2)

22 28

What about three variables (n = 3)

For m = 0 we get only the monomial xyz For m = 1 we get the following six monomials

f (xyz) f (xy)z xf (yz) f (x)yz xf (y)z xyf (z)

For m = 2 we get the following 20 monomials

f (f (xyz)) f (f (xy))z xf (f (yz)) f (f (x))yz xf (f (y))z

xyf (f (z)) f (x)f (y)z f (x)yf (z) xf (y)f (z) f (f (xy)z)

f (xf (yz)) f (f (x)yz) f (xyf (z)) f (xf (y)z) f (xy)f (z)

f (x)f (yz) f (f (x)y)z f (xf (y))z xf (f (y)z) xf (yf (z))

For m = 3 we get 50 monomials For m = 4 we get 105 monomials

This looks like the top of column 3 of the Narayana triangle (proof)

T (m 3) =1

m+3

(m+3

3

)(m+3

2

)= N(m+3 3)

1 6 20 50 105 196 336 540 825 1210 1716 2366 3185 4200 5440

23 28

Conjecture on enumeration of operator monomials

The number T (m n) of monomials with m operators and n variables

seems to appear in row m+n and column n of the Narayana triangle

T (m n) = N(m+n n) =1

m+n

(m+n

n

)(m+n

nminus1

)(1)

=1

m+n

(m+n)

mn

(m+n)

(m+1)(nminus1)(2)

=(m+n)(m+nminus1)

(m+1)mn(nminus1)(3)

=(m+1)n

m+n

((m+n)

(m+1)n

)2

(4)

It would be great to have a bijective proof that T (m n) = N(m+n n)

24 28

An interesting special case n = m+1

Assume the number of variables is one more than the number of operatorsSubstitute n = m+1 into formula (2) on the previous page

T (mm+1) =1

m+m+1middot (m+m+1)

m(m+1)middot (m+m+1)

(m+1)(m+1minus1)

=1

2m+1middot (2m+1)

m(m+1)middot (2m+1)

(m+1)m

=(2m)(2m+1)(m(m+1)

)2

The OEIS has a page for this sequence (oeisorgA000891)

1 3 20 175 1764 19404 226512 2760615 34763300 449141836

Its generating function is essentially the elliptic integral of the second kind

G (x) =1

4x

(1minus E (16x)

π2

)where E (x) =

int 1

0

radic1minus x2t2

1minus t2dt

25 28

What about three variables (n = 3)

For m = 0 we get only the monomial xyz For m = 1 we get the following six monomials

f (xyz) f (xy)z xf (yz) f (x)yz xf (y)z xyf (z)

For m = 2 we get the following 20 monomials

f (f (xyz)) f (f (xy))z xf (f (yz)) f (f (x))yz xf (f (y))z

xyf (f (z)) f (x)f (y)z f (x)yf (z) xf (y)f (z) f (f (xy)z)

f (xf (yz)) f (f (x)yz) f (xyf (z)) f (xf (y)z) f (xy)f (z)

f (x)f (yz) f (f (x)y)z f (xf (y))z xf (f (y)z) xf (yf (z))

For m = 3 we get 50 monomials For m = 4 we get 105 monomials

This looks like the top of column 3 of the Narayana triangle (proof)

T (m 3) =1

m+3

(m+3

3

)(m+3

2

)= N(m+3 3)

1 6 20 50 105 196 336 540 825 1210 1716 2366 3185 4200 5440

23 28

Conjecture on enumeration of operator monomials

The number T (m n) of monomials with m operators and n variables

seems to appear in row m+n and column n of the Narayana triangle

T (m n) = N(m+n n) =1

m+n

(m+n

n

)(m+n

nminus1

)(1)

=1

m+n

(m+n)

mn

(m+n)

(m+1)(nminus1)(2)

=(m+n)(m+nminus1)

(m+1)mn(nminus1)(3)

=(m+1)n

m+n

((m+n)

(m+1)n

)2

(4)

It would be great to have a bijective proof that T (m n) = N(m+n n)

24 28

An interesting special case n = m+1

Assume the number of variables is one more than the number of operatorsSubstitute n = m+1 into formula (2) on the previous page

T (mm+1) =1

m+m+1middot (m+m+1)

m(m+1)middot (m+m+1)

(m+1)(m+1minus1)

=1

2m+1middot (2m+1)

m(m+1)middot (2m+1)

(m+1)m

=(2m)(2m+1)(m(m+1)

)2

The OEIS has a page for this sequence (oeisorgA000891)

1 3 20 175 1764 19404 226512 2760615 34763300 449141836

Its generating function is essentially the elliptic integral of the second kind

G (x) =1

4x

(1minus E (16x)

π2

)where E (x) =

int 1

0

radic1minus x2t2

1minus t2dt

25 28

Conjecture on enumeration of operator monomials

The number T (m n) of monomials with m operators and n variables

seems to appear in row m+n and column n of the Narayana triangle

T (m n) = N(m+n n) =1

m+n

(m+n

n

)(m+n

nminus1

)(1)

=1

m+n

(m+n)

mn

(m+n)

(m+1)(nminus1)(2)

=(m+n)(m+nminus1)

(m+1)mn(nminus1)(3)

=(m+1)n

m+n

((m+n)

(m+1)n

)2

(4)

It would be great to have a bijective proof that T (m n) = N(m+n n)

24 28

An interesting special case n = m+1

Assume the number of variables is one more than the number of operatorsSubstitute n = m+1 into formula (2) on the previous page

T (mm+1) =1

m+m+1middot (m+m+1)

m(m+1)middot (m+m+1)

(m+1)(m+1minus1)

=1

2m+1middot (2m+1)

m(m+1)middot (2m+1)

(m+1)m

=(2m)(2m+1)(m(m+1)

)2

The OEIS has a page for this sequence (oeisorgA000891)

1 3 20 175 1764 19404 226512 2760615 34763300 449141836

Its generating function is essentially the elliptic integral of the second kind

G (x) =1

4x

(1minus E (16x)

π2

)where E (x) =

int 1

0

radic1minus x2t2

1minus t2dt

25 28

An interesting special case n = m+1

Assume the number of variables is one more than the number of operatorsSubstitute n = m+1 into formula (2) on the previous page

T (mm+1) =1

m+m+1middot (m+m+1)

m(m+1)middot (m+m+1)

(m+1)(m+1minus1)

=1

2m+1middot (2m+1)

m(m+1)middot (2m+1)

(m+1)m

=(2m)(2m+1)(m(m+1)

)2

The OEIS has a page for this sequence (oeisorgA000891)

1 3 20 175 1764 19404 226512 2760615 34763300 449141836

Its generating function is essentially the elliptic integral of the second kind

G (x) =1

4x

(1minus E (16x)

π2

)where E (x) =

int 1

0

radic1minus x2t2

1minus t2dt

25 28

Combinatorial interpretations of the special case n = m+1

Itrsquos known (oeisorgA000891) that T (mm+1) is the number of

bull Noncrossing partitions of 1 2m+1 into m+1 blocksbull Pairs of nonintersecting lattice paths (the only common vertices are

endpoints) from (0 0) to (m+1m+1) using steps (1 0) and (0 1)bull Returning walks of length 2m on the upper half of a square latticebull Proper mergings of two m-chainsbull Parallelogram polyominoes () with m+1 columns and m+1 rowsbull Tilings of an 〈m 2m〉 hexagon ()

bull E Barcucci A Frosini S Rinaldi On directed-convexpolyominoes in a rectangle Discrete Math 298 (2005) 62ndash78

bull W Y C Chen S X M Pang E X Y Qu R P StanleyPairs of noncrossing free Dyck paths and noncrossing partitionsDiscrete Math 309 (2009) 2834ndash2838

bull H Muhle Counting proper mergings of chains and antichainsDiscrete Math 327 (2014) 118ndash129

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Combinatorial interpretation of the general case

N(n+m n)= number of operator monomials (n variables m operators)

Consider n+m pairs of balanced parentheses with n nestings that is ()

For n = m = 2 we have seen these sequences before

()((())) (())(()) (()(())) ((()())) ((())()) ((()))()

In each sequence replace the two () from left to right by x and y

x((y)) (x)(y) (x(y)) ((xy)) ((x)y) ((x))y

Replace each left parenthesis rdquo(rdquo by the operator symbol rdquof (rdquo

xf 2(y) f (x)f (y) f (xf (y)) f 2(xy) f (f (x)y) f 2(x)y

We get the six operator monomials with n = 2 variables m = 2 operators

For n = 3 m = 1 we have also seen these sequences before

()()(()) ()(())() (())()() (()())() ()(()()) (()()())

Following the same procedure we obtain six operator monomials

xyf (z) xf (y)z f (x)yz f (xy)z xf (yz) f (xyz)

Does this give a bijective proof of the conjecture in the first line

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The End

Thank You for Your Attention

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The End

Thank You for Your Attention

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