Zeons, Permanents, the Johnson scheme, and Generalized Derangements
aa r X i v : . [ m a t h . C O ] O c t ZEONS, PERMANENTS, THE JOHNSON SCHEME, ANDGENERALIZED DERANGEMENTS
PHILIP FEINSILVER AND JOHN MCSORLEY
Abstract.
Starting with the zero-square “zeon algebra” the connection with per-manents is shown. Permanents of sub-matrices of a linear combination of the iden-tity matrix and all-ones matrix leads to moment polynomials with respect to theexponential distribution. A permanent trace formula analogous to MacMahon’sMaster Theorem is presented and applied. Connections with permutation groupsacting on sets and the Johnson association scheme arise. The families of num-bers appearing as matrix entries turn out to be related to interesting variationson derangements. These generalized derangements are considered in detail as anillustration of the theory. Introduction
Functions acting on a finite set can be conveniently expressed using matrices, wherebythe composition of functions corresponds to multiplication of the matrices. Essen-tially one is considering the induced action on the vector space with the elementsof the set acting as a basis. This action extends to tensor powers of the vectorspace. One can take symmetric powers, antisymmetric powers, etc., that yield rep-resentations of the multiplicative semigroup of functions. An especially interestingrepresentation occurs by taking non-reflexive, symmetric powers. Identifying the un-derlying set of cardinality n with { , , . . . , n } , the vector space has basis e , e , . . . .The action we are interested in may be found by saying that the elements e i generatea “zeon algebra”, the relations being that the e i commute, with e i = 0, 1 ≤ i ≤ n .To get a feeling for this, first we recall the action on Grassmann algebra where thematrix elements of the induced action arise as determinants. For the zeon case,permanents appear.An interesting connection with the centralizer algebra of the action of the symmetricgroup comes up. For the defining action on the set { , . . . , n } , represented as 0 -1permutation matrices, the centralizer algebra of n × n matrices commuting with theentire group is generated by I , the identity matrix, and J , the all-ones matrix. Thequestion was if they would help determine the centralizer algebra for the action onsubsets of a fixed size, ℓ -sets, for ℓ >
1. It is known that the basis for the centralizeralgebra is given by the adjacency matrices of the Johnson scheme. Could one find
PHILIP FEINSILVER AND JOHN MCSORLEY this working solely with I and J ? The result is that by computing the “zeon powers”,i.e., the action of sI + tJ , linear combinations of I and J , on ℓ -sets, the Johnsonscheme appears naturally. The coefficients are polynomials in s and t occurring asmoments of the exponential distribution. And they turn out to count derangements,and related generalized derangements. The occurrence of Laguerre polynomials in thecombinatorics of derangements is well-known. Here the F hypergeometric function,which is closely related to Poisson-Charlier polynomials, arises rather naturally.Here is an outline of the paper. § § sI + tJ are found in § § sI + tJ , including an interpretation of exponentialmoment polynomials by elementary subgraphs. In §
6, generalized derangement num-bers, specifically counting derangements and counting arrangements are consideredin detail. The Appendix has some derangement numbers and arrangement numbersfor reference, as well as a page of exponential polynomials. An example expressingexponential polynomials in terms of elementary subgraphs is given there.2.
Representations of functions acting on sets
Let V denote the vector space Q n or R n . We will look at the action of a linear mapon V extended to quotients of tensor powers V ⊗ ℓ . We work with coordinates ratherthan vectors. First, recall the Grassmann case. To find the action on V ∧ ℓ consideran algebra generated by n variables e i satisfying e i e j = − e j e i . In particular, e i = 0. Notation.
The standard n -set { , . . . , n } will be denoted [ n ]. Roman caps I, J,A, etc. denote subsets of [ n ]. We will identify them with the corresponding orderedtuples. Generally, given an n -tuple ( x , . . . , x n ) and a subset I ⊂ [ n ], we denoteproducts x I = Y j ∈ I x j where the indices are in increasing order if the variables are not assumed to commute.As an index we will use U to denote the full set [ n ].Italic I and J will denote the identity matrix and all-ones matrix respectively.For a matrix X IJ , say, where the labels are subsets of fixed size l , dictionary ordering isused. That is, convert to ordered tuples and use dictionary ordering. For example, for ERMANENTS AND GENERALIZED DERANGEMENTS 3 n = 4, l = 2, we have labels 12 , , , , ,
34 for rows one through six respectively.A basis for V ∧ ℓ is given by products e I = e i e i · · · e i ℓ I ⊂ [ n ], where we consider I as an ordered ℓ -tuple. Given a matrix X acting on V ,let y i = X j X ij e j with corresponding products y I . Then the matrix X ∧ ℓ has entries given by thecoefficients in the expansion y I = X J ( X ∧ ℓ ) IJ e J where the anticommutation rules are used to order the factors in e J . Note that thecoefficient of e j in y i is X ij itself. And for n >
3, the coefficient of e in y isdet (cid:18) X X X X (cid:19) We see that in general the IJ entry of X ∧ ℓ is the minor of X with row labels I andcolumn labels J. A standard term for the matrix X ∧ ℓ is a compound matrix . Notingthat X ∧ ℓ is (cid:0) nℓ (cid:1) × (cid:0) nℓ (cid:1) , in particular ℓ = n yields the one-by-one matrix with entryequal to ( X ∧ n ) UU = det X .In this work, we will use the algebra of zeons , standing for “zero-ons”, or morespecifically, “zero-square bosons”. That is, we assume that the variables e i satisfythe properties e i e j = e j e i and e i = 0A basis for the algebra is again given by e I , I ⊂ [ n ]. At level ℓ , the induced matrix X ∨ ℓ has IJ entries according to the expansion of y I y i = X j X ij e j −→ y I = X J ( X ∨ ℓ ) IJ e J similar to the Grassmann case. Since the variables commute, we see that the IJ entryof X ∨ ℓ is the permanent of the submatrix with rows I and columns J. In particular,( X ∨ n ) UU = per X . We refer to the matrix X ∨ ℓ as the “ ℓ th zeon power of X ”. PHILIP FEINSILVER AND JOHN MCSORLEY
Functions on the power set of [ n ] . Note that X ∨ ℓ is indexed by ℓ -sets.Suppose X f represents a function f : [ n ] → [ n ]. So it is a zero-one matrix with( X f ) ij = 1 the single entry in row i if f maps i to j . The ℓ th zeon power of X is thematrix of the induced map on ℓ -sets. If f maps an ℓ -set I to one of lower cardinality,then the corresponding row in X ∨ ℓ has all zero entries. Thus, the induced matricesin general correspond to “partial functions”.However, if X is a permutation matrix, then X ∨ ℓ is a permutation matrix for all0 ≤ ℓ ≤ n . So, given a group of permutation matrices, the map X → X ∨ ℓ is arepresentation of the group.2.2. Zeon powers of sI + tX . Our main theorem computes the ℓ th zeon power of sI + tX for an n × n matrix X , where s and t are scalar variables. Theorem 2.1.
For a given matrix X , for ≤ ℓ ≤ n , and indices | I | = | J | = ℓ , (cid:0) ( sI + tX ) ∨ ℓ (cid:1) IJ = X ≤ j ≤ ℓ s ℓ − j t j X A ⊂ I ∩ J | A | = ℓ − j (cid:0) X ∨ j (cid:1) I \ A , J \ A ℓ − jj − k kk JB CI AB ∩ C Figure 1.
Configuration of sets. B = I \ A, C = J \ A .
Proof.
Start with y i = se i + tξ i , where ξ i = P j X ij e j . Given I = ( i , . . . , i ℓ ), we wantthe coefficient of e J in the expansion of the product y I = y i · · · y i ℓ . Now, y I = ( se i + tξ i ) · · · ( se i ℓ + tξ i ℓ )Choose A ⊂ I with | A | = ℓ − j , 0 ≤ j ≤ ℓ . A typical term of the product has theform s ℓ − j t j e A ξ B ERMANENTS AND GENERALIZED DERANGEMENTS 5 where A ∩ B = ∅ , B = I \ A. ξ B denotes the product of terms ξ i with indices in B.Expanding, we have ξ B = X C (cid:0) X ∨ j (cid:1) BC e C and e A ξ B = X C (cid:0) X ∨ j (cid:1) BC e A e C Thus, for a contribution to the coefficient of e J , we have A ∪ C = J, where A ∩ C = ∅ .I.e., C = J \ A and A ⊂ I ∩ J. So the coefficient of s ℓ − j t j is as stated.2.3. Trace formula.
Another main feature is the trace formula which shows thepermanent of I + tX as the generating function for the traces of the zeon powersof X . This is the zeon analog of the theorem of MacMahon for representations onsymmetric tensors. Theorem 2.2.
We have the formula per ( sI + tX ) = X j s n − j t j tr X ∨ j Proof.
The permanent of sI + tX is the UU entry of ( sI + tX ) ∨ n . Specialize I = J = Uin Theorem 2.1. So A is any ( n − j )-set with I \ A = J \ A = A ′ , its complement in[ n ]. Thus per ( sI + tX ) = (( sI + tX ) ∨ n ) UU = X ≤ j ≤ n s n − j t j X | A | = n − j (cid:0) X ∨ j (cid:1) A ′ , A ′ = X ≤ j ≤ n s n − j t j tr X ∨ j as required.2.4. Permutation groups.
Let X be an n × n permutation matrix. We can expressper ( I + tX ) in terms of the cycle decomposition of the associated permutation. Proposition 2.3.
For a permutation matrix X , per ( I + tX ) = Y ≤ ℓ ≤ n (1 + t ℓ ) n X ( ℓ ) where n X ( ℓ ) is the number of cycles of length ℓ in the cycle decomposition of thecorresponding permutation. PHILIP FEINSILVER AND JOHN MCSORLEY
Proof.
Decomposing the permutation associated to X yields a decomposition intoinvariant subspaces of the underlying vector space V . So per ( I + tX ) will be theproduct of per ( I + tX c ) as c runs through the corresponding cycles with X c therestriction of X to the invariant subspace for each c . So we have to check that if X acts on V ℓ as a cycle of length ℓ , then per ( I + tX ) = 1 + t ℓ . For this, apply Theorem2.2. Apart from level zero, there is only one set fixed by any X ∨ j , namely when j = ℓ . So the trace of X ∨ j is zero unless j = ℓ and then it is one. The result follows.2.4.1. Cycle index. Orbits on ℓ -sets. Now consider a group, G , of permutation ma-trices. We have the cycle index Z G ( z , z , . . . , z n ) = 1 | G | X X ∈ G z n X (1)1 z n X (2)2 · · · z n X ( n ) n each z ℓ corresponding to ℓ -cycles in the cycle decomposition associated to the X ’s.From Proposition 2.3, we have an expression in terms of permanents. Combiningwith the trace formula, we get Theorem 2.4.
Let G be a permutation group of matrices. Then we have | G | X X ∈ G per ( I + tX ) = Z G (1 + t, t , . . . , t n )= X ℓ t ℓ ℓ -sets ) . Remark.
This result refers to three essential theorems in group theory acting onsets. Equality of the first and last expressions is the ‘permanent’ analog of Molien’sTheorem, which is the case for a group acting on the symmetric tensor algebra. Thatthe cycle index counts orbits on subsets is an instance of Polya Counting, with twocolors. The last expression follows by the Cauchy-Burnside Lemma applied to thegroups G ∨ ℓ = { X ∨ ℓ } X ∈ G .2.4.2. Centralizer algebra and Johnson scheme.
Given a group, G , of permutationmatrices, an important question is to determine the set (among all matrices) ofmatrices commuting with all of the matrices in G . This is the centralizer algebra ofthe group. For the symmetric group, the only matrices are I and J . For the actionof the symmetric group on ℓ -sets, a basis for the centralizer algebra is given by theincidence matrices for the Johnson distance. These are the same as the adjacencymatrices for the Johnson (association) scheme. Recall that the Johnson distancebetween two ℓ -sets I and J isdist JS (I , J) = | I ∆ J | = | I \ J | = | J \ I | ERMANENTS AND GENERALIZED DERANGEMENTS 7
The corresponding matrices JS nℓk are defined by(JS nℓk ) IJ = ( , if dist JS (I , J) = k G , acting on pairs, the Johnson basis is a basis for the centralizeralgebra. Since the Johnson distance is symmetric, it suffices to look at G ∨ .Now we come to the question that is one starting point for this work. If I and J are the only matrices commuting with all elements (as matrices) of the symmetricgroup, then since the map G → G ∨ ℓ is a homomorphism, we know that I ∨ ℓ and J ∨ ℓ are in the centralizer algebra of G ∨ ℓ . The question is: how to obtain the rest? The,perhaps surprising, answer is that in fact one can obtain the complete Johnson basisfrom I and J alone. This will be one of the main results, Theorem 4.1.2.4.3. Permanent of sI + tJ . First, let us consider sI + tJ . Proposition 2.5.
We have the formula per ( sI + tJ ) = n ! X ≤ ℓ ≤ n s ℓ t n − ℓ ℓ ! (1) Proof.
For X = J , we see directly, since all entries equal one in all submatrices, that( J ∨ ℓ ) IJ = ℓ !for all I and J. Taking traces tr J ∨ ℓ = (cid:18) nℓ (cid:19) ℓ !and by the trace formula, Theorem 2.2,per ( sI + tJ ) = X ℓ (cid:18) nℓ (cid:19) ℓ ! s n − ℓ t ℓ = X ℓ n !( n − ℓ )! s n − ℓ t ℓ . Reversing the order of summation yields the result stated.
Corollary 2.6.
For varying n , we will explicitly denote p n ( s, t ) = per ( sI n + tJ n ) .Then, with p ( s, t ) = 1 , ∞ X n =0 z n n ! p n ( s, t ) = e sz − tz . PHILIP FEINSILVER AND JOHN MCSORLEY
The Corollary exhibits the operational formula p n ( s, t ) = 11 − tD s s n where D s = d/ds . By inspection, this agrees with (1) as well.Observe that equation (1) can be rewritten asper ( sI + tJ ) = Z ∞ ( s + ty ) n e − y dy that is, these are “moment polynomials” for the exponential distribution with anadditional scale parameter.We proceed to examine these moment polynomials in detail.3. Exponential polynomials
For the exponential distribution, with density e − y on (0 , ∞ ), the moment polynomials are defined as h n ( x ) = Z ∞ ( x + y ) n e − y dy The exponential embeds naturally into the family of weights of the form x m e − x on(0 , ∞ ) as for generalized Laguerre polynomials. We define correspondingly h n,m ( x, t ) = Z ∞ ( x + ty ) n ( ty ) m e − y dy (2)for nonnegative integers n, m , introducing a factor of y m and a scale factor t . Werefer to these as exponential moment polynomials . Proposition 3.1.
Observe the following properties of the exponential moment poly-nomials.1. The generating function t m m ! ∞ X n =0 z n n ! h n,m ( x, t ) = e zx (1 − tz ) m for | tz | < .2. The operational formula t m m ! h n,m ( x, t ) = ( I − tD ) − ( m +1) x n where I is the identity operator and D = d/dx . ERMANENTS AND GENERALIZED DERANGEMENTS 9
3. The explicit form h n,m ( x, t ) = n X j =0 (cid:18) nj (cid:19) ( m + j )! x n − j t m + j Proof.
For the first formula, multiply the integral by z n /n ! and sum to get Z ∞ y m e zx + zty − y dy = e zx Z ∞ y m e − y (1 − tz ) dy which yields the stated result.For the second, write t m m ! ( I − tD ) − ( m +1) x n = t m Z ∞ y m e − ( I − tD ) y x n dy = Z ∞ ( ty ) m e − y ( x + ty ) n dy using the shift formula e aD f ( x ) = f ( x + a ).For the third, expand ( x + ty ) n by the binomial theorem and integrate.A variation we will encounter in the following is h n − m,m ( x, t ) = n − m X j =0 (cid:18) n − mj (cid:19) ( m + j )! x n − m − j t m + j (3)= n X j = m (cid:18) n − mj − m (cid:19) j ! x n − j t j (4)= n X j = m (cid:18) n − mn − j (cid:19) j ! x n − j t j (5)= n − m X j =0 (cid:18) n − mj (cid:19) ( n − j )! x j t n − j (6)replacing the index j ← j − m for (4) and reversing the order of summation for thelast line. And for future reference, the integral formula, h n − m,m ( x, t ) = Z ∞ ( x + ty ) n − m ( ty ) m e − y dy (7) Hypergeometric form.
Generalized hypergeometric functions provide expres-sions for the exponential moment polynomials that are often convenient. In thepresent context we will use F functions, defined by F (cid:18) a, b — (cid:12)(cid:12)(cid:12)(cid:12) x (cid:19) = ∞ X j =0 ( a ) j ( b ) j j ! x j where ( a ) j = Γ( a + j ) / Γ( a ) is the usual Pochhammer symbol. In particular, if a ,e.g., is a negative integer, the series reduces to a polynomial. Rearranging factors inthe expressions for h n,m , via h n − m,m , eq. (3), wecan formulate these as F hypergeometric functions. Proposition 3.2.
We have the following expressions for exponential moment poly-nomials. h n,m ( x, t ) = x n t m m ! F (cid:18) − n, m — (cid:12)(cid:12)(cid:12)(cid:12) − tx (cid:19) h n − m,m ( x, t ) = x n − m t m m ! F (cid:18) m − n, m — (cid:12)(cid:12)(cid:12)(cid:12) − tx (cid:19) . Zeon powers of sI + tJ We want to calculate ( sI + tJ ) ∨ ℓ , i.e., the (cid:0) nℓ (cid:1) × (cid:0) nℓ (cid:1) matrix with rows and columnslabelled by ℓ -subsets I , J ⊂ { , . . . , n } with the IJ entry equal to the permanent ofthe corresponding submatrix of sI + tJ . This is equivalent to the induced action ofthe original matrix sI + tJ on the ℓ th zeon space V ∨ ℓ . Theorem 4.1.
The ℓ th zeon power of sI + tJ is given by ( sI + tJ ) ∨ ℓ = X k ℓ X j = k (cid:18) ℓ − kℓ − j (cid:19) j ! s ℓ − j t j JS nℓk = X k h ℓ − k,k ( s, t ) JS nℓk where the h ’s are exponential moment polynomials.Proof. Choose I and J with | I | = | J | = ℓ . By Theorem 2.1, we have, using the factthat all of the entries of J ∨ j are equal to j !, (cid:0) ( sI + tJ ) ∨ ℓ (cid:1) IJ = X ≤ j ≤ ℓ s ℓ − j t j X A ⊂ I ∩ J | A | = ℓ − j (cid:0) J ∨ j (cid:1) I \ A , J \ A = X ≤ j ≤ ℓ s ℓ − j t j X A ⊂ I ∩ J | A | = ℓ − j j ! ERMANENTS AND GENERALIZED DERANGEMENTS 11
Now, if dist JS (I , J) = k , then | I ∩ J | = ℓ − k and there are (cid:18) ℓ − kℓ − j (cid:19) subsets A of I ∩ Jsatisfying the conditions of the sum. Hence the result.Note that the specialization ℓ = n , k = 0, recovers equation (1).We can write the above expansion using the hypergeometric form of the exponentialmoment polynomials, Proposition 3.2,( sI + tJ ) ∨ ℓ = X k s ℓ − k t k k ! F (cid:18) k − ℓ, k — (cid:12)(cid:12)(cid:12)(cid:12) − ts (cid:19) JS nℓk . Spectrum of the Johnson matrices.
Recall, e.g., [1, p. 220], that the spec-trum of the Johnson matrices for given n and ℓ are the numbersΛ nℓk ( α ) = X i (cid:18) ℓ − αi (cid:19)(cid:18) n − ℓ − α + ii (cid:19)(cid:18) ℓ − ik − i (cid:19) ( − k − i (8)where the eigenvalue for given α has multiplicity (cid:18) nα (cid:19) − (cid:18) nα − (cid:19) .For ℓ -sets, the Johnson distance takes values from 0 to min( ℓ, n − ℓ ), with α takingvalues from that same range.4.2. The spectrum of ( sI + tJ ) ∨ ℓ . Recall that as the Johnson matrices are sym-metric and generate a commutative algebra, they are simultaneously diagonalizableby an orthogonal transformation of the underlying vector space. Diagonalizing theequation in Theorem 4.1, we see that the spectrum of ( sI + tJ ) ∨ ℓ is given by X k h ℓ − k,k ( s, t ) Λ nℓk ( α ) . Proposition 4.2.
The spectrum of ( sI + tJ ) ∨ ℓ is given by s α t n − ℓ − α ( n − ℓ − α )! h ℓ − α,n − ℓ − α ( s, t ) = X i s ℓ − i t i (cid:18) ℓ − αi (cid:19)(cid:18) n − ℓ − α + ii (cid:19) i ! for ≤ α ≤ min( ℓ, n − ℓ ) , with respective multiplicities (cid:18) nα (cid:19) − (cid:18) nα − (cid:19) . Proof.
In the sum over i in equation (8), only the last two factors involve k . We have X k h ℓ − k,k ( s, t )( − k − i (cid:18) ℓ − ik − i (cid:19) = X k Z ∞ ( s + ty ) ℓ − k ( ty ) k ( − k − i (cid:18) ℓ − ik − i (cid:19) e − y dy setting k = i + m = X m Z ∞ ( s + ty ) ℓ − i − m ( ty ) i + m ( − m (cid:18) ℓ − im (cid:19) e − y dy = Z ∞ ( s + ty − ty ) ℓ − i ( ty ) i e − y dy = s ℓ − i t i i !using the binomial theorem to sum out m . Filling in the additional factors yields X k h ℓ − k,k ( s, t ) Λ nℓk ( α ) = X i s ℓ − i t i i ! (cid:18) ℓ − αi (cid:19)(cid:18) n − ℓ − α + ii (cid:19) . Taking out a denominator factor of ( n − ℓ − α )! and multiplying by s − α t n − ℓ − α gives X i s ℓ − α − i t n − ℓ − α + i (cid:18) ℓ − αi (cid:19) ( n − ℓ − α + i )!which is precisely h ℓ − α,n − ℓ − α as in the third statement of Proposition 3.1.As in Proposition 3.2, we can express the eigenvalues as follows. Corollary 4.3.
The spectrum of ( sI + tJ ) ∨ ℓ consists of the eigenvalues s ℓ F (cid:18) α − ℓ, n − ℓ − α — (cid:12)(cid:12)(cid:12)(cid:12) − ts (cid:19) for ≤ α ≤ min( ℓ, n − ℓ ) , with corresponding multiplicities as indicated above. Row-sums and trace identity.
For the row-sums, we know that the all-onesvector is a common eigenvector of the Johnson basis corresponding to α = 0. Theseare the valencies Λ k (0). For the Johnson scheme, we haveΛ nℓk (0) = (cid:18) ℓk (cid:19)(cid:18) n − ℓk (cid:19) e.g., see [1, p. 219], which can be checked directly from the formula for Λ nℓk ( α ),equation (8), with α set to zero. Setting α = 0 in Proposition 4.2 gives1 t n − ℓ ( n − ℓ )! h ℓ,n − ℓ ( s, t ) = X i (cid:18) ℓi (cid:19)(cid:18) n − ℓ + ii (cid:19) i ! s ℓ − i t i (9)for the row-sums of ( sI + tJ ) ∨ ℓ . ERMANENTS AND GENERALIZED DERANGEMENTS 13
Trace Identity.
Terms on the diagonal are the coefficient of JS nℓ , which is theidentity matrix. So the trace istr ( sI + tJ ) ∨ ℓ = (cid:18) nℓ (cid:19) h ℓ, ( s, t ) = (cid:18) nℓ (cid:19) X k (cid:18) ℓk (cid:19) k ! s ℓ − k t k Cancelling factorials and reversing the order of summation on k yields the formula: Proposition 4.4. tr ( sI + tJ ) ∨ ℓ = n !( n − ℓ )! X ≤ k ≤ ℓ s k t ℓ − k k !Now, Proposition 4.2 gives the tracetr ( sI + tJ ) ∨ ℓ = X ≤ α ≤ min( ℓ,n − ℓ ) (cid:20)(cid:18) nα (cid:19) − (cid:18) nα − (cid:19)(cid:21) X i s ℓ − i t i (cid:18) ℓ − αi (cid:19)(cid:18) n − ℓ − α + ii (cid:19) i !Equating the above expressions for the trace yields the identity X ≤ α ≤ min( ℓ,n − ℓ ) (cid:20)(cid:18) nα (cid:19) − (cid:18) nα − (cid:19)(cid:21) X i s ℓ − i t i (cid:18) ℓ − αi (cid:19)(cid:18) n − ℓ − α + ii (cid:19) i != n !( n − ℓ )! X ≤ j ≤ ℓ s j t ℓ − j j ! Example.
For n = 4, ℓ = 2 we have s + 2 st + 2 t st + 2 t st + 2 t st + 2 t st + 2 t t st + 2 t s + 2 st + 2 t st + 2 t st + 2 t t st + 2 t st + 2 t st + 2 t s + 2 st + 2 t t st + 2 t st + 2 t st + 2 t st + 2 t t s + 2 st + 2 t st + 2 t st + 2 t st + 2 t t st + 2 t st + 2 t s + 2 st + 2 t st + 2 t t st + 2 t st + 2 t st + 2 t st + 2 t s + 2 st + 2 t One can check that the entries are in agreement with Theorem 4.1. The trace is6 s + 12 st + 12 t . The spectrum iseigenvalue s + 6 st + 12 t , with multiplicity 1eigenvalue s + 2 st, with multiplicity 3eigenvalue s , with multiplicity 2and the trace can be verified from these as well. Remark.
What is interesting is that these matrices have polynomial entries with alleigenvalues polynomials as well and furthermore, the exact same set of polynomialsproduces the eigenvalues as well as the entries. Specializing s and t to integers, asimilar statement holds. All of these matrices will have integer entries with integereigenvalues all of which belong to closely related families of numbers. We will examineinteresting cases of this phenomenon later on in this paper.5. Permanents from sI + tJ Here we present a proof via recursion of the subpermanents of sI + tJ , therebyrecovering Theorem 4.1 from a different perspective. Remark.
For the remainder of this paper, we will work with an n × n matrixcorresponding to an ℓ × ℓ submatrix of the above discussion. Here we have blown upthe submatrix to full size as the object of consideration.Let M n,ℓ denote the n × n matrix with n − ℓ entries equal to s + t on the maindiagonal, and t ’s elsewhere. Note that M n, = sI + tJ and M n,n = tJ , where I and J are n × n . Define P n,ℓ = per ( M n,ℓ ) (10)to be the permanent of M n,ℓ .For ℓ = 0, define P , = 1, and, recalling equation (1), P n, = per ( sI + tJ ) = n X j =0 n ! j ! s j t n − j = n X j =0 n !( n − j )! s n − j t j . (11)We have also P n,n = per ( tJ ) = n ! t n for J of order n × n . These agree at P , = 1. Theorem 5.1.
For n ≥ , ≤ ℓ ≤ n , we have the recurrence P n,ℓ = P n,ℓ − − sP n − ,ℓ − . (12) Proof.
We have 0 ≤ ℓ ≤ n so n − ( ℓ −
1) = n − ℓ + 1 ≥
1, i.e., the matrix M n,ℓ − contains at least 1 entry on its main diagonal equal to s + t . Write the block form M n,ℓ − = (cid:20) s + t AA T M n − ,ℓ − (cid:21) with A = [ t, t, . . . t ] the 1 × ( n −
1) row vector of all t ’s, and A T its transpose. Nowcompute the permanent of M n,ℓ − expanding along the first row. We get P n,ℓ − = per ( M n,ℓ − ) = ( s + t ) per ( M n − ,ℓ − ) + F ( A, A T , M n − ,ℓ − ) (13) ERMANENTS AND GENERALIZED DERANGEMENTS 15 where F ( A, A T , M n − ,ℓ − ) is the contribution to P n,ℓ − involving A . Now t per ( M n − ,ℓ − ) + F ( A,A T , M n − ,ℓ − )= per (cid:0) (cid:20) t AA T M n − ,ℓ − (cid:21) (cid:1) = per (cid:0) (cid:20) A T M n − ,ℓ − t A (cid:21) (cid:1) = per (cid:0) (cid:20) M n − ,ℓ − A T A t (cid:21) (cid:1) = P n,ℓ . Thus from equation (13): P n,ℓ − = s per ( M n − ,ℓ − ) + t per ( M n − ,ℓ − ) + F ( A, A T , M n − ,ℓ − )= s P n − ,ℓ − + P n,ℓ and so the result.We arrange the polynomials P n,ℓ in a triangle, with the columns labelled by ℓ ≥ n ≥
0, starting with P , = 1 at the top vertex: P , P , P , P , P , P , ... . . . . . . P n − , . . . P n − ,n − P n − ,n − P n, P n, . . . P n,n − P n,n The recurrence says that to get the n, ℓ entry, you combine elements in column ℓ − n and n −
1, forming an L -shape. Thus, given the first column { P n, } n ≥ ,the table can be generated in full.Now we check that these are indeed our exponential moment polynomials. Addition-ally we derive an expression for P n,ℓ in terms of the initial sequence P n, . For clarity,we will explicitly denote the dependence of P n,ℓ on ( s, t ). Theorem 5.2.
For ℓ ≥ we have
1. The permanent of the n × n matrix with n − ℓ entries on the diagonal equal to s + t and all other entries equal to t is P n,ℓ ( s, t ) = h n − ℓ,ℓ ( s, t ) = n X j = ℓ (cid:18) n − ℓn − j (cid:19) j ! s n − j t j . (14) P n,ℓ ( s, t ) = ℓ X j =0 (cid:18) ℓj (cid:19) ( − j s j P n − j, ( s, t ) . (15)
3. And the complementary sum: s n = n X j =0 (cid:18) nℓ (cid:19) ( − ℓ P n,ℓ ( s, t ) . Proof.
The initial sequence P n, = h n, as noted in equation (11). We check that h n − ℓ,ℓ satisfies recurrence (12). Starting from the integral representation for h n − ℓ +1 ,ℓ − ,equation (2), we have h n − ℓ +1 ,ℓ − = Z ∞ ( s + ty ) n − ℓ +1 ( ty ) ℓ − e − y dy = Z ∞ ( s + ty )( s + ty ) n − ℓ ( ty ) ℓ − e − y dy = s h n − ℓ,ℓ − + h n − ℓ,ℓ as required, where we now identify h n − ℓ +1 ,ℓ − = P n,ℓ − , h n − ℓ,ℓ − = P n − ,ℓ − , and h n − ℓ,ℓ = P n,ℓ . And equation (5) gives an explicit form for P n,ℓ .For P n, = h n, , we get ℓ X j =0 (cid:18) ℓj (cid:19) ( − j s j Z ∞ ( s + ty ) n − j e − y dy = ℓ X j =0 (cid:18) ℓj (cid:19) ( − j s j Z ∞ ( s + ty ) n − ℓ ( s + ty ) ℓ − j e − y dy = Z ∞ ( s + ty ) n − ℓ ( s + ty − s ) ℓ e − y dy = h n − ℓ,ℓ as required. The proof for P n,ℓ = h n − ℓ,ℓ = Z ∞ ( s + ty ) n − ℓ ( ty ) ℓ e − y dy ERMANENTS AND GENERALIZED DERANGEMENTS 17 and the binomial theorem for the sum.5.1. ( sI + tJ ) ∨ ℓ revisited. Now we have an alternative proof of Theorem 4.1.
Lemma 5.3.
Let I and J be ℓ -subsets of [ n ] with dist JS (I , J) = k . Then per ( sI + tJ ) IJ = P ℓ,k ( s, t ) . Proof.
Now | I ∩ J | = ℓ − k so the submatrix ( sI + tJ ) IJ is permutationally equivalentto the ℓ × ℓ matrix with ℓ − k entries s + t on its main diagonal and t ’s elsewhere,i.e., to the matrix M ℓ,k . Hence, by definition of P ℓ,k ( s, t ), equation (10), we have theresult.Thus, the expansion in the Johnson basis:( sI + tJ ) ∨ ℓ = X k h ℓ − k,k ( s, t ) JS nℓk (16) Proof.
Let I and J be ℓ -subsets of [ n ] with Johnson-distance k . By definition, theIJ entry of the LHS of equation (16) equals the permanent of the submatrix fromrows I and columns J, per( sI + tJ ) IJ = P ℓ,k ( s, t ) = h ℓ − k,k ( s, t ), by the above Lemmaand Theorem 5.2, JS (I , J) = k , theonly nonzero contribution comes from the JS nℓk term. This yields h ℓ − k,k ( s, t ) × h ℓ − k,k ( s, t ) as required.5.2. Elementary subgraphs and permanents.
There is an approach to perma-nents of sI + tJ via elementary subgraphs, based on that of Biggs [2] for determinants.An elementary subgraph (see [2, p. 44]) of a graph G is a spanning subgraph of G allof whose components are 0, 1, or 2-regular, i.e., all of whose components are isolatedvertices, isolated edges, or cycles of length j ≥ K ( ℓ ) n be a copy of the complete graph K n with vertex set [ n ] in which the first n − ℓ vertices [ n − ℓ ] = { , , . . . , n − ℓ } are distinguished . We may now considerthe matrix M n,ℓ as the weighted adjacency matrix of K ( ℓ ) n in which the weights ofthe distinguished vertices are s + t , with all undistinguished vertices and all edgesassigned a weight of t .Let E be an elementary subgraph of K ( ℓ ) n . Then we describe E as having d ( E )distinguished isolated vertices and c ( E ) cycles. The weight of E , wt( E ), is definedas: wt( E ) = ( s + t ) d ( E ) t n − d ( E ) , (17) a homogeneous polynomial of degree n .This leads to an interpretation/derivation of P n,ℓ ( s, t ) as the permanent per ( M n,ℓ ). Theorem 5.4.
We have the expansion in elementary subgraphs P n,ℓ ( s, t ) = X E c ( E ) wt( E ) . Proof.
Assign weights to the components of E as follows:each distinguished isolated vertex will have weight s + t ;each undistinguished isolated vertex will have weight t ;each isolated edge will have weight t ;and each j -cycle, j ≥
3, will have weight t j .To obtain wt( E ) in agreement with equation (17) we form the product of theseweights over all components in E . The proof then follows along the lines of Propo-sition 7.2 of [2, p.44], slightly modified to incorporate isolated vertices and with de-terminant, ‘det’, replaced by permanent, ‘per ’, ignoring the minus signs. Effectively,each term in the permanent expansion thus corresponds to a weighted elementarysubgraph E of the weighted K ( ℓ ) n .An example with n = 3 is on the last page of the Appendix.5.3. Associated polynomials and some asymptotics.
Thinking of s and t asparameters, we define the associated polynomials Q n ( x ) = n X ℓ =0 (cid:18) nℓ (cid:19) x ℓ P n,ℓ . As in the proof of Q n ( x ) = Z ∞ ( s + ty + xty ) n e − y dy = X j (cid:18) nj (cid:19) s j (1 + x ) n − j t n − j ( n − j )!= n ! X j s j (1 + x ) n − j t n − j j ! . (18)Comparing with equation (11), we have ERMANENTS AND GENERALIZED DERANGEMENTS 19
Proposition 5.5. Q n ( x ) = n X ℓ =0 (cid:18) nℓ (cid:19) x ℓ P n,ℓ ( s, t ) = P n, ( s, t + xt ) . And we have
Proposition 5.6. As n → ∞ , for x = − , Q n ( x ) ∼ t n (1 + x ) n n ! e s/ ( t + tx ) with the special cases Q n ( −
1) = s n Q n (0) = P n, ∼ t n n ! e s/t Q n (1) = X ℓ (cid:18) nℓ (cid:19) P n,ℓ ∼ (2 t ) n n ! e s/ (2 t ) Proof.
From equation (18) Q n ( x ) = n ! X j s j (1 + x ) n − j t n − j j != t n (1 + x ) n n ! n X j =0 j ! (cid:18) s/t x (cid:19) j from which the result follows.6. Generalized derangement numbers
The formula (1) is suggestive of the derangement numbers (see, e.g., [4, p. 180]), d n = n ! n X j =0 ( − j j ! . This leads to
Definition.
A family of numbers, depending on n and ℓ , arising as the valuesof P n,ℓ ( s, t ) when s and t are assigned fixed integer values, are called generalizedderangement numbers . We have seen that the assignment s = − , t = 1 produces the usual derangementnumbers when ℓ = 0. In this section, we will examine in detail the cases s = − , t =1, generalized derangements , and s = t = 1, generalized arrangements . Remark.
Topics related to this material are discussed in Riordan, [6]. The article[7] is of related interest as well.6.1.
Generalized derangements of [ n ] . To start, define D n,ℓ = P n,ℓ ( − , . Equation (14) and Proposition 3.2 give: D n,ℓ = n X j = ℓ ( − n − j (cid:18) n − ℓn − j (cid:19) j ! = ( − n − ℓ ℓ ! F (cid:18) ℓ − n, ℓ — (cid:12)(cid:12)(cid:12)(cid:12) (cid:19) . (19)Equation (11) reads per ( J − I ) = D n, = d n the number derangements of [ n ]. So we have a combinatorial interpretation of D n, .6.1.1. Combinatorial interpretation of D n,ℓ . We now give a combinatorial interpre-tation of D n,ℓ for ℓ ≥ ℓ ≥
1, recurrence (12) for P n,ℓ ( − ,
1) gives: D n,ℓ = D n,ℓ − + D n − ,ℓ − . (20)We say that a subset I of [ n ] is deranged by a permutation if no point of I is fixedby the permutation. Proposition 6.1. D n, = d n , the number of derangements of [ n ] . In general, for ℓ ≥ , D n,ℓ is the number of permutations of [ n ] in which the set { , , . . . , n − ℓ } isderanged, with no restrictions on the ℓ -set { n − ℓ + 1 , . . . , n } .Proof. For ℓ ≥ D ∗ n,ℓ denote the set of permutations in the statement of theProposition. Let E n,ℓ = | D ∗ n,ℓ | . We claim that E n,ℓ = D n,ℓ .The case ℓ = 0 is immediate. We show that E n,ℓ satisfies recurrence (20).Now let ℓ >
0. Consider a permutation in D ∗ n,ℓ . The point n is either (1) deranged,or (2) not deranged ( i.e. , fixed).(1) If n is deranged, then the ( n − ℓ + 1)-set { , , . . . , n − ℓ, n } is deranged. Byswitching n ↔ n − ℓ +1 in all permutations of D ∗ n,ℓ we obtain a permutation in D ∗ n,ℓ − .Conversely, given any permutation of D ∗ n,ℓ − , we switch n ↔ n − ℓ + 1 to obtain a ERMANENTS AND GENERALIZED DERANGEMENTS 21 permutation in D ∗ n,ℓ where n is deranged. Hence the number of permutations in D ∗ n,ℓ with n deranged equals E n,ℓ − .(2) Here n is fixed so if we remove n from any permutation in D ∗ n,ℓ we obtain apermutation in D ∗ n − ,ℓ − . Conversely, given a permutation in D ∗ n − ,ℓ − we may include n as a fixed point to obtain a permutation in D ∗ n,ℓ with n fixed. Hence the numberof permutations in D ∗ n,ℓ with n fixed equals E n − ,ℓ − .Combining the above two paragraphs shows that E n,ℓ satisfies recurrence (20).And a quick check. D n,n = n !there being no restrictions at all in the combinatorial interpretation, in agreementwith (19) for ℓ = n . Example.
When n = 3 we have d = D , = 2 corresponding to the 2 permutationsof [3] in which { , , } is moved: 231 , D , = 3 corresponding to the 3 permutations of [3] in which { , } is moved:213 , , D , = 4 corresponding to the 4 permutations of [3] in which { } is moved:213 , , , D , = 3! = 6 corresponding to the 3 permutations of [3] in which ∅ is moved:123 , , , , , D n,ℓ = n − ℓ X j =0 ( − j (cid:18) n − ℓj (cid:19) ( n − j )! . (21) Remark.
Formulation (21) may be proved directly by inclusion-exclusion on per-mutations fixing given points.
Example. D , = X j =0 ( − j (cid:18) j (cid:19) (5 − j )! = (cid:18) (cid:19) − (cid:18) (cid:19)
4! + (cid:18) (cid:19) − (cid:18) (cid:19) −
72 + 18 − . Now, from s = − t = 1 we have: D n,ℓ = ℓ X j =0 (cid:18) ℓj (cid:19) d n − j . (22)Here is a combinatorial explanation. To obtain a permutation in D ∗ n,ℓ , we firstchoose j points from { n − ℓ + 1 , . . . , n } to be fixed. Then every derangement of theremaining ( n − j ) points will produce a permutation in D ∗ n,ℓ , and there are d n − j suchderangements. Example. D , = X j =0 (cid:18) j (cid:19) d − j = (cid:18) (cid:19) d + (cid:18) (cid:19) d + (cid:18) (cid:19) d = 1 ×
44 + 2 × × . Permanents from J − I . Theorem 4.1 specializes to( J − I ) ∨ ℓ = min( ℓ,n − ℓ ) X k =0 D ℓ,k JS nℓk . This can be written using the hypergeometric form:( J − I ) ∨ ℓ = min( ℓ,n − ℓ ) X k =0 ( − ℓ − k k ! F (cid:18) k − ℓ, k — (cid:12)(cid:12)(cid:12)(cid:12) (cid:19) JS nℓk . with spectrum eigenvalue ( − ℓ F (cid:18) α − ℓ, − α + n − ℓ + 1— (cid:12)(cid:12)(cid:12)(cid:12) (cid:19) occurring with multiplicity (cid:18) nα (cid:19) − (cid:18) nα − (cid:19) by Corollary 4.3 and Proposition 4.2.The entries of ( J − I ) ∨ ℓ are from the set of numbers D n,ℓ . For the spectrum, startwith α = 0. From equation (19), we have( − ℓ F (cid:18) − ℓ, n − ℓ + 1— (cid:12)(cid:12)(cid:12)(cid:12) (cid:19) = 1( n − ℓ )! D n,n − ℓ ERMANENTS AND GENERALIZED DERANGEMENTS 23 As α increases, we see that the spectrum consists of the numbers( − α ( n − ℓ − α )! D n − α,n − ℓ − α Think of moving in the derangement triangle, as in the Appendix, starting fromposition n, n − ℓ , rescaling the values by the factorial of the column at each step.Then the eigenvalues are found by successive knight’s moves, up 2 rows and onecolumn to the left, with alternating signs. Example.
For n = 5, ℓ = 3, we have( J − I ) ∨ = with characteristic polynomial λ ( λ −
32) ( λ + 3) Remark.
Except for ℓ = 2, the coefficients in the expansion of ( J − I ) ∨ ℓ in theJohnson basis will be distinct. Thus the Johnson basis itself can be read off directlyfrom ( J − I ) ∨ ℓ . In this sense, the centralizer algebra of the action of the symmetricgroup on ℓ -sets is determined by knowledge of the action of just J − I on ℓ -sets.6.2. Generalized arrangements of [ n ] . Given [ n ], 0 ≤ j ≤ n , a j -arrangement of[ n ] is a permutation of a j -subset of [ n ]. The number of j -arrangements of [ n ] is A ( n, j ) = n !( n − j )! . Note that there is a single 0-arrangement of [ n ], from the empty set. Define A n,ℓ = P n,ℓ (1 , A n,ℓ = n X j = ℓ (cid:18) n − ℓn − j (cid:19) j ! = ℓ ! F (cid:18) ℓ − n, ℓ — (cid:12)(cid:12)(cid:12)(cid:12) − (cid:19) . (23)Now define a n = A n, so a n = per ( I + J ) = n X j =0 n !( n − j )! = n X j =0 A ( n, j )is the total number of j -arrangements of [ n ] for j = 0 , , . . . , n . Thus we have acombinatorial interpretation of A n, .6.2.1. Combinatorial interpretation of A n,ℓ . We now give a combinatorial interpre-tation of A n,ℓ for ℓ ≥ ℓ ≥
1, recurrence (12) for P n,ℓ (1 ,
1) gives: A n,ℓ = A n,ℓ − − A n − ,ℓ − . (24) Proposition 6.2. A n, = a n , the total number of arrangements of [ n ] . In general,for ℓ ≥ , A n,ℓ is the number of arrangements of [ n ] which contain { , , . . . , ℓ } .Proof. For ℓ ≥
0, let A ∗ n,ℓ denote the set of arrangements of [ n ] which contain [ ℓ ].With [0] = ∅ , we note that A ∗ n, is the set of all arrangements. Let B n,ℓ = | A ∗ n,ℓ | . Weclaim that B n,ℓ = A n,ℓ .The initial values with ℓ = 0 are immediate. We show that B n,ℓ satisfies recurrence(24).Consider A ∗ n,ℓ − . Let A ∈ A ∗ n,ℓ − , so A is an arrangement of [ n ] containing [ ℓ − ℓ = 1, then A ∈ A ∗ n, is any arrangement. Now either ℓ ∈ A or ℓ A .If ℓ ∈ A , then A ∈ A ∗ n,ℓ , and so the number of arrangements in A ∗ n,ℓ − which contain ℓ equals B n,ℓ .If ℓ A , then by subtracting 1 from all parts of A which are ≥ ℓ + 1 we obtainan arrangement of [ n −
1] which contains [ ℓ − A ∗ n − ,ℓ − .Conversely, given an arrangement in A ∗ n − ,ℓ − , adding 1 to all parts ≥ ℓ yields anarrangement in A ∗ n,ℓ − which does not contain ℓ . Hence the number of arrangementsin A ∗ n,ℓ − which do not contain ℓ equals B n − ,ℓ − .We conclude that B n,ℓ − = B n,ℓ + B n − ,ℓ − , hence the result. ERMANENTS AND GENERALIZED DERANGEMENTS 25
Example.
When n = 3 we have a = A , = 16 corresponding to the 16 arrange-ments of [3]: [ ] , , , , , , , , , , , , , , , A , = 11 corresponding to the 11 arrangements of [3] which contain { } :1 , , , , , , , , , , A , = 8 corresponding to the 8 arrangements of [3] which contain { , } :12 , , , , , , , A , = 3! = 6 corresponding to the 6 arrangements of [3] which contain { , , } : 123 , , , , , P n,ℓ ( s, t ) = n X j = ℓ A ( j, ℓ ) A ( n − ℓ, j − ℓ ) s n − j t j . With s = t = 1 this gives: A n,ℓ = n X j = ℓ A ( j, ℓ ) A ( n − ℓ, j − ℓ ) . (25)Here is a combinatorial explanation:For any j ≥ ℓ , to obtain a j -arrangement A of [ n ] containing [ ℓ ] we may place the ℓ points of { , , . . . , ℓ } into these j positions in A ( j, ℓ ) ways. Then the remaining( j − ℓ ) positions in A can be filled in by a ( j − ℓ )-arrangement of the unused ( n − ℓ )points in A ( n − ℓ, j − ℓ ) ways. Hence (25). Example. A , = X j =2 A ( j, A (3 , j − A (2 , A (3 ,
0) + A (3 , A (3 ,
1) + A (4 , A (3 ,
2) + A (5 , A (3 , × × × ×
6= 2 + 18 + 72 + 120 = 212 . Finally, from s = 1 and t = 1 we have: A n,ℓ = ℓ X j =0 ( − j (cid:18) ℓj (cid:19) a n − j . Example. A , = X j =0 ( − j (cid:18) j (cid:19) a − j = (cid:18) (cid:19) a − (cid:18) (cid:19) a + (cid:18) (cid:19) a = 1 × − ×
65 + 1 ×
16 = 326 −
130 + 16 = 212 . Permanents from I + J . Theorem 4.1 specializes to( I + J ) ∨ ℓ = min( ℓ,n − ℓ ) X k =0 A ℓ,k JS nℓk . This can be written using the hypergeometric form:( I + J ) ∨ ℓ = min( ℓ,n − ℓ ) X k =0 k ! F (cid:18) k − ℓ, k — (cid:12)(cid:12)(cid:12)(cid:12) − (cid:19) JS nℓk . with spectrum eigenvalue F (cid:18) α − ℓ, − α + n − ℓ + 1— (cid:12)(cid:12)(cid:12)(cid:12) − (cid:19) occurring with multiplicity (cid:18) nα (cid:19) − (cid:18) nα − (cid:19) by Corollary 4.3 and Proposition 4.2. Example.
For n = 5, ℓ = 3, we have( I + J ) ∨ =
16 11 11 11 11 8 11 11 8 811 16 11 11 8 11 11 8 11 811 11 16 8 11 11 8 11 11 811 11 8 16 11 11 11 8 8 1111 8 11 11 16 11 8 11 8 118 11 11 11 11 16 8 8 11 1111 11 8 11 8 8 16 11 11 1111 8 11 8 11 8 11 16 11 118 11 11 8 8 11 11 11 16 118 8 8 11 11 11 11 11 11 16
ERMANENTS AND GENERALIZED DERANGEMENTS 27 with characteristic polynomial( λ − λ − ( λ − . As for the case of derangements, the Johnson basis can be read off directly from thematrix ( I + J ) ∨ ℓ . Acknowledgment.
We would like to thank Stacey Staples for discussions aboutzeons and trace formulas. Appendix
Generalized derangement numbers and integer sequences.
The first twocolumns of the D n,ℓ triangle, D n, and D n, , give sequences A000166 and A000255 inthe On-Line Encyclopedia of Integer Sequences [5]. The comments for A000255 donot contain our combinatorial interpretation.The first two columns of the A n,ℓ triangle, A n, and A n, , give sequences A000522and A001339. The comments contain our combinatorial interpretation. The next twocolumns, A n, and A n, , gives sequences A001340 and A00134; here our combinatorialinterpretation is not mentioned in the comments.eneralized Derangement Triangles ℓ = 0 is the leftmost column.The rows correspond to n from 0 to 9.Values of D n,ℓ . Values of A n,ℓ . Exponential polynomials h n,m ( s, t ). Note that, as is common for matrix indexing,we have dropped the commas in the numerical subscripts. n = 0 h = 1 , h = t , h = 2 t , h = 6 t , h = 24 t n = 1 h = s + t , h = st +2 t , h = 2 st +6 t , h = 6 st +24 t , h = 24 st +120 t n = 2 h = s + 2 st + 2 t , h = s t + 4 st + 6 t , h = 2 s t + 12 st + 24 t h = 6 s t + 48 st + 120 t , h = 24 s t + 240 st + 720 t n = 3 h = s + 3 s t + 6 st + 6 t h = s t + 6 s t + 18 st + 24 t , h = 2 s t + 18 s t + 72 st + 120 t h = 6 s t + 72 s t + 360 st + 720 t , h = 24 s t + 360 s t + 2160 st + 5040 t n = 4 h = s + 4 s t + 12 s t + 24 st + 24 t , h = s t + 8 s t + 36 s t + 96 st + 120 t h = 2 s t + 24 s t + 144 s t + 480 st + 720 t h = 6 s t + 96 s t + 720 s t + 2880 st + 5040 t h = 24 s t + 480 s t + 4320 s t + 20160 st + 40320 t ERMANENTS AND GENERALIZED DERANGEMENTS 31 s + t t tt s + t tt t s + t tt t ✁✁✁❆❆❆ s + tts + tts + tt P , = s + 3 s t + 6 st + 6 t ℓ = 0 tt t ( s + t ) tt t ❆❆❆ ( s + t ) t tt t ( s + t ) t tt t ✁✁✁ ( s + t ) t tt t ✁✁✁❆❆❆ t s + t t tt s + t tt t t t❞ t ✁✁✁❆❆❆ s + tts + ttt t P , = s t + 4 st + 6 t ℓ = 1 t❞ t ( s + t ) t t❞ t ❆❆❆ t t❞ t ( s + t ) t t❞ t ✁✁✁ ( s + t ) t t❞ t ✁✁✁❆❆❆ t s + t t tt t tt t t t❞ ❞ ✁✁✁❆❆❆ s + ttttt t P , = 2 st + 6 t ℓ = 2 t❞ ❞ ( s + t ) t t❞ ❞ ❆❆❆ t t❞ ❞ ( s + t ) t t❞ ❞ ✁✁✁ t t❞ ❞ ✁✁✁❆❆❆ t t t tt t tt t t ❞❞ ❞ ✁✁✁ ❆❆❆ t tttt t P , = 6 t ℓ = 3 ❞❞ ❞ t ❞❞ ❞ ❆❆❆ t ❞❞ ❞ t ❞❞ ❞ ✁✁✁ t ❞❞ ❞ ✁✁✁❆❆❆ t FIGURE 2. M ,ℓ , K ( ℓ )3 , P ,ℓ and the 5 weighted elementary subgraphs of K ( ℓ )3 for ℓ = 0 , ,
2, 3.Distinguished vertices are shown bold . References [1] Bannai, E., Ito, I.,
Algebraic combinatorics I: Association schemes , Benjamin-Cummings Pub-lishing Company, 1984.[2] Biggs, N.,
Algebraic Graph Theory , 2nd. ed, Cambridge Univ. Press, 1996.[3] Cameron, P. J.,
Permutation groups , London Mathematical Society Student Texts, , Cam-bridge University Press, 1999.[4] Comtet, L., Advanced Combinatorics, the Art of Finite and Infinite Expressions , D. Reidel,Dordrecht-Holland, 1974.[5]
Online Encyclopedia of Integer Sequences , http://oeis.org/ .[6] Riordan, J., Introduction to combinatorial analysis , Princeton U. Press, 1980.[7] Stonebridge, B. R.,
Derangements of a multiset , Bull. Inst. Math. & Appl., , 1992, 47–49. Department of MathematicsSouthern Illinois UniversityCarbondale, IL. 62901, U.S.A.
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