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Some easily derivable integer sequences

Journal of Integer Sequences, Vol. 3 (2000),

Article 00.2.2

Some Easily Derivable Integer Sequences

Valery A.Liskovets

Institute of Mathematics,National Academy of Sciences,Surganov str.11

220072,Minsk,BELARUS

Email address:liskov@im.bas-net.by

Abstract

We propose and discuss several simple ways of obtaining new enumerative sequences from existing ones.For instance,the number of graphs considered up to the action of an involutory transformation is expressible as the semi-sum of the total number of such graphs and the number of graphs invariant under the involution. Another,less familiar idea concerns even-and odd-edged graphs:the di?erence between their numbers often proves to be a very simple quantity(such as n!).More than30new sequences will be constructed by these methods.

Acknowledgement.This research was supported by INTAS(Grant INTAS-BELARUS97-0093).

Mathematics Subject Classi?cation(1991):05C30,05A19

Contents

1Introduction2

1.1De?nitions,classes of graphs (2)

1.2Enumerative functions (3)

2Subtraction3

2.1Disconnection (4)

2.2Weak and strong digraphs (4)

3Involutory equivalence4

3.1Complementarity (4)

3.1.1Double connection (5)

3.1.2Self-complementarity (6)

3.1.3A combination (7)

3.2Arc reversal (7)

3.3Planar maps (7)

3.3.1Duality and re?ection (8)

3.3.2Circular objects (8)

4Even-and odd-edged graphs9

4.1Labeled graphs (9)

4.1.1Connected graphs (9)

4.1.2Connected digraphs (10)

4.1.3Symmetric relations (10)

4.1.4Oriented graphs (10)

4.1.5Strongly connected digraphs (10)

4.1.6A digression:semi-strong digraphs (11)

4.1.7Eulerian digraphs (12)

4.2Unlabeled graphs (13)

5Concluding remark13 1Introduction

New realities set up new tasks.The On-line Encyclopedia of Integer Sequences[18](in the sequel referred to as the OEIS)is a rapidly growing facility,which has been playing a more and more important role in mathematical research.To be a comprehensive reference source,the OEIS needs to include as many naturally de?ned sequences as possible.The e?orts of numerous enthusiasts have been directed towards promoting this aim.The present work has been motivated by the same goals.

A fruitful idea is to generate new sequences from known ones.To implement it,various useful trans-formations of sequences have been proposed—see[4,5,19,20].In most cases discussed hitherto,these operations transform one sequence to another.

Here we consider some other operations of a similar type but which are less general,producing new enumerative sequences for graphs from two other sequences(in most cases,as their semi-sum).The cor-responding relations between the objects being counted are very simple,and,as a rule,already known. However,they have never been analyzed systematically(this can be partially explained just by their sim-plicity:serious researchers rarely considered them as deserving an independent formulation).As we will see,our operations do result in new and interesting sequences.In a sense,they might be considered as already implicitly present in the OEIS.However,they cannot be extracted by a formal rule and thus need to be presented in the OEIS explicitly.At the same time,we should avoid trivial sequences—not all new sequences deserve to be added to the OEIS.We will return to this question in Section5.

1.1De?nitions,classes of graphs

In what follows,n denotes the order of a graph,i.e.the number of nodes(or vertices).For uniformity, we always start with the case n=1,and usually n takes all natural values.In other words,we deal with sequences(or lists)of the form[a(1),a(2),a(3),...].N denotes the number of edges(in digraphs they are usually called arcs)and if there are n nodes and N edges we will sometimes speak of an(n,N)graph.

Φstands for an arbitrary class of graphs,undirected or directed.Graphs may have loops but not multiple edges(except for planar maps).The most important speci?c classes to be considered will be denoted by the following capital Greek letters,sometimes equipped with a symbolic subscript:

?Γ(simple)undirected graphs

?Γl(undirected)graphs with loops,i.e.symmetric re?exive relations

?Γe even(i.e.eulerian)graphs

?Γm median graphs,i.e.(n,N)-graphs with N=?n(n?1)/4?edges

?Γr regular graphs with unspeci?ed degrees

?Γt(vertex-)transitive graphs

?Γc circulant graphs(i.e.Cayley graphs of cyclic groups)

??digraphs

??l(binary)relations,i.e.digraphs with loops

??e balanced digraphs(i.e.eulerian digraphs:in-degree=out-degree for any vertex)

??c circulant digraphs

??oriented graphs,i.e.antisymmetric relations

?Θtournaments,https://www.wendangku.net/doc/f68809935.html,plete oriented graphs

?Λplanar maps(order=#(edges)).

1.2Enumerative functions

Lower case letters will be used for the cardinalities(denoted by#)of subsets of labeled graphs,and the corresponding capital letters will be used for unlabeled graphs of the same kind.The most important speci?c quantities to be mentioned are the following:

?a,A=#(all graphs)in a classΦ

?c,C=#(connected graphs)

?d,D=#(disconnected graphs)

?b,B=#(doubly connected graphs)(both the graph and its complement are connected)

?s,S=#(strongly connected digraphs,or strong digraphs)

?G=#(unlabeled self-complementary undirected graphs)

?K=#(unlabeled graphs up to complementarity)

?f E,F E=#(graphs with even number of edges(or arcs))and

?f O,F O=#(graphs with odd number of edges(or arcs)),where f=a,c,...,F=A,C,...

We denote the corresponding functions for n-graphs and(n,N)-graphs by f(Φ,n),F(Φ,n)and f(Φ,n,N),F(Φ,n,N)(or merely f(n),f(n,N),etc.if the class is understood),where f and F refer to labeled and unlabeled graphs respectively The corresponding exponential generating functions(e.g.f.)for labeled graphs and ordinary generating functions(o.g.f.)for unlabeled graphs are denoted by f(z),f(n,x), f(z,x)and F(z),F(n,x),F(z,x)),where the formal variable z corresponds to n and x corresponds to N. In particular,in the labeled case,

f(z,x)= n≥1f(n,x)z n n!

(so as not to confuse f(n,x)with f(z,x)|z=n,the latter expression will not be used here).

We identify any function f(n)with the sequence of its values[f(1),f(2),f(3),...].

Sequences in[18]will be referred to by their A-numbers.(Many of these sequences were added as a result of the present paper.)

2Subtraction

We begin with the most trivial case:the subtraction method for calculating objects that do not belong to a given subset of a set.In principle,this is an inexhaustible source of new sequences,but we restrict ourselves to several interesting classes,some of which will be used in what follows.

Consider an arbitrary class of graphsΦ.Using the above notation,we have for disconnected labeled graphs,

d(Φ,n)=a(Φ,n)?c(Φ,n)(1) and for disconnected unlabeled graphs,

D(Φ,n)=A(Φ,n)?C(Φ,n)(1?) Usually c(n)is expressible in terms of a(n)and C(n)in terms of A(n),and vice versa,in one of several ways depending on the labeling type and the repetition restrictions.See for example the transformations EULERi/EULER/WEIGH for unlabeled graphs and LOG/EXP for labeled ones[4,20].Therefore d(n) (and D(n))can usually be expressed solely in terms of a(n)or c(n)(resp.,in terms of A(n)or C(n)).In any case,(1)and(1?)are much easier for calculations if both a(n)and c(n)(resp.,A(n)and C(n))have already been calculated.

2.2Weak and strong digraphs

In the directed case(including the case of relations),connected digraphs are called weakly connected in order to distinguish them from strongly connected ones.As in Section2.1we may consider two further quantities:digraphs that are not strongly connected and(weakly)connected digraphs that are not strongly connected.Only the latter quantity makes sense for tournaments,because all tournaments are weakly connected.Neither notion makes sense for balanced digraphs,in which case weakly connected digraphs are all strongly connected.

This idea is quite fruitful not only for most of the classes of digraphs de?ned above but also for example for semi-regular digraphs:ones with the same out-degree at all vertices1.

One further notion,which we will use below(4.1.6),is that of a semi-strong digraph.A digraph is called semi-strong if all its weakly connected components are strongly connected(in particular,strong digraphs are semi-strong).In the unlabeled case,moreover,one should make a distinction between(at least)two kinds of semi-strong digraphs:with or without repetitions(i.e.isomorphic components).Again,using the ordinary enumerative relationship“connected–disconnected”,one can easily count semi-strong digraphs in any class for which the number of strongly connected ones is known.

In practice,these transformations are less productive since strongly connected digraphs(especially unla-beled ones)have been counted only for few types of digraphs(see,in particular,[26,11,12]);two of them will be discussed in4.1.5.

3Involutory equivalence

Diverse involutory operations on graphs serve as a source of new sequences.

3.1Complementarity

Several interesting enumerative sequences are related to the notion of complementary graph.

Many classes of graphs contain a uniquely de?ned complete graph(for every order).In particular,complete graphs exist in the families of ordinary undirected graphsΓ,undirected graphs with loopsΓl,directed graphs ?and relations?l.This notion allows us to introduce the complement of a graph.This is the graph on the same vertices in which the edges are those not in the complete graph.

It is clear that the complement of a disconnected graph is connected.This simple assertion allows us to easily count connected graphs(of given typeΦ)whose complement is also connected and belongs to the same class.We call them doubly connected.In the labeled case their number b(Φ,n)is given by

c(Φ,n)=b(Φ,n)+d(Φ,n),

whence by(1),

b(Φ,n)=2c(Φ,n)?a(Φ,n).(2) Likewise for unlabeled graphs,

B(Φ,n)=2C(Φ,n)?A(Φ,n).(2?) Now,for labeled simple undirected graphs,

a(Γ,n)=[1,2,8,64,1024,32768,2097152,...]=A006125and

c(Γ,n)=[1,1,4,38,728,26704,1866256,...]=A001187,resulting in

b(Γ,n)=[1,0,0,12,432,20640,1635360,...]=A054913.

For labeled digraphs,

a(?,n)=[1,4,64,4096,1048576,...]=A053763and

c(?,n)=[1,3,54,3834,1027080,...]=A003027,resulting in

b(?,n)=[1,2,44,3572,1005584,...]=A054914.

For unlabeled undirected graphs,

A(Γ,n)=[1,2,4,11,34,156,1044,12346,274668,...]=A000088,

C(Γ,n)=[1,1,2,6,21,112,853,11117,261080,...]=A001349,and we obtain

B(Γ,n)=[1,0,0,1,8,68,662,9888,247492,...]=A054915.

For unlabeled undirected regular graphs,

A(Γr,n)=[1,2,2,4,3,8,6,22,26,176,...]=A005176,

C(Γr,n)=[1,1,1,2,2,5,4,17,22,167,...]=A005177and

B(Γr,n)=[1,0,0,0,1,2,2,12,18,158,...]=A054916.

For vertex-transitive graphs,

A(Γt,n)=[2,2,4,3,8,4,14,9,22,...]=A006799,

C(Γt,n)=[1,1,2,2,5,3,10,7,18,...]=A006800and

B(Γt,n)=[0,0,0,1,2,2,6,5,14,...]=A054917.

For unlabeled digraphs,

A(?,n)=[1,3,16,218,9608,1540944,...]=A000273,

C(?,n)=[1,2,13,199,9364,1530843,...]=A003085and

B(?,n)=[1,1,10,180,9120,1520742,...]=A054918.

For unlabeled(re?exive)relations,

A(?l,n)=[2,10,104,3044,291968,...]=A000595,therefore,by the EULERi transformation[20],

C(?l,n)=[2,7,86,2818,285382,...]=A054919and

B(?l,n)=[2,4,68,2592,278796,...]=A054920.

For unlabeled symmetric relations(undirected graphs with loops),

A(Γl,n)=[2,6,20,90,544,5096,79264,...]=A000666,therefore,by the EULERi transformation,

C(Γl,n)=[2,3,10,50,354,3883,67994,...]=A054921and

B(Γl,n)=[2,0,0,10,164,2670,56724,...]=A054922.

Undirected graphs with the median number of edgesΓm need a slight modi?cation of the present approach. Nothing unusual arises for orders n=4k or4k+1.However for n≡2,3(mod4),the graph and its complement have di?erent numbers of edges,namely?n(n?1)/4?and?n(n?1)/4??1.We will use a prime ′in the symbols for the latter case.Now,in order to count doubly connected median graphs,one should,

instead of doubling C(Γm,n)as in(2?),take the sum C(Γm,n)+C′(Γm,n).In other words we have

B(Γm,n)=C(Γm,n)+C′(Γm,n)?A(Γm,n).(2′) Indeed,we have C=B+D′and A′=C′+D′.By de?nition,A′counts graphs that are complementary to ones counted by A,i.e.A=A′.These equalities give(2′).

Numerically,for unlabeled undirected graphs with n nodes and N=?n(n?1)/4?edges,

A(Γm,n)=[1,1,1,3,6,24,148,1646,34040,...]=A000717,

C(Γm,n)=[1,1,1,2,5,22,138,1579,33366,...]=A001437

and by the two-parameter table A054924,

C′(Γm,n)=[1,0,0,2,5,19,132,1579,33366,...]=A054926,whence

B(Γm,n)=[1,0,0,1,4,17,122,1512,32692,...]=A054927.

Of course,such a generalization can be applied to other similar classes of graphs(for example,regular of prescribed degree).

3.1.2Self-complementarity

Next we consider various classes of graphs that are invariant with respect to complementarity.Apart from the classes mentioned in3.1.1,complementarity is applicable,e.g.,to the class of regular graphs of unspeci?ed degreesΓr,regular undirected graphs of degree(n?1)/2(n odd),median n-graphs for n(n?1) divisible by4,undirected eulerian graphsΓe of odd order,balanced digraphs?e,arbitrary tournamentsΘand regular tournamentsΘr.On the other hand,e.g.,the following classes are not invariant with respect to complementarity:undirected eulerian graphs of even order,graphs with one cycle,graphs without1-valent nodes,regular undirected graphs of a given degree(not equal to(n?1)/2),oriented graphs(except for tournaments),functional digraphs,acyclic digraphs and so on.

For a class of unlabeled graphsΦcounted by A(Φ,n),let G(Φ,n)count self-complementary graphs (i.e.graphs isomorphic to their complements).We may ask:what is the number K(Φ,n)of graphs inΦconsidered up to complementarity?

The complement of a graph looks even more natural if one deals with the pair consisting of a graph and its complement:this may be interpreted as a complete graph with edges of two colors.In these terms, K(Φ,n)means the number of edge-2-colored unlabeled complete graphs whose colors are interchangeable and both one-colored edge subgraphs belong toΦ.The answer to the last question is now very simple:

A(Φ,n)+G(Φ,n)

K(Φ,n)=

G(?c,n)=[1,0,1,0,2,0,2,0,3,0,4,...]=A049309,resulting in

K(?c,n)=[1,1,2,3,4,10,8,23,27,70,56,...]=A054930.

In the last two cases,G(n)di?er from the corresponding sequences in OEIS by additional zeros inter-spersed appropriately in order to cover all orders.

One further class of graphs worth mentioning in this respect is that of bipartite graphs;we refer to[16] for enumerative results concerning the function G for such graphs.

In general,this idea can be productively applied to a class of graphs whenever we know any two out of the three corresponding sequences.

3.1.3A combination

Somewhat more arti?cially we can apply the same approach to connected graphs,i.e.we consider the number L(n)of unlabeled connected graphs up to https://www.wendangku.net/doc/f68809935.html,plementarity clearly preserves the subclass of connected graphs whose complement is also connected.Thus formula(3)is applicable,giving rise to L(n)=(B(n)+G(n))/2,where B(n)is determined by formula(2?).Thus

A(Φ,n)?G(Φ,n)

L(Φ,n)=C(Φ,n)?

,(3R)

where G R stands for the number of self-converse digraphs and K R for the number of(unlabeled)digraphs considered up to reversing the arcs.

For digraphs,A(?,n)=A000273(see3.1.1),

G R(?,n)=[1,3,10,70,708,15224,...]=A002499and we obtain

K R(?,n)=[1,3,13,144,5158,778084,...]=A054933.

For relations,A(?l,n)=A000595,

G R(?l,n)=[2,8,44,436,7176,222368,...]=A002500and

K R(?l,n)=[2,9,74,1740,149572,48575680,...]=A029849.

For oriented graphs,

A(?,n)=[1,2,7,42,582,21480,2142288,...]=A001174,

G R(?,n)=[1,2,5,18,102,848,12452,...]=A005639and we obtain

K R(?,n)=[1,2,6,30,342,11164,1077370,...]=A054934.

3.3Planar maps

Equation(3)has a form which is intrinsic for unlabeled objects possessing an additional involutory transfor-mation.Such transformations occur in particular for geometric and topological objects like planar maps.2

3.3.1Duality and re?ection

The idea can be applied to planar maps(or maps on other surfaces)with respect to topological duality. For the number A(Φ,n)of unrooted(=unlabeled)planar maps with n edges in a class of mapsΦand the corresponding number G D(Φ,n)of self-dual maps,we have,similarly to(3),

A(Φ,n)+G D(Φ,n)

K D(Φ,n)=

,(3D+)

where the superscript+means enumeration up to orientation-preserving transformations.Now,

A+(Λ,n)=[2,4,14,57,312,2071,15030,117735,967850,8268816,...]=A006384and

G+D(Λ,n)=[0,2,0,9,0,69,0,567,0,5112,...]=A006849interspersed with0s.Hence

K+D(Λ,n)=[1,3,7,33,156,1070,7515,59151,483925,4136964,...]=A054935.

Instead of duality,let us consider re?ections.We obtain the formula

A+(Λ,n)+G ach(Λ,n)

A(Λ,n)=

180?rotation combined with the transposition of the colors.According to[14],the number of such self-dual necklaces is given by the expression

Q(n)=

h(n)+2?(n?1)/2?

k|n,k odd

φ(k)2n/k

involving the Euler totient functionφ(n).This is the sequence

Q(n)=Q(Ψ,n)=[1,1,2,2,4,5,9,12,23,34,63,...]=A007147.

At the same time,

h(n)=h(Θ,n)=[1,1,2,2,4,6,10,16,30,52,94,...]=A000016enumerates so-called vortex-free labeled tournaments(see in particular[8],p.14).It is curious to notice that there is also a sensible shift transformation of Q(n):according to[1],

Q(n)?[n2/12]?1

enumerates a class of polytopal spheres,where square brackets mean the nearest integer.Numerically this is

[0,0,0,0,1,1,4,6,15,25,52,...]=A059736.

Other speci?c examples of self-dual necklaces can be found,e.g.,in[14,17].Instead of discussing them here,we turn to an important but less familiar classΞof circular object called chord diagrams.A chord diagram is a set of chords between pairwise di?erent nodes lying on an oriented circle.Chords may intersect and their sets are considered up to an isotopy transforming the circle to itself.If no restrictions are imposed, the number of chord diagrams A+(Ξ,n)with n chords and the number of reversible(achiral)chord diagrams G ach(Ξ,n)can easily be evaluated(see details in[25,2]).The corresponding(3a)-type formula has A(Ξ,n) on the left-hand side,where A(Ξ,n)denotes the number of chord diagrams considered up to re?ection.

Numerically,

A+(Ξ,n)=[1,2,5,18,105,902,9749,127072,1915951,...]=A007769and

G ach(Ξ,n)=[1,2,5,16,53,206,817,3620,16361,...]=A018191,therefore

A(Ξ,n)=[1,2,5,17,79,554,5283,65346,966156,...]=A054499.So,for the complementary sequence of chiral chord diagrams G ch(Ξ,n)=A(Ξ,n)?G ach(Ξ,n)we obtain

G ch(Ξ,n)=[0,0,0,1,26,348,4466,61726,949795,...]=A054938.

4Even-and odd-edged graphs

Consider a speci?c type of sequence:the numbers f E(n)and f O(n)of graphs(of a given class with unspeci?ed numbers of edges)with even and odd numbers of edges.In some non-trivial cases one can easily express both numbers in terms of the numbers of the corresponding graphs.We use a formal approach based on generating functions.The formulae arising in this way are fairly uniform,but require individual proofs. The general idea(going back to[6])is to evaluate the di?erence f E(Φ,n)?f O(Φ,n)(in other words,this is a weighted enumeration of graphs,where an(n,N)-graph gets the weight(?1)N).It is clearly equal to f(Φ,n,?1)and often turns out to be a very simple function.

We also consider analogous sequences F E(n)and F O(n)for unlabeled graphs,but here fewer results have been obtained.

4.1Labeled graphs

4.1.1Connected graphs

For the classΓ,as we know,the e.g.f.of the number c(n,N)of labeled connected(n,N)-graphs satis?es the equation

c(z,x)=log(1+a(z,x)),

where the corresponding o.g.f.for n-graphs for varying N are a(n,x)=(1+x)n(n?1)/2and

c(n,x)= N c(n,N)x N(Γis dropped everywhere for simplicity).Thus a(n,?1)=0for n>1, a(1,?1)=1and a(z,?1)=z.Hence c(z,?1)=log(1+z)and

c E(n)?c O(n)=c(n,?1)=?(?1)n(n?1)!.

This is Amer.Math.Monthly problem#6673,and in[22]one can?nd another proof and a generalization to k-component graphs.We notice also that(?1)n?1(n?1)!is the M¨o bius function of the lattice of set partitions.

Finally,c E(n)+c O(n)=c(n),hence

c(Γ,n)?(?1)n(n?1)!

c E(Γ,n)=

Numerically(with c(Γ,n)=[1,1,4,38,728,26704,1866256...]=A001187,

c E(Γ,n)=[1,0,3,16,376,13292,933488,...]=A054939and

c O(Γ,n)=[0,1,1,22,352,13412,932768,...]=A054940.

4.1.2Connected digraphs

The same result is valid for(weakly)connected labeled digraphs?(see my comment in[22]);in the proof we need only use the generating function(1+x)n(n?1)instead of(1+x)n(n?1)/2.

4.1.3Symmetric relations

For the class of graphs with loopsΓl,the same proof with(1+x)n(n+1)/2instead of(1+x)n(n?1)/2results in a(z,?1)=0and c(n,?1)=0.Hence

c E(Γl,n)=c O(Γl,n)=c(Γl,n)/2

(by complementarity,this is evident for n≡1,2(mod4)).

4.1.4Oriented graphs

For oriented graphs?,we work with the polynomials a(n,x)=(1+2x)n(n?1)/2,so that a(n,?1)=(?1)n(n?1)/2.Now a(z,?1)=cos(z)+sin(z)?1and

c(?,z,?1)=log(cos(z)+sin(z)).

Therefore

c E(?,n)?c O(?,n)=[1,-2,4,-16,80,-512,3904,-34816,...],which is A000831(the expansion of (1+tan x)/(1?tan x))up to alternating signs.

c E(?,n)+c O(?,n)=c(?,n)=[1,2,20,624,55248,13982208,...]=A054941.Thus

c E(?,n)=[1,0,12,304,27664,6990848,...]=A054942and

c O(?,n)=[0,2,8,320,27584,6991360,...]=A054943.

4.1.5Strongly connected digraphs

Proposition.For labeled strong digraphs,

s E(?,n)?s O(?,n)=(n?1)!.(5)

Remark.This is the Amer.Math.Monthly problem[15]mentioned earlier without proof in[22]. Proof.Let s(n,N)=s(?,n,N).The left-hand di?erence in(5)is s(n,?1).According to[11](cf. also[26]),

s(z,x)=?log(1?v(z,x)),

where v(z,x)= n≥1v(n,x)z n/n!,v(n,x)=a(n,x)u(n,x),a(n,x)=(1+x)n(n?1)/2,

a(z,x)= n≥1a(n,x)z n/n!(hence a(n,N)=a(Γ,n,N)is the number of all labeled undirected graphs) and

u(z,x)= n≥1u(n,x)z n1+a(z,x).(6)

As we saw in4.1.1,a(n,?1)=0for n>1.Moreover,a(1,?1)=u(1,?1)=1.Therefore

v(z,?1)=z,whence s(z,?1)=?log(1?z)and s(n,?1)=(n?1)!.

and

s(?,n)?(n?1)!

s O(?,n)=

V(n)=[1,1,4,78,4960,1041872,...]=A054951.Now S O(?,n)+S E(?,n)=S W(?,n),the number of semi-strong digraphs with pairwise di?erent components.We have1+ n S W(?,n)z n= n(1+z n)S(?,n) (this series corresponds to the WEIGH transformation[4,5,20]).Therefore

S W(?,n)=[1,1,6,88,5136,1052154,...]=A054952.Thus

S O(?,n)=[1,1,5,83,5048,1047013,...]=A054953and

S E(?,n)=[0,0,1,5,88,5141,...]=A054954.

Evidently,other types of disconnected(di)graphs,labeled or unlabeled,speci?ed by the parity of the number of components are also worth considering.

4.1.7Eulerian digraphs

The next assertion is new.

Proposition.For labeled balanced digraphs,

a E(?e,n)=

a(?e,n)+n!

.(7O) For labeled Eulerian(i.e.connected balanced)digraphs,

c E(?e,n)=

c(?e,n)+(n?1)!

.(8O)

Proof.According to Theorem2of[10],the o.g.f.a(?e,n,x)of balanced digraphs can be expressed by a formula in terms of m-roots of unity,m≥n.Choosing m=n,and putting x:=?1,we have from that formula,

a(?e,n,?1)=n?n n!

1≤k=l≤n

(1?w k?l),

where w is a primitive n-root of unity.Thus

a(?e,n,?1)=n?n n!

r=1

(1?w r)n.

But r(1?w r)=n,since this is merely the polynomial(z n?1)/(z?1)evaluated at z=1.Thus,

a(?e,n,?1)=n!

This implies formulae(7E)and(7O).

Now,for connected balanced digraphs,c E(?e,n)?c O(?e,n)=c(?e,n,?1).As usual, c(?e,z,x)=log(1+a(?e,z,x)).By the above formulae,a(?e,z,?1)=z/(1?z),thus we have log(1+z/(1?z))= n≥1z n/n and c(?e,n,?1)=(n?1)!.

4.2Unlabeled graphs

Here we restrict ourselves to one class of graphs,Γ(but compare also 4.1.6).Consider the dif-ference A E(Γ,n)?A O(Γ,n).This is clearly the value at x=?1of the corresponding o.g.f. A(Γ,n,x)= N A(Γ,n,N)x N.According to the P′o lya enumeration theorem(see for example[7],(4.1.8)),

A(Γ,n,x)=Z(S(2)n,1+x,1+x2,...),

where Z(S(2)n,z1,z2,...)denotes the cycle index of the symmetric group S n in its induced action on the 2-subsets of vertices.Thus

A E(Γ,n)?A O(Γ,n)=Z(S(2)n,0,2,0,2,...).(9) We see that the right-hand side coincides with the formula(6.2.3)in[7]for the number G(Γ,n)of self-complementary graphs.Thus[23],A E(Γ,n)?A O(Γ,n)=G(Γ,n).But

A E(Γ,n)+A O(Γ,n)=A(Γ,n).Therefore

A(Γ,n)+G(Γ,n)

A E(Γ,n)=

.(10O)

So,comparing formulae(10E)and(3),we obtain the following identity:

A E(Γ,n)=K(Γ,n).

We note also that A E(Γ,n)=A O(Γ,n)=A(Γ,n)/2if n=4k+2or4k+3.

From the numerical data for A(Γ,n)and G(Γ,n)(or,instead,K(Γ,n))presented in3.1.1,one gets

A O(Γ,n)=[0,1,2,5,16,78,522,6168,137316,...]=A054960.

Similar assertions are valid for arbitrary digraphs and some other classes of graphs.

5Concluding remark

In principle,there is an easy way to obtain numerous new sequences from known https://www.wendangku.net/doc/f68809935.html,ly,if a(n)and b(n)count objects of two types,then of course their product a(n)b(n)counts ordered pairs of objects,and their sum a(n)+b(n)counts objects of their disjoint union.As a rule this can hardly be considered as a really fruitful idea:in general,such pairs and the union are unnatural.But sometimes,the term-by-term product(and,still more often,the sum)of two sequences turns out to have a natural interpretation,though possibly unexpected.In this work we encountered various sequences that can be presented as the semi-sum or sum of two other sequences.Only one sequence(namely,v(n)in4.1.5)was presented as the product of two sequences(one of which is,moreover,primitive).Several more such examples can be found in[9].As far as I know,no systematic investigations of such meaningful operations has been undertaken so far.

References

[1]B.Bagchi and B.Datta,A structure theorem for pseudomanifolds,Discr.Math.,188(1998),41–60.

[2]D.Bar-Natan,On the Vassiliev knot invariants,Topology,34(1995),423–472.

[3]E.A.Bender and E.R.Can?eld,Enumeration of connected invariant graphs,https://www.wendangku.net/doc/f68809935.html,bin.Th.,B34

(1983),268–278.

[4]M.Bernstein and N.J.A.Sloane,Some canonical sequences of integers,Linear Alg.&Its Appl.,226–

228(1995),57–72.

[5]P.J.Cameron,Some sequences of integers,Discr.Math.,75(1989),89–102.

[6]R.Frucht and F.Harary,Self-complementary generalized orbits of a permutation group,Canad.Math.

Bull.,17(1974),203–208.

[7]F.Harary,E.M.Palmer,Graphical Enumeration,Acad.Press,N.Y.(1973).

[8]D.E.Knuth.Axioms and Hulls.Lect.Notes Comput.Sci.,606,Springer-Verlag,Berlin(1992).

[9]L.M.Koganov,V.A.Liskovets and T.R.S.Walsh,Total vertex enumeration in rooted planar maps,Ars

Combin.54(2000),149–160.

[10]V.A.Liskovets,On the number of Eulerian digraphs and homogeneous tournaments,Vesci AN BSSR

(ser.?z.-mat.n.),No1(1971),22–27(in Russian).

[11]V.A.Liskovets,A contribution to the enumeration of strongly connected digraphs,Dokl.AN BSSR,

17,No12(1973),1077–1080(in Russian).

[12]V.A.Liskovets,On a general enumerative scheme for labeled graphs,Dokl.AN BSSR,21,No6(1977),

496–499(in Russian).

[13]V.A.Liskovets,Enumeration of nonisomorphic planar maps,Selecta Math.Soviet.,4(1985),304–323.

[14]E.M.Palmer and R.W.Robinson,Enumeration of self-dual con?gurations,Pacif.J.Math.,110(1984),

203–221.

[15]J.Propp,Problem#10620,Amer.Math.Monthly,104(1997),870.

[16]S.J.Quinn,Factorisation of complete bipartite graphs into two isomorphic subgraphs,Combinatorial

Mathematics VI(A.Horadam and W.D.Wallis eds.),Lect.Notes in Math.,748,Springer,Berlin(1979), 98–111.

[17]R.W.Robinson,Counting graphs with a duality property,Combinatorics(Proc.8th https://www.wendangku.net/doc/f68809935.html,b.Conf.,

Swansea,1981),Lond.Math.Soc.Lect.Notes Ser.,52(1981),156–186.

[18]N.J.A.Sloane,The On-Line Encyclopedia of Integer Sequences,published electronically at

https://www.wendangku.net/doc/f68809935.html,/～njas/sequences/

[19]N.J.A.Sloane,Help File for Superseeker,a sub-page of[18](1999).

[20]N.J.A.Sloane,Transformations of Integer Sequences,a sub-page of[18](2001).

[21]N.J.A.Sloane and S.Plou?e,The Encyclopedia of Integer Sequences,Academic Press,San Diego(1995).

[22]Solution of problem#6673,Amer.Math.Monthly,101(1994),686–687.

[23]Solution of problem#10285,Amer.Math.Monthly,103(1996),268–269.

[24]Solution of problem#10620,Amer.Math.Monthly106(1999),865–867.

[25]A.Stoimenow,On the number of chord diagrams,Discr.Math.218(2000),209–233.

[26]E.M.Wright,The number of strong digraphs,Bull.London Math.Soc.,3(1971),348–350.

A018191A029849A035512A049287A049289A049297A049309A053763A054499A054913A054914A054915 A054916A054917A054918A054919A054920A054921A054922A054924A054926A054927A054928A054929 A054930A054931A054932A054933A054934A054935A054936A054937A054938A054939A054940A054941 A054942A054943A054944A054945A054946A054947A054948A054949A054950A054951A054952A054953 A054954A054955A054956A054957A054958A054959A054960A059735A059736)

Return to Journal of Integer Sequences home page.

初中英语介词用法归纳总结

初中英语介词用法归纳总结常用介词基本用法辨析表示方位的介词：in, to, on 1. in 表示在某地范围之内。 Shanghai is/lies in the east of China. 上海在中国的东部。 2. to 表示在某地范围之外。 Japan is/lies to the east of China. 日本位于中国的东面。 3. on 表示与某地相邻或接壤。 Mongolia is/lies on the north of China. 蒙古国位于中国北边。表示计量的介词：at, for, by 1. at 表示“以……速度”“以……价格”。 It flies at about 900 kilometers an hour. 它以每小时900公里的速度飞行。 I sold my car at a high price. 我以高价出售了我的汽车。 2. for 表示“用……交换，以……为代价”。 He sold his car for 500 dollars. 他以五百元把车卖了。

注意：at表示单价(price) ，for表示总钱数。 3. by 表示“以……计”，后跟度量单位。 They paid him by the month. 他们按月给他计酬。 Here eggs are sold by weight. 在这里鸡蛋是按重量卖的。表示材料的介词：of, from, in 1. of 成品仍可看出原料。 This box is made of paper. 这个盒子是纸做的。 2. from 成品已看不出原料。 Wine is made from grapes. 葡萄酒是葡萄酿成的。 3. in 表示用某种材料或语言。 Please fill in the form in pencil first. 请先用铅笔填写这个表格。They talk in English. 他们用英语交谈。表示工具或手段的介词：by, with, on 1. by 用某种方式，多用于交通。 I went there by bus. 我坐公共汽车去那儿。 2. with表示“用某种工具”。

some和any的用法与练习题

some和 any 的用法及练习题( 一) 一、用法： some意思为：一些。可用来修饰可数名词和不可数名词，常常用于肯定句 . any 意思为：任何一些。它可以修饰可数名词和不可数名词，当修饰可数名词时要用复数形式。常用于否定句和疑问句。注意： 1、在表示请求和邀请时，some也可以用在疑问句中。 2、表示“任何”或“任何一个”时，也可以用在肯定句中。 3、和后没有名词时，用作代词，也可用作副词。二、练习题： 1.There are ()newspapers on the table. 2.Is there ( )bread on the plate. 3.Are there () boats on the river? 4.---Do you have () brothers ?---Yes ,I have two brothers. 5.---Is there () tea in the cup? --- Yes,there is () tea in it ,but there isn’t milk. 6.I want to ask you() questions. 7.My little boy wants ()water to drink. 8.There are () tables in the room ,but there aren’t ( )chairs. 9.Would you like () milk? 10.Will you give me () paper? 复合不定代词的用法及练习一.定义: 由 some，any，no，every 加上 -body,-one,-thing,-where构成的不定代词，叫做复合不定代词 . 二. 分类： 1.指人：含 -body 或 -one 的复合不定代词指人 . 2.含-thing 的复合不定代词指物。 3.含-where 的复合不定代词指地点。三：复合不定代词： somebody =someone某人 something 某物，某事，某东西 somewhere在某处，到某处 anybody= anyone 任何人，无论谁 anything任何事物，无论何事，任何东西 anywhere 在任何地方 nobody=no one 无一人 nothing 无一物，没有任何东西 everybody =everyone每人，大家，人人 everything每一个事物，一切 everywhere 到处 , 处处 , 每一处

初中英语名词练习题与详解

名词判断对错 1、［误］ Please give me a paper. ［正］ Please give me a piece of paper. ［析］不要认为可以数的名词就是可数名词，这种原因是对英语中可数与不可数名词的概念与中文中的能数与不能数相混淆了，所以造成了这样的错误，因paper 在英语中是属于物质名词一类，是不可数名词。而不可数名词要表达数量时，要用与之相关的量词来表达，如： two pieces of paper. 2、［误］ Please give me two letter papers. ［正］ Please give me two pieces of letter paper. ［析］ paper 作为纸讲是不可数名词，而作为报纸、考卷、文章讲时则是可数名词，如：Each student should write a paper on what he has learnt. 3、［误］ My glasses is broken. ［正］ My glasses are broken. 4、［误］ I want to buy two shoes. ［正］ I want to buy two pairs of shoes. ［析］英语中glasses—眼镜， shoes—鞋， trousers—裤子等由两部分组成的名词一般要用复数形式。如果要表示一副眼镜应用 a pair of glasses 而这时的谓语动词应与量词相一致。如：5、This pair of glasses is very good. ［误］ May I borrow two radioes? ［正］ May I borrow two radios? ［析］以o 结尾的名词大都是用加es 来表示其复数形式，但如果 o 前面是一个元音字母或外来语时则只加s 就可以了。这样的词有zoo— zoos,piano—pianos. 6、［误］ This is a Mary's dictionary. ［正］ This is Mary's dictionary. ［析］如名词前有指示代词this, that, these those, 及其他修饰词our,some, every, which,或所有格时，则不要再加冠词。 7、［误］ There are much people in the garden. ［正］ There are many people in the garden. ［析］可数名词前应用 many, few, a few, a lot of 来修饰，而 people 是可数名词，而且是复数名词，如： The people are planting trees here. 8、［误］ I want a few water. ［正］ I want a little water. ［析］不可数名词前可以用 a little, little, a lot of, some来修饰，但不可用many,few 来修饰。 9、［误］ Thank you very much. Y our family is very kind to me. ［正］ Thank you very much. Y our family are very kind to me. 10、［误］ Tom's and Mary's family are waiting for us. ［正］ Tom's and Mary's families are waiting for us. 11、［误］ I'm sorry . I have to go. Tom's families are waiting for me. ［正］ I'm sorry. I have to go. Tom's family are waiting for me. ［析］集合名词如果指某个集合的整体，则应视为单数，如指某个集合体中的个体则应视为复数。如 :My family is a big family. When I came in, Tom's family were watching TV. 即汤姆一家人正在看电视。这样的集合名词有：family class, team 等。

初中英语介词用法总结

初中英语介词用法总结介词(preposition)：也叫前置词。在英语里，它的搭配能力最强。但不能单独做句子成分需要和名词或代词（或相当于名词的其他词类、短语及从句）构成介词短语，才能在句中充当成分。介词是一种虚词，不能独立充当句子成分，需与动词、形容词和名词搭配，才能在句子中充当成分。介词是用于名词或代词之前，表示词与词之间关系的词类，介词常与动词、形容词和名词搭配表示不同意义。介词短语中介词后接名词、代词或可以替代名词的词（如：动名词v-ing）.介词后的代词永远为宾格形式。介词的种类：（1）简单介词：about, across, after, against, among, around, at, before, behind, below, beside, but, by, down, during, for, from, in, of, on, over, near, round, since, to, under, up, with等等。（2）合成介词：inside, into, outside, throughout, upon, without, within （3）短语介词：according to, along with, apart from, because of, in front of, in spite of, instead of, owing to, up to, with reguard to （4）分词介词：considering, reguarding, including, concerning 介词短语：构成介词+名词We go to school from Monday to Saturday. 介词+代词Could you look for it instead of me? 介词+动名词He insisted on staying home. 介词+连接代/副词I was thinking of how we could get there. 介词+不定式/从句He gives us some advice on how to finish it. 介词的用法：一、介词to的常见用法 1.动词+to a)动词+ to adjust to适应, attend to处理；照料, agree to赞同,

some和any的用法

some和any的用法：（1）两者修饰可数单数名词，表某一个；任何一个；修饰可数复数名词和不可数名词，表一些；有些。〔2）一般的用法：some用于肯定句；any用于疑问句，否定句或条件句。 I am looking for some matches. Do you have any matches? I do not have any matches. （3）特殊的用法： (A) 在期望对方肯定的回答时，问句也用some。 Will you lend me some money? (=Please lend me some money.) (B) any表任何或任何一个时，也可用于肯定句。 Come any day you like. （4）some和any后没有名词时，当做代名词，此外两者也可做副词。 Some of them are my students.〔代名词） Is your mother any better?（副词） 3. many和much的用法：（1）many修饰复数可数名词，表许多； much修饰不可数名词，表量或程度。 He has many friends, but few true ones. There hasn't been much good weather recently. （2）many a: many a和many同义，但语气比较强，并且要与单数名词及单数形动词连用。 Many a prisoner has been set free. (=Many prisoners have been set free.) （3）as many和so many均等于the same number of。前有as, like时, 只用so many。 These are not all the books I have. These are as many more upstairs.

some与any的用法区别教案资料

s o m e与a n y的用法区别

some与any的用法区别一、一般说来，some用于肯定句，any用于否定句和疑问句。例如： She wants some chalk. She doesn’t want any chalk. Here are some beautiful flowers for you. Here aren’t any beautiful flowers. 二、any可与not以外其他有否定含义的词连用，表达否定概念。例如： He never had any regular schooling. In no case should any such idea be allowed to spread unchecked. The young accountant seldom (rarely, hardly, scarcely) makes any error in his books. I can answer your questions without any hesitation. 三、any可以用于表达疑问概念的条件句中。例如： If you are looking for any stamps, you can find them in my drawer. If there are any good apples in the shop, bring me two pounds of them. If you have any trouble, please let me know. 四、在下列场合，some也可用于疑问句。 1、说话人认为对方的答复将是肯定的。例如： Are you expecting some visitors this afternoon?（说话人认为下午有人要求，所以用some）

50套初中英语数词

50套初中英语数词一、初中英语数词 1．We throw rubbish every year. A. ton of B. tons C. tons of D. a ton of 【答案】 C 【解析】【分析】句意：我们每年扔大量的垃圾。ton，吨，前面没有具体数字，因此用tons of，大量的，故选C。【点评】考查固定搭配，注意平时识记。 2．Two students to the opening ceremony last Friday. A. hundreds; were invited B. hundred; were invited C. hundreds of; invited 【答案】 B 【解析】【分析】句意：上周五有200名学生被邀请参加开幕式。根据题干中的two与选项中的hundred可知此题考查确切数量的表达方式，hundred要用单数形式；students与invite存在动宾关系，此处要用被动语态，由last Friday，可知要用一般过去时，故选B。【点评】考查数量的表达方式以及被动语态。注意确切数量与不确切数量在表达上的不同。 3．—Do you know the boy is sitting next to Peter？ —Yes. He is Peter's friend. They are celebrating his birthday. A. who; ninth B. that; nine C. which; ninth 【答案】 A 【解析】【分析】句意：——你知道那个坐在彼得旁边的男孩吗?——是的。他是彼得的朋友。他们正在庆祝他的九岁生日。分析句子结构可知，第一空所在句子是定语从句，先行词是人，连接词在从句中作主语，所以应该用who/that引导，which连接定语从句时先行词应该是物，故排除C；nine九，基数词；ninth第九，序数词；第二空根据空后的birthday为名词单数可知，此处需要序数词，表示某人几岁生日应该用序数词表示第几个生日，故选A。【点评】考查定语从句的连接词的辨析和序数词。注意区别定语从句的连接词的使用原则，理解单词词义。 4．There are ________________ months in a year. My birthday is in the ________________ month. A. twelve; twelve B. twelfth; twelfth C. twelve; twelfth D. twelfth; twelve 【答案】 C 【解析】【分析】句意：一年有12个月，我的生日在第12个月。名词复数months前是基数词，twelve是基数词，the定冠词后是序数词，twelfth是序数词，故选C。【点评】考查数词，注意名词复数前是基数词，定冠词后是序数词的用法。

some和any的用法

some和any的用法 1.some adj.一些；某些；某个pron. 某些；若干；某些人 a.adj. some可以修饰可数名词或不可数名词，意为“某些”。 Some people are playing football. (some+可数名词) I ate some bread. (some+不可数名词) b.adj. some后面可以修饰可数名词的单数，意为“某(个)”。 Some day you will know. (some+可数名词的单数) 有一天你会知道的。 Some student cheated in the exam.(some+可数名词的单数) 有个学生考试作弊。对比：Some students cheated in the exam.有些学生考试作弊。 c.pron. some此时作代词，后面不需要再加名词就可以表示“有些(人)”的意思。 All students are in the classroom, and some are doing their homework. d.pron. some作代词，意为“若干(…)”。 There are 10 apples on the table. You can take some. 桌上10个苹果，你可以拿走一些。 2.any adj.任何的；所有的pron.任何一个；任何 a.adj. any可以修饰可数名词和不可数名词，意为“任何的，所有的任何一（…）”。（用于否定意义的陈述句、疑问句、条件状语从句if中） Do you have any ideas?(any+可数名词复数)(疑问句) 你有什么想法吗？ I don’t have any bread.(any+不可数名词)(否定意义的陈述句) Please tell me if you have any problem.(if引导的条件状语从句) b.any后面可以加可数名词的单数，意为“任何一（…）”。 Any error would lead to failure.(any+可数名词单数) 任何（一个）错误都会导致失败。 c.pron. any此时作代词，与some里面c点的用法相似，只是表示这个意义的时候，any多用于否定句和疑问句中。比较：There are 10 apples on the table. You can take some. 桌子上有10个苹果，你可以拿走一些。 There are 10 apples on the table, but you can’t take any. 桌子上有10个苹果，但是你不能拿。 There are some apples on the table.桌上有些苹果。 There aren’t any apples on the table.桌上没有苹果。由此，把陈述句变为否定句/一般疑问句的时候，要把some改成any。思考：some只用于肯定句，any只用于否定句和疑问句中吗吗？不一定，要看句子本身想表达的意思。 1.some可以用于肯定句和疑问句中。在表示请求、邀请、提建议等带有委婉语气的疑问句中，用some表示说话人希望得到肯定的回答。例如： Would you like some coffee?你想喝咖啡吗? 这里用some而不用any，是因为说话人期待得到对方肯定的回答。（因此Would you like…?你想要…吗?这个句型中多用some而不用any）比较： Do you have any books?这里用any而不用some，说明这只是因为这只是纯粹的疑问。

初中英语最全英语介词用法

表示方位的介词：in, to, on 1. in 表示在某地范围之内。如：Shanghai is/lies in the east of China.上海在中国的东部。 2. to 表示在某地范围之外。如：Japan is/lies to the east of China. 日本位于中国的东面。 3. on 表示与某地相邻或接壤。如：Mongolia is/lies on the north of China.蒙古国位于中国北边。表示计量的介词：at, for, by 1. at表示“以……速度”“以……价格” 如：It flies at about 900 kilometers a hour.它以每小时900公里的速度飞行。 I sold my car at a high price. 我以高价出售了我的汽车。 2. for表示“用……交换，以……为代价”如：He sold his car for 500 dollars. 他以五百元把车卖了。注意：at表示单价(price) ，for表示总钱数。

3. by表示“以……计”，后跟度量单位如：They paid him by the month. 他们按月给他计酬。 Here eggs are sold by weight. 在这里鸡蛋是按重量卖的。表示材料的介词：of, from, in 1. of成品仍可看出原料如：This box is made of paper. 这个盒子是纸做的。 2. from成品已看不出原料如：Wine is made from grapes. 葡萄酒是葡萄酿成的。 3. in 表示用某种材料或语言如：Please fill in the form in pencil first. 请先用铅笔填写这个表格。 They talk in English. 他们用英语交谈。注意：in指用材料，不用冠词；而with指用工具，要用冠词。如：draw in pencil/draw with a pencil 表示工具或手段的介词：by, with, on 1、by用某种方式，多用于交通如by bus乘公共汽车，by e-mail. 通过电子邮件。

人教版初中英语短语大全最全

人教版初中英语短语大全 1)be back/in/out 回来/在家/外出 2)be at home/work 在家/上班 3)be good at 善于，擅长于 4)be careful of 当心，注意，仔细 5)be covered with 被……复盖 6)be ready for 为……作好准备 7)be surprised (at) 对……感到惊讶 8)be interested in 对……感到举 9)be born 出生 10)be on 在进行，在上演，（灯）亮着 11)be able to do sth. 能够做…… 12)be afraid of (to do sth. that…)害怕……(不敢做……，恐怕……) 13)be angry with sb. 生(某人)的气 14)be pleased (with) 对……感到高兴(满意) 15)be famous for 以……而著名 16)be strict in (with) (对工作、对人)严格要求 17)be from 来自……，什么地方人 18)be hungry/thirsty/tired 饿了/渴了/累了 19)be worried 担忧 20)be (well) worth doing (非常)值得做…… 21)be covered with 被……所覆盖…… 22)be in (great) need of (很)需要 23)be in trouble 处于困境中 24)be glad to do sth. 很高兴做…… 25)be late for ……迟到 26)be made of (from) 由……制成 27)be satisfied with 对……感到满意 28)be free 空闲的，有空 29)be (ill) in bed 卧病在床 30)be busy doing (with) 忙于做……(忙于……) (二)由come、do、get、give、go、have、help、keep、make、looke、put、set、send、take、turn、play等动词构成的词组 1)come back 回来 2)come down 下来 3)come in 进入，进来 4)come on 快，走吧，跟我来 5)come out出来 6)come out of 从……出来 7)come up 上来 8)come from 来自…… 9)do one's lessons/homework 做功课/回家作业 10)do more speaking/reading 多做口头练习/朗读 11)do one's best 尽力 12)do some shopping (cooking reading, clean ing)买东西(做饭菜，读点书，大扫除) 13)do a good deed (good deeds)做一件好事(做好事) 14)do morning exercises 做早操 15)do eye exercises 做眼保健操 16)do well in 在……某方面干得好 17)get up 起身 18)get everything ready 把一切都准备好 19)get ready for (=be ready for) 为……作好准备 20)get on (well) with 与……相处(融洽) 21)get back 返回 22)get rid of 除掉，去除 23)get in 进入，收集 24)get on/off 上/下车 25)get to 到达 26)get there 到达那里 27)give sb. a call 给……打电话 28)give a talk 作报告 29)give a lecture (a piano concert)作讲座(举行钢琴音乐会) 30)give back 归还，送回 31)give……some advice on 给……一些忠告 32)give lessons to 给……上课 33)give in 屈服 34)give up 放弃 35)give sb. a chance 给……一次机会 36)give a message to……给……一个口信 37)go ahead 先走，向前走，去吧，干吧 38)go to the cinema 看电影 39)go go bed 睡觉(make the bed 整理床铺) 40)go to school (college) 上学(上大学) 41)go to (the) hospital 去医院看病

初中英语介词用法归纳整理

初中英语介词用法归纳整理表示时间的介词 at：用于表示时刻，时间的某一点。 on:用于星期，某天，某一天的上午，下午，晚上指具体的某一天时，一律用on in:用于表示周，月，季节，年，泛指上午，下午，晚上 before:在...之前 after:在...之后 by:在....前时间截止到... untiltill：直到.....为止 for：达...之久表示过了多少时间 during：在....期间 through：一直..从开始到结束 from：从...起时间 since:自从...以来表示从以前某时一直到现在仍在继续 in：过...后未来时间 within：不超过...的范围表示场所，方向的介词 at :在某地点表示比较狭窄的场所 in:在某地表示比较宽敞的场所 on:在...上面，有接触面 above：在...上方 over：在...正上方，是under的反义词 under：在..下面，在...之内 below :在...下方不一定是正下方

near：近的，不远的 by:在...的旁边，比near的距离要近 between：在两者之间 among：在三者或者更多的之中 around：环绕，在...的周围，在....的四周 in front of:在...的前面 behind：在...后边 in：在..之内，用于表示静止的位置 into：进入 out of :和into一样，也表示有一定的运动方向 along：沿着 across：横过平面物体 through：贯通，通过 to :达到..地点目的地或方向 for：表示目的，为了..... from：从...地点起其他介词 with：和..在一起; 具有，带有; 用某种工具或方法 in：表示用什么材料例如：墨水，铅笔等或用什么语言。表示衣着.声调特点时，不用with而用in。 by:通过...方法，手段 of:属于...的，表示...的数量或种类 from：来自某地，某人，以...起始 without：没有，是with的反义词 like :像...一样

some和any地用法

(1)some和any 的用法： some一般用于肯定句中,意思是“几个”、“一些”、“某个”作定语时可修饰可数名词或不可数名词。如：I have some work to do today. (今天我有些事情要做)/ They will go there some day.(他们有朝一日会去那儿) some 用于疑问句时,表示建议、请求或希望得到肯定回答。如：Would you like some coffee with sugar?(你要加糖的咖啡吗？) any 一般用于疑问句或否定句中,意思是“任何一些”、“任何一个”,作定语时可修饰可数或不可数名词。如：They didn’t have any friends here. (他们在这里没有朋友)/ Have you got any questions to ask?(你有问题要问吗？) any 用于肯定句时,意思是“任何的”。Come here with any friend.(随便带什么朋友来吧。) (2)no和none的用法： no是形容词,只能作定语表示,意思是“没有”,修饰可数名词(单数或复数)或不可数名词。如：There is no time left. Please hurry up.(没有时间了,请快点) / They had no reading books to lend.(他们没有阅读用书可以出借) none只能独立使用,在句子中可作主语、宾语和表语,意思是“没有一个人(或事物)”,表示复数或单数。如：None of them is/are in the

classroom.(他们当中没有一个在教室里) / I have many books, but none is interesting.(我有很多的书,但没有一本是有趣的) (3)all和both的用法： all指三者或三者以上的人或物,用来代替或修饰可数名词；也可用来代替或修饰不可数名词。 both指两个人或物,用来代替或修饰可数名词。all和both在句子中作主语、宾语、表语、定语等。如：I know all of the four British students in their school.(他们学校里四个英国学生我全认识) / --Would you like this one or that one? –Both.(你要这个还是那个？两个都要。) all和both既可以修饰名词(all/both+(the)+名词),也可以独立使用,采用“all/both + of the +名词(复数)”的形式,其中的of 可以省略。如：All (of) (the) boys are naughty.(是男孩都调皮) (4)every和each用法： every是形容词,只能作定语修饰单数名词,意思是“每一个”,表示整体概念； each是形容词、代词,可用作主语、宾语、定语等,意思是“每个”或者“各个”,表示单个概念；each可以放在名词前,可以后跟of 短语,与动词同时出现时要放在“be动词、助动词、情态动词”之后或者行为动词之前 every和each都用作单数理解,但是下文中既可以用单数的代词(如he/him/his)也可以用复数的代词(如they/them/their)替代。如：Every one of the students in his class studies very hard.(他班上每个学生学

some,any,one ones those that的区别和用法

some和any的区别和用法要表示"一些"的意思，可用some, any。 some 是肯定词，常用于肯定句；any是非肯定词，常用于否定句或疑问句。例如： There are some letters for me. There aren''t any letters for me. Are there any letters for me? I seldom get any sleep these days. any也常用于条件分句以及带有否定含义的句子中： If you have any trouble, please let me know. 如果你有任何麻烦，请让我知道。 I forgot to ask for any change. 我忘了要一些零钱。当说话人期待肯定回答时，some也可用于疑问句, 比如当说话人期待来信时，他可以问道：Are there some letters for me? 当购物时向售货员提问或者主人向客人表示款待时，也可在疑问句中用some: Could I have some of these apples? some和any 既可以修饰可数名词又可以修饰不可数名词，some常用在肯定句中，而any则常用在否定和疑问句中。因此 some和any 的用法主要是考虑用在肯定句、疑问句还是否定句中，与名词的可数与否无关。 some意为“一些”，可作形容词和代词。它常修饰可数名词复数。如：some books一些书，some boys一些男孩，也可修饰不可数名词，如：some water一些水，some tea一些茶叶，some常用在肯定句中。any意为“任何一些”，它也可修饰可数名词复数或不可数名词，常用于疑问句和否定句。如： --I have some tea here. 我这儿有些茶叶。 --I can’t see any tea. 我没看见茶叶。 --Do you have any friends at school? 你在学校有些朋友吗? --I have some English books, they are my best friends. 我有英语书，它们是我最好的朋友。但在表示建议，反问，请求的疑问句中，或期望得到肯定回答时，多用some而不用any。如：Would you like some coffee? 你要不要来点咖啡? What about some fruit juice? 来点水果汁如何? 当any表示“任何”的意义，起强调作用时，它可以用在肯定句中； Any student can answer this question.任何学生都可以回答这个问题。选题角度：辨析some和any的不同用法：some 常用在肯定句中，而any 则常用在否定和疑问句中。在表示建议，反问，请求的疑问句中，或期望得到肯定回答时，多用some而不用any。一般用于疑问句或否定句中，用于never, hardly, without等词之后，用于if / whether 之后。而some则用于肯定句中，用于建议或请求的疑问句中，用于预料会作肯定回答的疑问句中，用于表示反问的否定的疑句中。如： 1. I’d been expecting ________ letters the whole morning, but there weren’t ________ for me. (全国卷)卷 A. some; any B. many; a few C. some; one D. a few; none

英语量词用法大全

英语量词用法大全常用语法场景1：当和外国友人走在公园，突然看到一群鸽子。这时候你是不是就会说:There are many pigeons 或者a lot of呢？想再高大上一点吗？Sam, look! There is a flock of pigeons over there. 场景2:当说三张纸的时候，是不是很习惯说：“three papers”？错了哦。paper 是不可数名词。正确说法是：three pieces of paper。 ◆◆量词用法整理+讲解◆◆ 量词可以用于描述可数名词比如a herd of elephants(一群大象）, 也可以描述不可数名词；比如three pieces of paper（三张纸） 1、描述一群...量词+群+of+名词一群人a/an crowd／group／army／team／of people；一群牛、象、马、天鹅a herd of cattle／elephants／horses／swans

一群鸟、鹅、母鸡、羊、燕子a flock of birds／geese／hens／goats／swallows 一群猎狗、狼a pack of hounds／wolves 2、描述一丝／点／层一丝怀疑a shadow of doubt 一线未来之光a glimpse of future 一缕月光a streak of moonlight 一层霜／雪／糖霜a layer of frost／snow／cream 3. piece块；片；段；项；件；篇；首；幅；张 a piece of bread／paper／wood／furniture／land／advice／news／meat ／cloth／music... 4、英译“一阵” 一阵哭泣／喝彩／炮击／雷声

some与any的用法区别

初中英语名词用法讲解

一、名词的分类名词可分为普通名词和专有名词两大类。 1. 普通名词又可分为：（1）个体名词。如：cup，desk，student等。一般可数，有单复数形式。（2）集体名词。如：class，team，family等。一般可数，有单复数形式。（3）物质名词。如：rice，water，cotton等。一般不可数，没有单复数之分。（4）抽象名词。如：love，work，life等。一般不可数，没有单复数之分。 2. 专有名词：如：China，Newton，London等。二、名词的数（一）可数名词的复数形式的构成规则 1. 一般情况下在名词的词尾加s，如：book books，pencil pencils. 2. 以-s，-x，-ch，-sh结尾的名词加-es，其读音为［iz］。如：bus buses，box boxes，watch watches，dish dishes等。 3. 以-y结尾的名词：（1）以“辅音字母+y”结尾的名词，把y改为i再加es，读音为［iz］，如：factory factories，company companies等。（2）以“元音字母+y”结尾的名词，或专有名词以y结尾，直接在词尾加-s，读音为［z］。如：key keys，Henry Henrys等。 4. 以-f和-fe结尾的名词：（1）变-f或-fe为v再加-es，读音为［vz］。如：thief thieves，wife wives，half halves等。（2）直接在词尾加-s，如：roof roofs，gulf gulfs，chief chiefs，proof proofs等。（3）两者均可。如：handkerchief handkerchiefs或handkerchieves. 5. 以-o结尾的名词：