Displaying 1-10 of 30 results found.
Number of labeled n-node graphs with at most one cycle in each connected component.
+10
90
1, 1, 2, 8, 57, 608, 8524, 145800, 2918123, 66617234, 1704913434, 48300128696, 1499864341015, 50648006463048, 1847622972848648, 72406232075624192, 3033607843748296089, 135313823447621913500, 6402077421524339766058, 320237988317922139148736
COMMENTS
The total number of those graphs of order 5 is 608. The number of forests of trees on n labeled nodes of order 5 is 291, so the majority of the graphs of that kind have one or more unicycles.
Also the number of labeled graphs with n vertices satisfying a strict version of the axiom of choice. The axiom of choice says that, given any set of nonempty sets Y, it is possible to choose a set containing an element from each. The strict version requires this set to have the same cardinality as Y, meaning no element is chosen more than once. The connected case is A129271, complement A140638. The unlabeled version is A134964. - Gus Wiseman, Dec 22 2023
FORMULA
a(0) = 1; for n >=1, a(n) = Sum of n!prod_{j=1}^n\{ frac{ A129271(j)^{c_j} } { j!^{c_j}c_j! } } over all the partitions of n, c_1 + 2c_2 + ... + nc_n; c_1, c_2, ..., c_n >= 0. E.g.f.: sqrt(-LambertW(-x)/(x*(1+LambertW(-x))))*exp(-3/4 * LambertW(-x)^2). - Vladeta Jovovic, Sep 16 2008 a(n) ~ 2^(-1/4) * Gamma(3/4) * exp(-1/4) * n^(n-1/4) / sqrt(Pi) * (1-7*Pi/(12*Gamma(3/4)^2*sqrt(n))). - Vaclav Kotesovec, Oct 08 2013
EXAMPLE
Below we see the 7 partitions of n=5 in the form c_1 + 2c_2 + ... + nc_n followed by the corresponding number of graphs. We consider the values of A129271(j) given by the table j|1|2|3| 4| 5|
----+-+-+-+--+---+
a(j)|1|1|4|31|347|
1*5 -> 5!1^5 / (1!^5 * 5!) = 1
2*1 + 1*3 -> 5!1^1 * 1^3 / (2!^1 * 1! * 1!^3 * 3!) = 10
2*2 + 1*1 -> 5!1^2 * 1^1 / (2!^2 * 2! * 1!^1 * 1!) = 15
3*1 + 1*2 -> 5!4^1 * 1^2 / (3!^1 * 1! * 1!^2 * 2!) = 40
3*1 + 2*1 -> 5!4^1 * 1^1 / (3!^1 * 1! * 2!^1 * 1!) = 40
4*1 + 1*1 -> 5!31^1 * 1^1 / (4!^1 * 1! * 1!^1 * 1!) = 155
5*1 -> 5!347^1 / (5!^1 * 1!) = 347
Total 608
MAPLE
cy:= proc(n) option remember; binomial(n-1, 2)*
add((n-3)!/(n-2-t)! *n^(n-2-t), t=1..n-2)
end:
T:= proc(n, k) option remember;
if k=0 then 1
elif k<0 or n<k then 0
else add(binomial(n-1, j)*((j+1)^(j-1)*T(n-j-1, k-j)
+cy(j+1)*T(n-j-1, k-j-1)), j=0..k)
fi
end:
a:= n-> add(T(n, k), k=0..n):
MATHEMATICA
nn=20; t=Sum[n^(n-1)x^n/n!, {n, 1, nn}]; Range[0, nn]!CoefficientList[ Series[ Exp[t/2-3t^2/4]/(1-t)^(1/2), {x, 0, nn}], x] (* Geoffrey Critzer, Sep 05 2012 *) Table[Length[Select[Subsets[Subsets[Range[n], {2}]], Select[Tuples[#], UnsameQ@@#&]!={}&]], {n, 0, 5}] (* Gus Wiseman, Dec 22 2023 *)
PROG
(PARI) x='x+O('x^50); Vec(serlaplace(sqrt(-lambertw(-x)/(x*(1+ lambertw(-x))))*exp(-(3/4)*lambertw(-x)^2))) \\ G. C. Greubel, Nov 16 2017
CROSSREFS
A143543 counts graphs by number of connected components.
Number of sets of nonempty subsets of {1..n} contradicting a strict version of the axiom of choice.
+10
66
0, 0, 1, 67, 30997, 2147296425, 9223372036784737528, 170141183460469231731687303625772608225, 57896044618658097711785492504343953926634992332820282019728791606173188627779
COMMENTS
The axiom of choice says that, given any set of nonempty sets Y, it is possible to choose a set containing an element from each. The strict version requires this set to have the same cardinality as Y, meaning no element is chosen more than once.
EXAMPLE
The a(2) = 1 set-system is {{1},{2},{1,2}}.
The a(3) = 67 set-systems have following 21 non-isomorphic representatives:
{{1},{2},{1,2}}
{{1},{2},{3},{1,2}}
{{1},{2},{3},{1,2,3}}
{{1},{2},{1,2},{1,3}}
{{1},{2},{1,2},{1,2,3}}
{{1},{2},{1,3},{2,3}}
{{1},{2},{1,3},{1,2,3}}
{{1},{1,2},{1,3},{2,3}}
{{1},{1,2},{1,3},{1,2,3}}
{{1},{1,2},{2,3},{1,2,3}}
{{1,2},{1,3},{2,3},{1,2,3}}
{{1},{2},{3},{1,2},{1,3}}
{{1},{2},{3},{1,2},{1,2,3}}
{{1},{2},{1,2},{1,3},{2,3}}
{{1},{2},{1,2},{1,3},{1,2,3}}
{{1},{2},{1,3},{2,3},{1,2,3}}
{{1},{1,2},{1,3},{2,3},{1,2,3}}
{{1},{2},{3},{1,2},{1,3},{2,3}}
{{1},{2},{3},{1,2},{1,3},{1,2,3}}
{{1},{2},{1,2},{1,3},{2,3},{1,2,3}}
{{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}
MATHEMATICA
Table[Length[Select[Subsets[Rest[Subsets[Range[n]]]], Select[Tuples[#], UnsameQ@@#&]=={}&]], {n, 0, 3}]
CROSSREFS
Multisets of multisets of this type are ranked by A355529. The version without singletons is A367769. The version allowing empty edges is A367901. These set-systems have ranks A367907. A059201 counts covering T_0 set-systems.
A323818 counts covering connected set-systems.
A326031 gives weight of the set-system with BII-number n.
Cf. A007716, A083323, A092918, A102896, A283877, A306445, A355739, A355740, A367862, A367905, A368409, A368413.
Number of sets of nonempty subsets of {1..n} satisfying a strict version of the axiom of choice.
+10
64
1, 2, 7, 61, 1771, 187223, 70038280, 90111497503, 397783376192189
COMMENTS
The axiom of choice says that, given any set of nonempty sets Y, it is possible to choose a set containing an element from each. The strict version requires this set to have the same cardinality as Y, meaning no element is chosen more than once.
EXAMPLE
The a(2) = 7 set-systems:
{}
{{1}}
{{2}}
{{1,2}}
{{1},{2}}
{{1},{1,2}}
{{2},{1,2}}
MATHEMATICA
Table[Length[Select[Subsets[Subsets[Range[n]]], Select[Tuples[#], UnsameQ@@#&]!={}&]], {n, 0, 3}]
CROSSREFS
The version without singletons is A367770. The complement allowing empty edges is A367901. These set-systems have ranks A367906. A059201 counts covering T_0 set-systems.
A323818 counts covering connected set-systems.
A326031 gives weight of the set-system with BII-number n.
Cf. A007716, A083323, A092918, A102896, A283877, A306445, A355739, A355740, A367862, A367905, A370636.
Number of labeled simple graphs with n vertices contradicting a strict version of the axiom of choice.
+10
61
0, 0, 0, 0, 7, 416, 24244, 1951352, 265517333, 68652859502, 35182667175398, 36028748718835272, 73786974794973865449, 302231454853009287213496, 2475880078568912926825399800, 40564819207303268441662426947840, 1329227995784915869870199216532048487
COMMENTS
The axiom of choice says that, given any set of nonempty sets Y, it is possible to choose a set containing an element from each. The strict version requires this set to have the same cardinality as Y, meaning no element is chosen more than once.
In the connected case, these are just graphs with more than one cycle.
EXAMPLE
Non-isomorphic representatives of the a(4) = 7 graphs:
{{1,2},{1,3},{1,4},{2,3},{2,4}}
{{1,2},{1,3},{1,4},{2,3},{2,4},{3,4}}
MATHEMATICA
Table[Length[Select[Subsets[Subsets[Range[n], {2}]], Select[Tuples[#], UnsameQ@@#&]=={}&]], {n, 0, 5}]
CROSSREFS
The connected case is A140638 (graphs with more than one cycle). A143543 counts simple labeled graphs by number of connected components.
Cf. A057500, A116508, A326754, A355739, A355740, A367769, A367770, A367863, A367901, A367902, A367904.
Numbers n such that it is not possible to choose a different binary index of each binary index of n.
+10
60
7, 15, 23, 25, 27, 29, 30, 31, 39, 42, 43, 45, 46, 47, 51, 53, 54, 55, 57, 58, 59, 60, 61, 62, 63, 71, 75, 77, 78, 79, 83, 85, 86, 87, 89, 90, 91, 92, 93, 94, 95, 99, 101, 102, 103, 105, 106, 107, 108, 109, 110, 111, 113, 114, 115, 116, 117, 118, 119, 120, 121
COMMENTS
Also BII-numbers of set-systems (sets of nonempty sets) contradicting a strict version of the axiom of choice.
A binary index of n (row n of A048793) is any position of a 1 in its reversed binary expansion. A set-system is a finite set of finite nonempty sets. We define the set-system with BII-number n to be obtained by taking the binary indices of each binary index of n. Every finite set of finite nonempty sets has a different BII-number. For example, 18 has reversed binary digits (0,1,0,0,1), and since the binary indices of 2 and 5 are {2} and {1,3} respectively, the BII-number of {{2},{1,3}} is 18. The axiom of choice says that, given any set of nonempty sets Y, it is possible to choose a set containing an element from each. The strict version requires this set to have the same cardinality as Y, meaning no element is chosen more than once.
EXAMPLE
The set-system {{1},{2},{1,2},{1,3}} with BII-number 23 has choices (1,2,1,1), (1,2,1,3), (1,2,2,1), (1,2,2,3), but none of these has all different elements, so 23 is in the sequence.
The terms together with the corresponding set-systems begin:
7: {{1},{2},{1,2}}
15: {{1},{2},{1,2},{3}}
23: {{1},{2},{1,2},{1,3}}
25: {{1},{3},{1,3}}
27: {{1},{2},{3},{1,3}}
29: {{1},{1,2},{3},{1,3}}
30: {{2},{1,2},{3},{1,3}}
31: {{1},{2},{1,2},{3},{1,3}}
39: {{1},{2},{1,2},{2,3}}
42: {{2},{3},{2,3}}
43: {{1},{2},{3},{2,3}}
45: {{1},{1,2},{3},{2,3}}
46: {{2},{1,2},{3},{2,3}}
47: {{1},{2},{1,2},{3},{2,3}}
51: {{1},{2},{1,3},{2,3}}
MATHEMATICA
bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n, 2]], 1];
Select[Range[100], Select[Tuples[bpe/@bpe[#]], UnsameQ@@#&]=={}&]
PROG
(Python)
from itertools import count, islice, product
def bin_i(n): #binary indices
return([(i+1) for i, x in enumerate(bin(n)[2:][::-1]) if x =='1'])
def a_gen(): #generator of terms
for n in count(1):
p = list(product(*[bin_i(k) for k in bin_i(n)]))
x = len(p)
for j in range(x):
if len(set(p[j])) == len(p[j]): break
if j+1 == x: yield(n)
CROSSREFS
A070939 gives length of binary expansion.
A096111 gives product of binary indices.
A326031 gives weight of the set-system with BII-number n.
Cf. A000612, A055621, A072639, A083323, A309326, A326702, A326753, A367769, A367901, A367902, A367912. BII-numbers: A309314 (hyperforests), A326701 (set partitions), A326703 (chains), A326704 (antichains), A326749 (connected), A326750 (clutters), A326751 (blobs), A326752 (hypertrees), A326754 (covers), A326783 (uniform), A326784 (regular), A326788 (simple), A330217 (achiral).
Number of n-vertex labeled simple graphs with n edges and no isolated vertices.
+10
52
1, 0, 0, 1, 15, 222, 3760, 73755, 1657845, 42143500, 1197163134, 37613828070, 1295741321875, 48577055308320, 1969293264235635, 85852853154670693, 4005625283891276535, 199166987259400191480, 10513996906985414443720, 587316057411626070658200, 34612299496604684775762261
FORMULA
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n,k) * binomial(binomial(k,2), n). - Andrew Howroyd, Dec 29 2023
EXAMPLE
Non-isomorphic representatives of the a(4) = 15 graphs:
{{1,2},{1,3},{1,4},{2,3}}
{{1,2},{1,3},{2,4},{3,4}}
MATHEMATICA
Table[Length[Select[Subsets[Subsets[Range[n], {2}]], Union@@#==Range[n]&&Length[#]==n&]], {n, 0, 5}]
PROG
(PARI) a(n) = sum(k=0, n, (-1)^(n-k) * binomial(n, k) * binomial(binomial(k, 2), n)) \\ Andrew Howroyd, Dec 29 2023
CROSSREFS
The non-covering version is A116508. A143543 counts simple labeled graphs by number of connected components.
Cf. A003465, A006126, A305000, A316983, A319559, A323817, A326754, A367769, A367901, A367902, A367903.
Number of labeled simple graphs covering n vertices and contradicting a strict version of the axiom of choice.
+10
41
0, 0, 0, 0, 7, 381, 21853, 1790135, 250562543, 66331467215, 34507857686001, 35645472109753873, 73356936892660012513, 301275024409580265134121, 2471655539736293803311467943, 40527712706903494712385171632959, 1328579255614092966328511889576785109
COMMENTS
The axiom of choice says that, given any set of nonempty sets Y, it is possible to choose a set containing an element from each. The strict version requires this set to have the same cardinality as Y, meaning no element is chosen more than once.
EXAMPLE
The a(4) = 7 graphs:
{{1,2},{1,3},{1,4},{2,3},{2,4}}
{{1,2},{1,3},{1,4},{2,3},{3,4}}
{{1,2},{1,3},{1,4},{2,4},{3,4}}
{{1,2},{1,3},{2,3},{2,4},{3,4}}
{{1,2},{1,4},{2,3},{2,4},{3,4}}
{{1,3},{1,4},{2,3},{2,4},{3,4}}
{{1,2},{1,3},{1,4},{2,3},{2,4},{3,4}}
MATHEMATICA
Table[Length[Select[Subsets[Subsets[Range[n], {2}]], Union@@#==Range[n]&&Select[Tuples[#], UnsameQ@@#&]=={}&]], {n, 0, 5}]
CROSSREFS
A143543 counts simple labeled graphs by number of connected components.
Number of factorizations of n into positive integers > 1 such that it is not possible to choose a different prime factor of each factor.
+10
41
0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 1, 0, 0, 0, 4, 0, 1, 0, 1, 0, 0, 0, 3, 1, 0, 2, 1, 0, 0, 0, 6, 0, 0, 0, 4, 0, 0, 0, 3, 0, 0, 0, 1, 1, 0, 0, 7, 1, 1, 0, 1, 0, 3, 0, 3, 0, 0, 0, 2, 0, 0, 1, 10, 0, 0, 0, 1, 0, 0, 0, 10, 0, 0, 1, 1, 0, 0, 0, 7, 4, 0, 0, 2, 0, 0
COMMENTS
For example, the factorization f = 2*3*6 has two ways to choose a prime factor of each factor, namely (2,3,2) and (2,3,3), but neither of these has all different elements, so f is counted under a(36).
EXAMPLE
The a(1) = 0 through a(24) = 3 factorizations:
... 2*2 ... 2*4 3*3 .. 2*2*3 ... 2*8 . 2*3*3 . 2*2*5 ... 2*2*6
2*2*2 4*4 2*3*4
2*2*4 2*2*2*3
2*2*2*2
MATHEMATICA
facs[n_]:=If[n<=1, {{}}, Join@@Table[Map[Prepend[#, d]&, Select[facs[n/d], Min@@#>=d&]], {d, Rest[Divisors[n]]}]];
Table[Length[Select[facs[n], Select[Tuples[First/@FactorInteger[#]&/@#], UnsameQ@@#&]=={}&]], {n, 100}]
CROSSREFS
The complement is counted by A368414.
Number of unlabeled graphs of positive excess with n nodes.
+10
37
0, 0, 0, 2, 15, 110, 936, 12073, 273972, 12003332, 1018992968, 165091159269, 50502031331411, 29054155657134165, 31426485969804026075, 64001015704527557101231, 245935864153532932681481794, 1787577725145611700547871854870, 24637809253125004524383007473440146
COMMENTS
We can find in "The Birth of the Giant Component" p. 53, see the link, the following: "The excess of a graph or multigraph is the number of edges plus the number of acyclic components, minus the number of vertices."
If G has just one complex component with 4 nodes, the "non-complex part" of G can be,
a) One forest of order 4. There are 6 forests, so 2*6=12 graphs.
b) One triangle and one isolated vertex, or 2*1=2 graphs.
c) One unicyclic graph of order 4, so 2*2=4 graphs.
Also the number of unchoosable unlabeled graphs with up to n vertices, where a graph is choosable iff it is possible to choose a different vertex from each edge. The labeled version is A367867, covering A367868, connected A140638. - Gus Wiseman, Feb 13 2024
EXAMPLE
Below we show that a(8) = 12073. Note that A140636(n) is the number of connected graphs of positive excess with n nodes. Let G be a disconnected graph of positive excess with 8 nodes. In this case, G has one or two complex components. We have 3 graphs of order 8 with two complex components. One of those graphs is depicted in the figure below:
O---O...O---O
|\..|...|\./|
|.\.|...|.X.|
|..\|...|/.\|
O---O...O---O
If G has one complex component with 5 nodes, the non-complex part of G can be,
a) One forest of order 3. There are 3 forests, so A140636(5) * 3 = 39 graphs. b) One triangle, so A140636(5) = 13 graphs. If G has one complex component with 6 nodes, the non-complex part of G is a forest of order 2. There are 2 forests. We have A140636(6) * 2, or 186 graphs. If G has one complex component with 7 nodes, the non-complex part of G is one isolated vertex. We have A140636(7), or 809 graphs. Finally if G is connected, we have A140636(8), or 11005 graphs. The total is 3 + 12 + 2 + 4 + 39 + 13 + 186 + 809 + 11005 = 12073.
MATHEMATICA
brute[m_]:=First[Sort[Table[Sort[Sort /@ (m/.Rule@@@Table[{(Union@@m)[[i]], p[[i]]}, {i, Length[p]}])], {p, Permutations[Range[Length[Union@@m]]]}]]];
Table[Length[Union[brute /@ Select[Subsets[Subsets[Range[n], {2}]], Select[Tuples[#], UnsameQ@@#&]=={}&]]], {n, 0, 5}] (* Gus Wiseman, Feb 14 2024 *)
CROSSREFS
The connected covering case is A140636.
Number of factorizations of n into positive integers > 1 such that it is possible to choose a different prime factor of each factor.
+10
37
1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 2, 2, 1, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 1, 3, 1, 5, 1, 1, 2, 2, 2, 5, 1, 2, 2, 4, 1, 5, 1, 3, 3, 2, 1, 5, 1, 3, 2, 3, 1, 4, 2, 4, 2, 2, 1, 9, 1, 2, 3, 1, 2, 5, 1, 3, 2, 5, 1, 6, 1, 2, 3, 3, 2, 5, 1, 5, 1, 2, 1, 9, 2, 2, 2
COMMENTS
For example, the factorization f = 2*3*6 has two ways to choose a prime factor of each factor, namely (2,3,2) and (2,3,3), but neither of these has all different elements, so f is not counted under a(36).
EXAMPLE
The a(n) factorizations for selected n:
1 6 12 24 30 60 72 120
2*3 2*6 2*12 2*15 2*30 2*36 2*60
3*4 3*8 3*10 3*20 3*24 3*40
4*6 5*6 4*15 4*18 4*30
2*3*5 5*12 6*12 5*24
6*10 8*9 6*20
2*3*10 8*15
2*5*6 10*12
3*4*5 2*3*20
2*5*12
2*6*10
3*4*10
3*5*8
4*5*6
MATHEMATICA
facs[n_]:=If[n<=1, {{}}, Join @@ Table[Map[Prepend[#, d]&, Select[facs[n/d], Min@@#>=d&]], {d, Rest[Divisors[n]]}]];
Table[Length[Select[facs[n], Select[Tuples[First/@FactorInteger[#]&/@#], UnsameQ@@#&]!={}&]], {n, 100}]
CROSSREFS
The complement is counted by A368413.
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