cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Feryal Alayont

Feryal Alayont's wiki page.

Feryal Alayont has authored 6 sequences.

A380461 Number of edge covers of fan graph F_{2,n}.

Original entry on oeis.org

1, 16, 154, 1289, 10180, 78372, 596337, 4512900, 34064998, 256825009, 1935169456, 14577526976, 109797758833, 826945679592, 6227993359362, 46904386459065, 353244994467916, 2660340755025580, 20035394638446769, 150889230111278492, 1136366561949728110
Offset: 1

Author

Feryal Alayont, Jun 22 2025

Keywords

Comments

The fan graph F_{2,n} is the join of the path graph P_n with two isolated vertices. That is, two new vertices are added and each is connected to every vertex in P_n. a(n) is the number of edge covers of F_{2,n}.

Examples

			For n = 1, the fan graph F_{2,1} becomes a path with three vertices. There is exactly 1 edge cover.
For n = 2, the base graph P_2 has endpoints v and w, and two added vertices a and b each connect to both v and w. Each of a and b has 3 edge-covering options: use the edge to v, the edge to w, or both. This gives 3 × 3 = 9 combinations. For each, the edge (v,w) from P_2 may be included or not, giving 18 configurations. However, in 2 of these, omitting the edge (v,w) leaves v or w uncovered. Therefore, the number of valid edge covers is 18 - 2 = 16.
		

Formula

a(n) = 11*a(n-1) - 24*a(n-2) - 21*a(n-3) + 33*a(n-4) + 34*a(n-5) + 8*a(n-6), for n >= 7.

A374096 Number of edge covers of fan graph F_{1,n}.

Original entry on oeis.org

1, 4, 16, 59, 214, 768, 2745, 9792, 34900, 124339, 442906, 1577540, 5618665, 20011452, 71272296, 253840779, 904068526, 3219889720, 11467810393, 40843217384, 145465283884, 518082304131, 1845177508818, 6571697181084, 23405446635913, 83359734391300, 296890096642144
Offset: 1

Author

Feryal Alayont, Jun 28 2024

Keywords

Examples

			The fan graph F_{1,2} is the cycle with three vertices and has 4 edge covers.
The graph F_{1,3} is formed by adding a chord to the cycle with four vertices. The cycle C_4 has 7 edge covers, so there are 7 edge covers of F_{1,3} without the chord. If the chord is there, the two endpoints are covered. To cover the remaining two vertices, we need at least one of the two edges on each side of each vertex, giving us 3*3 choices total. So we have 16 edge covers for F_{1,3}.
We interpret F_{1,1} to be the path with two vertices with one edge cover.
		

Programs

  • Mathematica
    LinearRecurrence[{4, 0, -5, -2}, {1, 4, 16, 59}, 30] (* Paolo Xausa, Jan 22 2025 *)

Formula

a(n) = 4*a(n-1)-5*a(n-3)-2*a(n-4).
G.f.: x/((1 - x - x^2)*(1 - 3*x - 2*x^2)). - Stefano Spezia, Jun 29 2024
a(n) = A007483(n-1) - A212804(n). - R. J. Mathar, Jun 29 2024

A373293 Number of edge covers of the fan graph F_{n,3}.

Original entry on oeis.org

16, 154, 1240, 9202, 66016, 466954, 3283240, 23026402, 161316016, 1129605754, 7908421240, 55362491602, 387548070016, 2712868376554, 18990174295240, 132931507044802, 930521410248016, 6513652454539354, 45595574930185240, 319169047756526002
Offset: 1

Author

Feryal Alayont, Jun 22 2024

Keywords

Comments

Label vertices of F_{n,3} with v1, ..., v_n, a, b, c, with b adjacent to both a and c. An edge cover is a subset of the edges so that each vertex is the endpoint of at least one edge. Each of the n vertices has 7 ways to connect to vertices a, b, c. Once we connect all v_i's, we then have 4 options for the edges ab and bc to exist or not, giving 4*7^m options. But not all of these will be an edge cover. For example, if all v_i's connected to a and b only, we have to add edge bc in the second step. So 5*3^m are removed. But we removed 3 cases where all v_i's connected to only a, or only b, or only c too many times.

Crossrefs

Cf. A100774 (in F_{n,2}).

Programs

  • Mathematica
    A373293[n_] := 4*7^n - 5*3^n + 3; Array[A373293, 25] (* or *)
    LinearRecurrence[{11, -31, 21}, {16, 154, 1240}, 25] (* Paolo Xausa, Jun 24 2024 *)
  • Python
    def a_n(n):
        return 4 * 7**n - 5 * 3**n + 3

Formula

a(n) = 4*7^n - 5*3^n + 3.
From Stefano Spezia, Jun 24 2024: (Start)
G.f.: 2*x*(8 - 11*x + 21*x^2)/((1 - x)*(1 - 3*x)*(1 - 7*x)).
E.g.f.: 4*exp(7*x) - 5*exp(3*x) + 3*exp(x) - 2. (End)

A363476 a(n) = Fibonacci(n)^2 * Fibonacci(n+1)^3.

Original entry on oeis.org

0, 1, 8, 108, 1125, 12800, 140608, 1565109, 17333064, 192329500, 2132531225, 23651979264, 262296652032, 2908947562937, 32260582549000, 357775937196300, 3967793428038237, 44003514081895936, 488006404120114496, 5412074146674562125, 60020821224245910600
Offset: 0

Author

Feryal Alayont, Jun 03 2023

Keywords

Comments

For n>1, a(n) is the number of edge covers of a caterpillar graph with spine P_(5n-4), one pendant attached at vertex n counting from the left end of the spine, a second one at 2n-1, a third at 3n-2 and a fourth at 4n-3. The caterpillar graph for n=3 is as follows:
* * * *
| | | |
*--*--*--v1--*--v2--*--v3--*--*--*
Each pendant edge must be included in an edge cover and hence allows the left and right sides of a vertex adjacent to a pendant to be independent. Therefore, the caterpillar can be split into independent paths, two P_4's (ends) and three P_5's (middle). Each P_n has F_{n-1} edge covers, resulting in the a(n) expression.
The sequence also counts number of subsets of {1, 2, ..., 5n-7} which do not contain two numbers whose difference is 5, a special instance of a general result given in Math. Mag. Problem 1854 (see Links). The equivalence to the above description can be seen as follows. Every vertex except v1, v2 and v3 is incident with at least one of the pendant edges. Therefore, if we label the middle eight edges in the spine with numbers 4, 1, 6, 2, 7, 3, 8, 5 (starting from the left), the edges have to be chosen so that both 1,6, both 2,7, and both 3,8 cannot be missing. This corresponds to choosing subsets of {1, 2, ..., 8} which do not contain two numbers whose difference is 5.

References

  • F. Alayont and E. Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs; submitted.

Crossrefs

Programs

  • Mathematica
    a[n_] := Fibonacci[n]^2 * Fibonacci[n+1]^3; Array[a, 21, 0] (* Amiram Eldar, Jun 06 2023 *)

Formula

G.f.: x*(1+4*x^2+x^3) / ((1-x-x^2)*(1-11*x-x^2)*(1+4*x-x^2)).

A358934 a(n) = Fibonacci(n+1)^5 - Fibonacci(n-1)^5.

Original entry on oeis.org

0, 1, 31, 242, 3093, 32525, 368168, 4051333, 45064131, 499200274, 5538624025, 61414079849, 681135796944, 7553728681433, 83772910243607, 929052526388050, 10303364319347757, 114266002348885717, 1267229634537217144, 14053790947047408701, 155858934437282250075
Offset: 0

Author

Feryal Alayont, Dec 06 2022

Keywords

Comments

a(n) is the number of edge covers of a spider graph with five branches where each branch has n vertices besides the center vertex. The idea is each branch is treated as a path P_(n+2). Each branch acts independently then and has F_(n+1) covers (P_n has F(n-1) covers), hence F_(n+1)^5 total. Except we remove the cases where each branch is missing the connecting edge to the center, which is when that edge cover comes from P_n , hence the minus F_(n-1)^5.
An edge cover of a graph is a subset of edges for which each vertex is incident to at least one edge in the subset.

Examples

			Case n=1 is a star graph with five branches and one edge cover (all edges).
   *   *
    \ /
  *__C__*
     |
     *
For n=2, there are 31 edge covers of the graph obtained by gluing five P_3 paths at one single vertex. Each of the pendant edges of the P_3's have to be in the edge cover for the pendants to be incident with an edge. The middle vertices are then automatically incident with at least one edge. There remains the center vertex. We then need at least one of the remaining five edges to be in the subset, giving us 2^5-1 choices.
   *__ *   *__*
       \ /
  *__*__C__*__*
        |
        *__*
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{8, 40, -60, -40, 8, 1}, {0, 1, 31, 242, 3093, 32525}, 20] (* Amiram Eldar, Dec 07 2022 *)
    Join[{0},#[[3]]-#[[1]]&/@Partition[Fibonacci[Range[0,30]]^5,3,1]] (* Harvey P. Dale, Aug 05 2024 *)
  • Python
    from sympy import fibonacci
    def a(n):
        return fibonacci(n+1)**5-fibonacci(n-1)**5
    
  • Python
    from gmpy2 import fib2
    def A358934(n): return sum(f:=fib2(n))**5-f[1]**5 # Chai Wah Wu, Jan 04 2023

Formula

From Stefano Spezia, Dec 07 2022: (Start)
G.f.: x*(1 + 23*x - 46*x^2 - 23*x^3 + x^4)/((1 + 4*x - x^2)*(1 - x - x^2)*(1 - 11*x - x^2)).
a(n) = 8*a(n-1) + 40*a(n-2) - 60*a(n-3) - 40*a(n-4) + 8*a(n-5) + a(n-6) for n > 5. (End)
5*a(n) = 2*A000045(n)+11*A049666(n)+8*(-1)^n*A001076(n). - R. J. Mathar, May 07 2024

A358917 a(n) = Fibonacci(n+1)^4 - Fibonacci(n-1)^4.

Original entry on oeis.org

0, 1, 15, 80, 609, 4015, 27936, 190385, 1307775, 8956144, 61405905, 420831071, 2884553280, 19770670945, 135511114479, 928804587920, 6366127657281, 43634071586575, 299072419071840, 2049872742473489, 14050037090947935, 96300386075488816, 660052667580788145
Offset: 0

Author

Feryal Alayont, Dec 05 2022

Keywords

Comments

a(n) is the number of edge covers of a spider graph with four branches where each branch has n vertices besides the center vertex.
An edge cover of a graph is a subset of edges for which each vertex is incident to at least one edge in the subset.
The idea is each branch is treated as a path P_(n+2). Each branch acts independently then and has F(n+1) covers (P_n has F(n-1) covers), hence F(n+1)^4 total. Except we remove the cases where each branch is missing the connecting edge to the center, which is when that edge cover comes from P_n , hence the minus F(n-1)^4.

Examples

			Case n=1 is a star graph with four branches and one edge cover (all edges).
   *   *
    \ /
  *__C__*
For n=2, there are 15 edge covers of the graph obtained by gluing four P_3 paths at one single vertex. Each of the pendant edges of the P_3's have to be in the edge cover for the pendants to be incident with an edge. The middle vertices are then automatically incident with at least one edge. There remains the center vertex. We then need at least one of the remaining four edges to be in the subset, giving us 2^4-1 choices.
    *       *
     \     /
      *   *
       \ /
  *__*__C__*__*
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{4, 19, 4, -1}, {0, 1, 15, 80}, 25] (* Amiram Eldar, Dec 06 2022 *)
  • Python
    from sympy import fibonacci
    def a(n): return fibonacci(n+1)**4-fibonacci(n-1)**4

Formula

From Stefano Spezia, Dec 06 2022: (Start)
G.f.: x*(1 + 11*x + x^2)/((1 - 7*x + x^2)*(1 + 3*x + x^2)).
a(n) = 4*a(n-1) + 19*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. (End)
5*a(n) = 9*A004187(n) + 4*(-1)^n*A001906(n) . - R. J. Mathar, May 07 2024