A142979 -id:A142979 - cp's OEIS frontend

A317057 a(n) is the number of time-dependent assembly trees satisfying the connected gluing rule for a cycle on n vertices.

1, 1, 4, 23, 166, 1437, 14512, 167491, 2174746, 31374953, 497909380, 8619976719, 161667969646, 3265326093109, 70663046421208, 1631123626335707, 40004637435452866, 1038860856732399105, 28476428717448349996

Offset: 1

Nick Mayers, Robert Short, Aria Dougherty, Jul 20 2018

A time-dependent assembly tree for a connected graph G = (V, E) on n vertices is a rooted tree, each node of which is labeled with a subset U of V and a nonnegative integer i such that:
1) each internal node has at least two children,
2) there are leaves labeled (v, 0) for each vertex v in V,
3) the label on the root is (V, m) for 1 <= m <= n-1,
4) for each node (U, i) with i < m, U is the union of the {u} for the children (u, 0) of (U, i),
5) if (U, i) and (U', i') are adjacent nodes with U a subset of U', then i < i',
6) for each 0 <= i <= m, there exists a node (U, i) with U a subset of V.
A time-dependent assembly tree is said to satisfy the connected gluing rule if each internal vertex v of G, the graph induced by the vertices in the labels is connected.
Essentially the same as A053525. - R. J. Mathar, Aug 20 2018

M. Bona and A. Vince, The Number of Ways to Assemble a Graph, arXiv preprint arXiv:1204.3842 [math.CO], 2012.
A. Dougherty, N. Mayers, and R. Short, How to Build a Graph in n Days: Some Variants on Graph Assembly, arXiv preprint arXiv:1807.08079 [math.CO], 2018.

Cf. A047781, A317058, A317059, A142979, A317060.

a:=[1,1];; for n in [3..20] do a[n]:=1+Sum([2..n-1],j->Binomial(n,j)*a[j]); od; a; # Muniru A Asiru, Jul 25 2018

A317057 := proc(n)
    option remember;
    if n <=2 then
        1;
    else
        1+add(binomial(n,j)*procname(j), j=2..n-1) ;
    end if;
end proc:
seq(A317057(n),n=1..30) ; # R. J. Mathar, Aug 08 2018

Nest[Function[{a, n}, Append[a, 1 + Sum[Binomial[n, j] a[[j]], {j, 2, n - 1}]]][#, Length@ # + 1] &, {1, 1}, 17] (* Michael De Vlieger, Jul 26 2018 *)

lista(nn) = my(v = vector(nn)); for (n=1, nn, if (n<=2, v[n] = 1, v[n] = 1 + sum(j=2, n-1, binomial(n, j)*v[j]))); v; \\ Michel Marcus, Aug 08 2018

@cached_function
def TimeDepenConCycle(n):
    if (n==1):
        return 1
    elif (n==2):
        return 1
    else:
        return sum([binomial(n, j)*TimeDepenConCycle(j) for j in range(2, n)])+1
print(','.join(str(TimeDepenConCycle(i)) for i in range(1, 20)))

a(n) = 1 + Sum_{j = 2..n-1} binomial(n, j)*a(j), a(1) = a(2) = 1.
E.g.f.: (x - x*e^x + e^x - 1)/(2 - e^x).
a(n+1) = Sum_{k = 1..n} Stirling_2(n, k) * A142979(k). - Peter Bala, Dec 09 2024

A142980 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1) + (n + 1)^2a(n).

Original entry on oeis.org

1, 5, 29, 190, 1414, 11820, 110004, 1129200, 12686256, 154896480, 2043108000, 28958014080, 438997622400, 7088892491520, 121487996448000, 2202440792832000, 42113131054848000, 847071044402688000, 17880009683784192000, 395192695448291328000, 9127967350755133440000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 2 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

Seiichi Manyama, Table of n, a(n) for n = 1..448

Cf. A024167, A135148, A142979, A142981, A142982.

p := n -> 2*n^2+ 2*n+1: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)

Module[{a, n}, RecurrenceTable[{a[n+2] == 5*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 5}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n) is the Ehrhart polynomial for the 2-dimensional cross polytope (a square).
Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n+1)^2*a(n).
The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 5, b(2) = 26.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + (n-1)^2/5)))), for n >= 2.
The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = 1/(5 + 1^2/(5 + 2^2/(5 + 3^2/(5 + ... + n^2/(5 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4*k^4 + 1)) = log(2) - (1 - 1/2); the final equality is a result of Glaisher.
Thus a(n) ~ c*n^2*n! as n -> oo, where c = 2*log(2) - 1.
From Peter Bala, Dec 09 2024: (Start)
E.g.f.: A(x) = ((1 + x)^2 *log(1 + x) - 2*x^2)/(1 - x)^3 satisfies the differential equation 1 + (x + 5)*A(x) + (x^2 - 1)*A(x)' with A(0) = 0.
Sum_{k = 1..n} Stirling2(n, k)*a(k) = A135148(n+1). (End)

A142981 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1) + (n+1)^2a(n).

Original entry on oeis.org

1, 7, 53, 434, 3886, 38052, 406260, 4708368, 58959216, 794092320, 11454567840, 176267145600, 2883327788160, 49972442123520, 914939341344000, 17648374867200000, 357763095454464000, 7604722004802048000, 169148296960860672000, 3929342722459564032000, 95164717841561217024000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 3 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Seiichi Manyama, Table of n, a(n) for n = 1..447

Cf. A001845, A024167, A142979, A142980, A142982.

p := n -> (4*n^3+6*n^2+8*n+3)/3: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)

Module[{a, n}, RecurrenceTable[{a[n+2] == 7*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 7}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = (4*n^3 + 6*n^2 + 8*n + 3)/3 = A001845(n) is the Ehrhart polynomial for the 3-dimensional cross polytope (the octahedron).
Recurrence: a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1) + (n + 1)^2*a(n). The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 7, b(2) = 50.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7 + 1^2/(7 + 2^2/(7 + 3^2/(7 + ... + (n-1)^2/7)))), for n >= 2.
The behavior of a(n) for large n is given by limit_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p(k-1)*p(k)) = 1/(7 + 1^2/(7 + 2^2/(7 + 3^2/(7 + ... + n^2/(7 + ...))))) = (1 - 1/2 + 1/3) - log(2); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]). Thus a(n) ~ c*n^3*n! as n -> oo, where c = (10 - 12*log(2))/9.
E.g.f.: A(x) = (2*x*(4*x^2 + 3*x + 3) - 3*(x + 1)^3*log(1 + x) )/(3*(1 - x)^4) satisfies the differential equation 1 + (x + 7)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0. - Peter Bala, Dec 09 2024

A142982 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1) + (n + 1)^2a(n).

Original entry on oeis.org

1, 9, 85, 846, 8974, 101916, 1240308, 16156656, 224789616, 3331795680, 52465122720, 875333381760, 15432978107520, 286828144485120, 5606317009440000, 114993185594112000, 2470155824763648000, 55464433059571200000, 1299510384759562752000, 31718253797341267968000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 4 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n), which arises when accelerating the convergence of Mercator's series for the constant log(2). See A142979 for remarks on the general case.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Seiichi Manyama, Table of n, a(n) for n = 1..446

Cf. A024167, A142979, A142980, A142981.

p := n -> (2*n^4+4*n^3+10*n^2+8*n+3)/3: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20);

Module[{a, n}, RecurrenceTable[{a[n+2] == 9*a[n+1] + (n+1)^2*a[n], a[1] == 1, a[2] == 9}, a, {n, 25}]] (* Paolo Xausa, Dec 12 2024 *)

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = (2*n^4 + 4*n^3 + 10*n^2 + 8*n + 3)/3 = A001846(n) is the Ehrhart polynomial for the 4-dimensional cross polytope (the 16-cell).
Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1) + (n + 1)^2*a(n).
The sequence b(n) := n!*p(n) satisfies the same recurrence with b(1) = 9, b(2) = 82.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + (n-1)^2/9)))), for n >= 2.
The behavior of a(n) for large n is given by limit_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p(k-1)*p(k)) = 1/(9 + 1^2/(9 + 2^2/(9 + 3^2/(9 + ... + n^2/(9 + ...))))) = log(2) - (1 - 1/2 + 1/3 - 1/4); the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
Thus a(n) ~ c*n^4*n! as n -> oo, where c = (12*log(2) - 7)/18.
E.g.f.: A(x) = (3*(x + 1)^4*log(1 + x) - 4*x^2*(2*x^2 + 2*x + 3))/(3*(1 - x)^5) satisfies the differential equation 1 + (x + 9)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0. - Peter Bala, Dec 09 2024

A142989 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1)+(n+1)(n+3)*a(n).

Original entry on oeis.org

1, 5, 33, 240, 1992, 18360, 187416, 2093760, 25462080, 334592640, 4728412800, 71488811520, 1151817408000, 19699405286400, 356504125824000, 6805868977152000, 136702533123072000, 2881808345235456000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 1 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1)+(n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142990 (m=2) and A142991 (m=3).

Harvey P. Dale, Table of n, a(n) for n = 1..447

Cf. A142979, A142983, A142988, A142990, A142991.

p := n -> (2*n-1)/3: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);

RecurrenceTable[{a[1]==1,a[2]==5,a[n]==5a[n-1]+(n-1)(n+1)a[n-2]},a,{n,20}] (* Harvey P. Dale, Jun 17 2013 *)

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (2*n-1)/3. Recurrence: a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 5, b(2) = 28. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5+1*3/(5+2*4/(5+3*5/(5+...+(n-1)*(n+1)/5)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(5+1*3/(5+2*4/(5+3*5/(5+...+(n-1)*(n+1)/(5+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 17/2-12*log(2).

A142990 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1)+(n+1)(n+3)*a(n).

Original entry on oeis.org

1, 7, 57, 504, 4896, 51912, 598392, 7459200, 100085760, 1439061120, 22083719040, 360371773440, 6232667212800, 113901166310400, 2193425619840000, 44398776748032000, 942498015750144000, 20938290999865344000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 2 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1)+(n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142989 (m=1) and A142991 (m=3).

Seiichi Manyama, Table of n, a(n) for n = 1..446

Cf. A142979, A142983, A142988, A142989, A142991.

p := n -> (n^2-n+1)/3: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (n^2-n+1)/3. Recurrence: a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 7, b(2) = 52. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/7)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(7+1*3/(7+2*4/(7+3*5/(7+...+(n-1)*(n+1)/(7+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 24*log(2)-33/2.

A142991 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1)+(n+1)(n+3)*a(n).

Original entry on oeis.org

1, 9, 89, 936, 10560, 127800, 1657080, 22965120, 339252480, 5326819200, 88651670400, 1559600179200, 28929882240000, 564490975104000, 11560712397696000, 247991610230784000, 5561409662613504000

Offset: 1

Peter Bala, Jul 17 2008

This is the case m = 3 of the general recurrence a(1) = 1, a(2) = 2*m+3, a(n+2) = (2*m+3)*a(n+1) + (n+1)*(n+3)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). For remarks on the general case see A142988 (m=0). For other cases see A142989 (m=1) and A142990 (m=2).

Seiichi Manyama, Table of n, a(n) for n = 1..445

Cf. A142979, A142983, A142988, A142989, A142990.

p := n -> (2*n^3-3*n^2+7*n-3)/15: a := n -> (n+2)!*p(n+2)*sum ((-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), k = 1..n): seq(a(n), n = 1..20);

RecurrenceTable[{a[1]==1,a[2]==9,a[n+2]==9a[n+1]+(n+1)(n+3)a[n]},a,{n,20}] (* Harvey P. Dale, Jul 18 2020 *)

a(n) = (n+2)!*p(n+2)*sum {k = 1..n} (-1)^(k+1)/(k*(k+1)*(k+2)*p(k+1)*p(k+2)), where p(n) = (2*n^3-3*n^2+7*n-3)/15. Recurrence: a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n). The sequence b(n) := 1/2*(n+2)!*p(n+2) satisfies the same recurrence with b(1) = 9, b(2) = 84. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/9)))), for n >=2. Lim n -> infinity a(n)/b(n) = 1/(9+1*3/(9+2*4/(9+3*5/(9+...+(n-1)*(n+1)/(9+...))))) = 2*sum {k = 1..inf} (-1)^(k+1)/ (k*(k+1)*(k+2)*p(k+1)*p(k+2)) = 167/6 - 40*log(2).

cp's OEIS Frontend

A317057 a(n) is the number of time-dependent assembly trees satisfying the connected gluing rule for a cycle on n vertices.

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A142980 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1) + (n + 1)^2*a(n).

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A142981 a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1) + (n+1)^2*a(n).

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A142982 a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1) + (n + 1)^2*a(n).

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A142989 a(1) = 1, a(2) = 5, a(n+2) = 5*a(n+1)+(n+1)*(n+3)*a(n).

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Formula

A142990 a(1) = 1, a(2) = 7, a(n+2) = 7*a(n+1)+(n+1)*(n+3)*a(n).

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Formula

A142991 a(1) = 1, a(2) = 9, a(n+2) = 9*a(n+1)+(n+1)*(n+3)*a(n).

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A142980 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1) + (n + 1)^2a(n).

A142981 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1) + (n+1)^2a(n).

A142982 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1) + (n + 1)^2a(n).

A142989 a(1) = 1, a(2) = 5, a(n+2) = 5a(n+1)+(n+1)(n+3)*a(n).

A142990 a(1) = 1, a(2) = 7, a(n+2) = 7a(n+1)+(n+1)(n+3)*a(n).

A142991 a(1) = 1, a(2) = 9, a(n+2) = 9a(n+1)+(n+1)(n+3)*a(n).