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.

Showing 1-4 of 4 results.

A331505 Number of labeled graphs with n nodes and floor(n/2) edges.

Original entry on oeis.org

1, 1, 3, 15, 45, 455, 1330, 20475, 58905, 1221759, 3478761, 90858768, 256851595, 8093990190, 22760723700, 840261910995, 2353351951665, 99615373765775, 278110855548955, 13278694407181203, 36976937738226486
Offset: 1

Views

Author

Washington Bomfim, Jan 18 2020

Keywords

Comments

Considering the permutation model of graph evolution (see the Flajolet reference) with 2n vertices initially isolated, the probability of the occurrence of an acyclic graph at the critical point n is Pp(n) = A302112(n)/a(2n). Note that a(2n) is the number of labeled graphs with 2n nodes and n edges.
Since a(2n) = C(C(2n, 2), n) we have Pp(n)= A302112(n)/C(C(2n, 2), n).
Therefore, by Vaclav Kotesovec's approximation in A302112, Pp(n) ~ e^(3/4) * P(n), where P(n) = c1 / n^(1/6) is the corresponding probability in the uniform model. Cf. A331500.
If t < n, P(n) is a lower bound of P(t). If t > n, P(n) is an upper bound of P(t). Here P(t) is the probability of an acyclic graph in time t.
Concerning the permutation model, the presence of cycles in graphs evolving near the critical time should be estimated by the above approximation.

Examples

			a(4) is 15 because for n = 4, floor(n/2) = 2, and there are two graphs with four points and two edges. See the figure below or the J. Riordan reference.
The non-isomorphic graphs with four nodes and two edges along with the corresponding number of labeled graphs are as follows:
.
  *--*     *  *
  |        |  |
  |        |  |
  *  *     *  *
   12        3
Pp(2) = A302112(2)/a(4) = 15/15 = 1. All the graphs with four nodes and two edges are acyclic.
		

References

  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 109.

Crossrefs

Cf. A000717, A084546, A331504, A302112 (numerators of Pp(n)), A331500, A331501.

Programs

  • PARI
    C(x, y) = binomial(x, y);
    a(n) = C(C(n,2), n\2);
    A302112(n)={my(S=0, j); /* From Jon E. Schoenfield's formula in A302112. */
    for(j = 0, n,
       S+=(-1/2)^j* C(n, j) * C(2*n-1, n+j-1) * (2*n)^(n-j) * (n+j)!
       );
    (1/n!)*S
    }; /* end A302112(n) */
    c1 = (2/3)^(1/3) * sqrt(Pi) / gamma(1/3);
    UpBoundP(n) = c1 / n^(1/6);            /* Approximation for P(n) */
    UpBoundPp(n) = exp(3/4) * UpBoundP(n); /* Approximation for Pp(n) */
    Pp(n) = A302112(n)/a(2*n);
    Ratio(n) = UpBoundPp(n) / Pp(n);

Formula

a(n) = C( C(n,2), floor(n/2) ).

A379411 a(n) = n + floor(n*s/r) + floor(n*t/r), where r = e^(1/4), s = e^(1/2), t = e^(3/4).

Original entry on oeis.org

3, 7, 10, 15, 19, 22, 26, 31, 34, 38, 43, 46, 50, 54, 58, 62, 66, 70, 74, 77, 81, 86, 89, 93, 98, 101, 105, 109, 113, 117, 121, 125, 129, 133, 136, 141, 145, 148, 153, 156, 160, 164, 168, 172, 176, 180, 184, 188, 191, 196, 200, 203, 208, 212, 215, 219, 223
Offset: 1

Views

Author

Clark Kimberling, Jan 18 2025

Keywords

Comments

This sequence and A379412 and A379413 partition the positive integers; see A184812 for a proof. For each k in A000027, write "a" if k=A379411(n) for some n, "b" if k=A379412(n) for some n, and "c" if k=A379413(n) for some n. Concatenating these letters for k = 1,2,3,... spells the following infinite word:
cbacbcabcacbcbacbcabcacbcabcbcacbacbcabcbcacbacbcabccabcbacbcabccabcbacbcacbacbcabcbcacbacbcabccbacbacbcabccabcbacbcacbcabcbacbcacbcabcabccbacb...

Crossrefs

Programs

  • Mathematica
    r = E^(1/4); s = E^(1/2); t = E^(3/4);
    Table[n + Floor[n*s/r] + Floor[n*t/r], {n, 1, 120}]  (* A379411 *)
    Table[n + Floor[n*r/s] + Floor[n*t/s], {n, 1, 120}]  (* A379412 *)
    Table[n + Floor[n*r/t] + Floor[n*s/t], {n, 1, 120}]  (* A379413 *)

Formula

a(n) = n + floor(n*r) + floor(n*r^2), where r = e^(1/4).

A379413 a(n) = n + floor(n*r/t) + floor(n*s/t), where r = e^(1/4), s = e^(1/2), t = e^(3/4).

Original entry on oeis.org

1, 4, 6, 9, 11, 13, 16, 18, 21, 23, 25, 28, 30, 32, 35, 37, 40, 42, 44, 47, 49, 52, 53, 56, 59, 61, 64, 65, 68, 71, 73, 75, 78, 80, 83, 85, 87, 90, 92, 95, 96, 99, 102, 104, 107, 108, 111, 114, 116, 118, 120, 123, 126, 128, 130, 132, 135, 138, 139, 142, 144
Offset: 1

Views

Author

Clark Kimberling, Jan 18 2025

Keywords

Comments

This sequence and A379411 and A379412 partition the positive integers; see A378142 for a proof.

Crossrefs

Programs

  • Mathematica
    r = E^(1/4); s = E^(1/2); t = E^(3/4);
    Table[n + Floor[n*s/r] + Floor[n*t/r], {n, 1, 120}]  (* A379411 *)
    Table[n + Floor[n*r/s] + Floor[n*t/s], {n, 1, 120}]  (* A379412 *)
    Table[n + Floor[n*r/t] + Floor[n*s/t], {n, 1, 120}]  (* A379413 *)

Formula

a(n) = n + floor(n/r) + floor(n*r^2), where r = e^(1/4).

A331502 Decimal expansion of exp(4/9).

Original entry on oeis.org

1, 5, 5, 9, 6, 2, 3, 4, 9, 7, 6, 0, 6, 7, 8, 0, 7, 1, 5, 5, 3, 3, 7, 0, 9, 2, 8, 6, 9, 7, 9, 4, 7, 1, 1, 8, 6, 3, 9, 4, 8, 2, 4, 0, 1, 1, 4, 2, 2, 1, 4, 2, 3, 5, 4, 3, 9, 0, 2, 7, 8, 4, 3, 1, 5, 4, 3, 5, 6, 3, 8, 5, 0, 1, 3, 3, 1, 0, 6, 3, 2, 6, 4, 2, 7, 5, 8, 1, 6, 1, 2, 4, 9, 2, 9, 9, 4, 0, 1, 5, 4, 2, 9, 1, 6, 9
Offset: 1

Views

Author

Washington Bomfim, Mar 04 2020

Keywords

Comments

Considering graph evolutions (see the Flajolet link) with 3n vertices initially isolated, the probability of the occurrence of an acyclic graph at the point n, (n = 1/3 * 3n), in the uniform model, will be denoted by P13(n). In the case of the permutation model, the respective probability will be denoted by Pp13(n).
Pp13(n) / P13(n) ~ exp(4/9) since Pp13(n) = f(n) / C(N,n), where f(n) is the number of labeled forests with 3n nodes and n edges, and C(N,n), N = 3n *(3n-1)/2 (see the Lucatero link) is the number of labeled graphs with 3n nodes and n edges.
Because P13(n) = f(n)* n!* 2^n / (3n)^(2n), Pp13(n) / P13(n) = (3n)^(2n) / (C(N,n)* n! *2^n), and Lim_{n->oo} Pp13(n) / P13(n) = exp(4/9).

Examples

			1.55962349760678071553370928697947118639482401142214...
		

Crossrefs

Programs

  • Maple
    evalf(exp(4/9), 134);
  • Mathematica
    RealDigits[Exp[4/9],10,120][[1]] (* Harvey P. Dale, Jun 05 2023 *)
  • PARI
    exp(4/9)

Formula

Equals Lim_{n->oo} Pp13(n) / P13(n) = Lim_{n->oo} (3*n)^(2*n) / (binomial((3*n *(3*n-1)/2), n) * n! * 2^n).
Showing 1-4 of 4 results.