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.

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A331563 Number of labeled cyclic graphs with n edges and 2n vertices.

Original entry on oeis.org

0, 0, 20, 1610, 129654, 11688369, 1194822915, 137766789810, 17758192128830, 2535895233070628, 397875362655895761, 68087081506276861665, 12626853606957534296975, 2523446241515288646389325
Offset: 1

Views

Author

Washington Bomfim, Jan 20 2020

Keywords

Examples

			a(4) = 1610 since we have 3 non-isomorphic cyclic graphs with 4 edges and 8 nodes. (See illustration below.)
To compute a(4) we can consult A057500, which provides the number of labeled connected unicycles. Because A057500(4)=15, and knowing that there are 3 labeled squares, we have 15-3 = 12 Paw Graphs [see Weisstein link]. So graph 1 is labeled in 12 * C(8,4) = 840 ways. Graph 2 is labeled in 3* C(8,4) = 210 ways. A105599 gives 10 as the number of labeled forests with 5 nodes and 4 components, so graph 3 is labeled in 10 * C(8,3) = 560 ways. We have 840 + 210 + 560 = 1610.
.
  graph 1    graph 2    graph 3 (triangle + forest with
                                 5 nodes and 4 components)
   *--*       *--*       *--* *
   | /|       |  |       | /  |
   |/ |       |  |       |/   |
   *  *       *--*       *    *
  * * * *    * * * *      * * *
		

Crossrefs

Formula

a(n) = A331505(2n) - A302112(n).

A332957 Number of labeled forests with n trees and 2*n nodes without trees of order exceeding n - floor(n/4).

Original entry on oeis.org

0, 3, 195, 5145, 504945, 47896695, 4900634739, 432180333585, 57753467156385, 8322385282168815, 1297830988347128235, 195029484510994774641, 36350420770495795788825, 7233322663694914389377775, 1534057832485607065840824075
Offset: 1

Views

Author

Washington Bomfim, Apr 09 2020

Keywords

Comments

When n increases practically all forests do not have trees of order exceeding n - floor (n/4). See table below.
n = 1 4 8 16 28 44 48 83
a(n)/A302112(n)*100 = 0.0% 27.3% 61.5% 85.8% 95.6% 98.8% 99.1% 99.94%

Examples

			E.g., a(4) = 5145. The partitions of 2*n = 8 with four parts, no parts exceeding n - floor(n/4) = 3 are [1*2 + 3*2], [1*1 + 2*2 + 3*1], and [2*4].
The first partition corresponds to f(1)^2*f(3)^2 / ((2!*1!^2)*(2!*3!^2)) = 1^2*3^2 / (2*2*36) = 1/16, the second corresponds to f(1)^1 * f(2)^2 * f(3)^1 / ((1!*1!^1)*(2!*2!^2)*(1!*3!^1)) = 1/16, and the last corresponds to f(2)^4 / (4!*2!^4) = 1/384. Finally 8! * (1/16 + 1/16 + 1/384)= 5145. (See formula).
		

References

  • D. E. Knuth, The Art of Computer Programming, Volume 4, Fascicle 3: Generating All Combinations and Partitions, Addison-Wesley, 2005, pp. 39, 47.

Crossrefs

Programs

  • PARI
    a(n) = { my(S=0); forpart(aa=2*n, my(D = Set(aa)); S += prod(j= 1, #D, my(p=D[j], c=#select(x-> x==p, Vec(aa))); (p^(p-2))^c / (c!* p!^c)), [1, n-n\4], [n, n]); (2*n)! * S};
    
  • PARI
    a(n)={my(p=sum(k=1, n-n\4, k^(k-2)*x^k/k!)); (2*n)!*polcoef( polcoef( exp(y*p + O(x*x^(2*n))), 2*n, x), n, y)} \\ Andrew Howroyd, Apr 10 2020

Formula

a(n) = (2*n)! * Sum_{P(2*n,n)} Product_{p=1..2*n} f(p)^c_p / (c_p! * p!^c_p), where f(n) = A000272(n) = n^(n-2), and P(2*n,n) are the partitions of 2*n with n parts, without parts exceeding n - floor(n/4):
c_1 + 2*c_2 + ... + (2*n)*c_(2*n) = 2n; c_1, c_2, ..., c_(2*n) >= 0.

A332961 Triangle read by rows: T(n,k) is the number of labeled forests with n trees and 2n nodes and with the largest tree having at most k nodes, (n>=1, 2<=k<=n+1).

Original entry on oeis.org

1, 3, 15, 15, 195, 435, 105, 5145, 11865, 18865, 945, 152145, 504945, 819945, 1092105, 10395, 6039495, 27106695, 47896695, 65859255, 79170399, 135135, 276351075, 1663737075, 3429501075, 4900634739, 6111948843, 6899167275, 2027025, 14985795825, 120931635825, 286156695825, 432180333585
Offset: 1

Views

Author

Washington Bomfim, May 11 2020

Keywords

Comments

The first formula is based on Kolchin's formula (1.4.2) [see the Kolchin reference].
Let S be the set of labeled forests with n trees and 2n nodes.
We know that the largest trees in S have n+1 nodes. It follows from line n=6 of the triangle that more than 33% of the forests in S do not have trees with more than 4/7*(n+1) nodes.
The percentages goes to 61%, 83%, and 100% respectively for (5/7)*(n+1) nodes, (6/7)*(n+1) nodes, and n+1 nodes.

Examples

			Triangle T(n,k) begins:
      1;
      3,      15;
     15,     195,      435;
    105,    5145,    11865,    18865;
    945,  152145,   504945,   819945,  1092105;
  10395, 6039495, 27106695, 47896695, 65859255, 79170399;
  ...
The graphs for T(2,2) and T(2,3) are illustrated below:
   o---o   :   o---o     o   o
           :             |
   o---o   :   o---o     o---o
T(2,2) = 3 since the first graph on the left has 3 labelings.
T(2,3) = 15 since the first graph has 3 labelings, and the second has 12 labelings.
		

References

  • V. F. Kolchin, Random Graphs. Encyclopedia of Mathematics and Its Applications 53. Cambridge Univ. Press, Cambridge, 1999, pp 30-31.

Crossrefs

Diagonal is A302112.
Columns k=2..3 are A001147, A001147 + A332960.

Programs

  • PARI
    T(n,k) = { my(S = 0, D, p, c);
    forpart(a = 2*n, D = Set(a);
       S += prod(j=1,#D, p=D[j]; c=#select(x-> x==p,Vec(a)); (p^(p-2)/p!)^c /c!)
    , [1, k], [n, n]); (2*n)! * S };

Formula

T[n,k] = Sum_{j=2..k} (T1[n,j]), where T1[n,j] is the triangle A332959.
T(n,k) = ((2*n)!/n!) * Sum_{compositions p_1 + ... + p_n = 2*n, 1 <= p_i <= k}
Product_{j=1..n} f(p_j) / p_j!, where f(p_j) = A000272(p_j) = p_j^(p_j-2).
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