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 1-2 of 2 results.

A191371 Number of simple labeled graphs with (at most) 3-colored nodes such that no edge connects two nodes of the same color.

Original entry on oeis.org

1, 3, 15, 123, 1635, 35043, 1206915, 66622083, 5884188675, 830476531203, 187106645932035, 67241729173555203, 38521811621470420995, 35161184767296890265603, 51112793797110111859802115, 118291368253025368001553530883, 435713124846749574718274002747395, 2553666761436949125065383836043837443
Offset: 0

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Author

Geoffrey Critzer, Jun 01 2011

Keywords

Comments

Cf. A213442, which counts colorings of labeled graphs on n vertices using exactly 3 colors. For 3-colorable labeled graphs on n vertices see A076315. - Peter Bala, Apr 12 2013

Examples

			a(2) = 15: There are two labeled 3-colorable graphs on 2 nodes, namely
A)  1    2   B)  1    2
    o    o       o----o
Using 3 colors there are 9 ways to color the graph of type A and 3*2 = 6 ways to color the graph of type B so that adjacent vertices do not share the same color. Thus there are in total 15 labeled 3-colored graphs on 2 vertices.
		

References

  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, 1973, 16-18.

Crossrefs

Column k=3 of A322280.

Programs

  • Mathematica
    f[{k_,r_,m_}]:= Binomial[m+r+k,k] Binomial[m+r,r] 2^ (k r + k m + r m); Table[Total[Map[f,Compositions[n,3]]],{n,0,20}]
  • PARI
    seq(n)={my(p=(sum(j=0, n, x^j/(j!*2^binomial(j, 2))) + O(x*x^n))^3); Vecrev(sum(j=0, n, j!*2^binomial(j,2)*polcoef(p,j)*x^j))} \\ Andrew Howroyd, Dec 03 2018
  • Python
    from sympy import binomial
    def a047863(n): return sum([binomial(n, k)*2**(k*(n - k)) for k in range(n + 1)])
    def a(n): return sum([binomial(n, k)*2**(k*(n - k))*a047863(k) for k in range(n + 1)]) # Indranil Ghosh, Jun 03 2017
    

Formula

a(n) = Sum(C(n,k)*C(n-k,r)*2^(k*r+k*m+r*m)) where the sum is taken over all nonnegative solutions to k + r + m = n.
From Peter Bala, Apr 12 2013: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*2^(k*(n-k))*A047863(k).
a(n) = 3*A000685(n) for n >= 1.
Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)). Then a generating function for this sequence is E(x)^3 = sum {n >= 0} a(n)*x^n/(n!*2^C(n,2)) = 1 + 3*x + 15*x^2/(2!*2) + 123*x^3/(3!*2^3) + 1635*x^4/(4!*2^6) + ....
In general, for k = 1, 2, ..., E(x)^k is a generating function for labeled k-colored graphs (see Read). For other examples see A047863 (k = 2) and A223887 (k = 4). (End)

A000686 Number of 4-colored labeled graphs on n nodes, divided by 4.

Original entry on oeis.org

1, 7, 85, 1777, 63601, 3882817, 403308865, 71139019777, 21276992674561, 10778161937857537, 9238819435213784065, 13390649605615389843457, 32796747486424209782108161, 135669064080920007649863745537, 947468281528010179181982467702785, 11166618111585805201637975219611631617
Offset: 1

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Author

Keywords

Comments

Sequence represents 1/4 of the number of 4-colored labeled graphs on n nodes. Indeed, on p. 413 of the Read paper, column 4 is 4, 28, 340, 7108, ... - Emeric Deutsch, May 06 2004

References

  • R. C. Read, personal communication.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Mathematica
    b[n_] := Sum[ 2^((i-j)*j + i*(n-i))*Binomial[n, i]*Binomial[i, j], {i, 0, n}, {j, 0, i}]; a[n_] := 1/4*Sum[ Binomial[n, k]*2^(k*(n-k))*b[k], {k, 0, n}]; Table[a[n], {n, 1, 14}] (* Jean-François Alcover, Dec 07 2011, after Emeric Deutsch *)
  • PARI
    N=66;  x='x+O('x^N);
    E=sum(n=0, N, x^n/(n!*2^binomial(n,2)) );
    tgf=E^4-1;  v=Vec(tgf);
    v=vector(#v, n, v[n] * n! * 2^(n*(n-1)/2) ) / 4
    /* Joerg Arndt, Apr 10 2013 */

Formula

a(n) = (1/4)*Sum_{k=0..n} binomial(n, k)*2^(k(n-k))*b(k), where b(0)=1 and b(k) = 3*A000685(k) for k > 0. - Emeric Deutsch, May 06 2004
From Peter Bala, Apr 12 2013: (Start)
a(n) = (1/4)*A223887(n).
a(n) = (1/4)*Sum_{k = 0..n} binomial(n,k)*2^(k*(n-k))*b(k)*b(n-k), where b(n) := Sum_{k = 0..n} binomial(n,k)*2^(k*(n-k)).
Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)). Then a generating function for this sequence is (1/4)*(E(x)^4 - 1) = Sum_{n >= 0} a(n)*x^n/(n!*2^C(n,2)) = x + 7*x^2/(2!*2) + 85*x^3/(3!*2^3) + .... (End)

Extensions

More terms from Pab Ter (pabrlos(AT)yahoo.com) and Emeric Deutsch, May 05 2004
Showing 1-2 of 2 results.