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

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

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

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).

Links

• S. R. Finch, Bipartite, k-colorable and k-colored graphs. (4*{a(n)})
• S. R. Finch, Bipartite, k-colorable and k-colored graphs, June 5, 2003. [Cached copy, with permission of the author]
• R. C. Read, The number of k-colored graphs on labelled nodes, Canad. J. Math., 12 (1960), 410-414.
• R. C. Read, Letter to N. J. A. Sloane, Oct. 29, 1976.

Crossrefs

Cf. A000683, A000685, A223887.

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