A058875 Triangle T(n,k) = C_n(k)/2^(k*(k-1)/2) where C_n(k) = number of k-colored labeled graphs with n nodes (n >= 1, 1 <= k <= n).

1, 1, 1, 1, 6, 1, 1, 40, 24, 1, 1, 360, 640, 80, 1, 1, 4576, 24000, 7040, 240, 1, 1, 82656, 1367296, 878080, 62720, 672, 1, 1, 2122240, 122056704, 169967616, 23224320, 487424, 1792, 1, 1, 77366400, 17282252800, 53247344640, 13440516096

Offset

1,5

Comments

From Peter Bala, Apr 12 2013: (Start)

A coloring of a simple graph G is a choice of color for each graph vertex such that no two vertices sharing the same edge have the same color.

Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)) = 1 + x + x^2/(2*2!) + x^3/(2^3*3!) + .... Read has shown that (E(x) - 1)^k is a generating function for labeled graphs on n nodes that can be colored using exactly k colors. Cases include A213441 (k = 2), A213442 (k = 3) and A224068 (k = 4).

If colorings of a graph using k colors are counted as the same if they differ only by a permutation of the colors then a generating function is 1/k!*(E(x) - 1)^k , which is a generating function for the k-th column of A058843. Removing a further factor of 2^C(k,2) gives 1/(k!*2^C(k,2))*(E(x) - 1)^k as a generating function for the k-th column of this triangle. (End)

References

F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 18, Table 1.5.1.

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1275

S. R. Finch, Bipartite, k-colorable and k-colored graphs

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.

Eric Weisstein's World of Mathematics, k-Colorable Graph

Formula

C_n(k) = Sum_{i=1..n-1} binomial(n, i)*2^(i*(n-i))*C_i(k-1)/k.

From Peter Bala, Apr 12 2013: (Start)

Recurrence equation: T(n,k) = 1/2^(k-1)*Sum_{i = 1..n-1} binomial(n-1,i)*2^(i*(n-i))*T(i,k-1).

Let E(x) = Sum_{n >= 0} x^n/(n!*2^C(n,2)) = 1 + x + x^2/(2!*2) + x^3/(3!*2^3) + .... Then a generating function for this triangle is E(x*(E(z) - 1)) = 1 + x*z + (x + x^2 )*z^2/(2!*2) + (x + 6*x^2 + x^3)*z^3/(3!*2^3) + (x + 40*x^2 + 24*x^3 + x^4)*z^4/(4!*2^6) + .... Cf. A008277 with e.g.f. exp(x*(exp(z) - 1)).

The row polynomials R(n,x) satisfy the recurrence equation R(n,x) = x*sum {k = 0..n-1} binomial(n-1,k)*2^(k*(n-k))*R(k,x/2) with R(0,x) = 1. The row polynomials appear to have only real zeros.

Column 2 = 1/(2!*2)*A213441; column 3 = 1/(3!*2^3)*A213442; column 4 = 1/(4!*2^6)*A224068. (End)

T(n,k) = A058843(n,k)/2^binomial(k,2). - Andrew Howroyd, Nov 30 2018

Example

Triangle begins:
  1;
  1,     1;
  1,     6,       1;
  1,    40,      24,      1;
  1,   360,     640,     80,     1;
  1,  4576,   24000,   7040,   240,   1;
  1, 82656, 1367296, 878080, 62720, 672, 1;
  ...

Mathematica

maxn=8; t[_, 1]=1; t[n_, k_]:=t[n, k]=Sum[Binomial[n, j]*2^(j*(n-j))*t[j, k-1]/k, {j, 1, n-1}]; Flatten[Table[t[n, k]/2^Binomial[k, 2], {n, 1, maxn}, {k, 1, n}]]  (* Geoffrey Critzer, Oct 06 2012, after code from Jean-François Alcover in A058843 *)

Prog

(PARI)
b(n)={n!*2^binomial(n, 2)}
T(n, k)={b(n)*polcoef((sum(j=1, n, x^j/b(j)) + O(x*x^n))^k, n)/b(k)} \\ Andrew Howroyd, Nov 30 2018

Crossrefs

Apart from scaling, same as A058843.

Columns give A058872 and A000683, A058873 and A006201, A058874 and A006202, also A006218.

Cf. A213441, A213442, A224068.

Sequence in context: A203338 A158116 A172343 * A156764 A156765 A015117

Adjacent sequences: A058872 A058873 A058874 * A058876 A058877 A058878

Keyword

nonn,easy,tabl

Author

N. J. A. Sloane, Jan 07 2001

Status

approved