A058843 Triangle T(n,k) = C_n(k) where C_n(k) = number of k-colored labeled graphs with n nodes (n >= 1, 1<=k<=n).

1, 1, 2, 1, 12, 8, 1, 80, 192, 64, 1, 720, 5120, 5120, 1024, 1, 9152, 192000, 450560, 245760, 32768, 1, 165312, 10938368, 56197120, 64225280, 22020096, 2097152, 1, 4244480, 976453632, 10877927424, 23781703680, 15971909632, 3758096384

Offset: 1

Author

N. J. A. Sloane, Jan 07 2001

Keywords

• nonn
• easy
• tabl

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).

In this triangle, colorings of a labeled graph using k colors that differ only by a permutation of the k colors are treated as the same giving 1/k!*(E(x) - 1)^k as a generating function function for the k-th column. (End)

Examples

			Triangle begins:
  1;
  1,    2;
  1,   12,      8;
  1,   80,    192,     64;
  1,  720,   5120,   5120,   1024;
  1, 9152, 192000, 450560, 245760, 32768;
  ...
		

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
• R. C. Read, The number of k-colored graphs on labelled nodes, Canad. J. Math., 12 (1960), 410-414.

Crossrefs

Apart from scaling, same as A058875.

Columns give A058872 and A000683, A058873 and A006201, A058874 and A006202, also A006218. A213441, A213442, A224068.

Row sums give A240936.

Programs

Maple

for p from 1 to 20 do C[p,1] := 1; od: for k from 2 to 20 do for p from 1 to k-1 do C[p,k] := 0; od: od: for k from 2 to 10 do for p from k to 10 do C[p,k] := add( binomial(p,n)*2^(n*(p-n))*C[n,k-1]/k,n=1..p-1); od: od:

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], {n, 1, maxn}, {k, 1, n}]] (* Jean-François Alcover, Sep 21 2011 *)

PARI

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

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) = sum {i = 1..n-1} binomial(n-1,i)*2^(i*(n-i))*T(i,k-1).

A generating function: exp(x*(E(z) - 1)) = 1 + x*z + (x + 2*x^2)*z^2/(2!*2) + (x + 12*x^2 + 8*x^3)*z^3/(3!*2^3) + .... Cf. A008277 with e.g.f. exp(x*(exp(z) - 1)).

A generating function for column k: 1/k!*(E(x) - 1)^k = sum {n>=k} T(n,k)x^n/(n!*2^C(n,2)).

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

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