A006201 Number of colorings of labeled graphs on n nodes using exactly 3 colors, divided by 48.

0, 0, 1, 24, 640, 24000, 1367296, 122056704, 17282252800, 3897054412800, 1400795928395776, 802530102499344384, 732523556206878392320, 1064849635418836398243840, 2464403435614136308036796416

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

Equals 1/48*A213442. - Peter Bala, Apr 12 2013

References

• F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 18, table 1.5.1, column 3 (divided by 8).
• R. C. Read, personal communication.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..75
• 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. A diagonal of A058843. A213442.

Programs

Mathematica

F2[n_] := Sum[Binomial[n, r]*2^(r*(n-r)), {r, 1, n-1}]; F3[n_] := Sum[Binomial[n, r]*2^(r*(n-r))*F2[r], {r, 1, n-1}]; a[n_] := F3[n]/48; Table[a[n], {n, 1, 15}] (* Jean-François Alcover, Mar 06 2014, after Maple code in A213442 *)

PARI

seq(n)={Vec(serconvol(sum(j=1, n, x^j*j!*2^binomial(j,2)) + O(x*x^n), (sum(j=1, n, x^j/(j!*2^binomial(j,2))) + O(x*x^n))^3)/48, -n)} \\ Andrew Howroyd, Nov 30 2018

Formula

Let E(x) = sum {n >= 0} x^n/(n!*2^C(n,2)) = 1 + x + x^2/(2!*2) + x^3/(3!*2^3) + x^4/(4!*2^6) + .... Then a generating function is 1/48*(E(x) - 1)^3 = x^3/(3!*2^3) + 24*x^4/(4!*2^6) + 640*x^6/(5!*2^10) + ... (see Read). - Peter Bala, Apr 12 2013

Extensions

More terms from Vladeta Jovovic, Feb 03 2000