A179824 Chromatic polynomial of the star graph on 4 vertices (claw graph) and the path graph on 4 vertices.

2, 24, 108, 320, 750, 1512, 2744, 4608, 7290, 11000, 15972, 22464, 30758, 41160, 54000, 69632, 88434, 110808, 137180, 168000, 203742, 244904, 292008, 345600, 406250, 474552, 551124, 636608, 731670, 837000, 953312, 1081344, 1221858

Offset: 2

Author

Jaime Soffer (jaime.soffer(AT)gmail.com), Jul 28 2010

Keywords

• nonn
• easy

Comments

To generate a Pythagorean triangle one uses (a,b) to get sides b^2-a^2, 2*a*b, and a^2+b^2, having a perimeter of 2*b*(a+b). If for a one uses the triangular number n*(n+1)/2 and for b the next triangular number (n+1)*(n+2)/2, the perimeter of the triangle so formed is (n+1)^3 * (n+2), which will give the same results as this sequence starting at the second term. - J. M. Bergot, Apr 01 2012

Define b(0)=0 and b(n)=A179824(n+1) for n > 0. Then b(n) is the number of 4-tuples (w,x,y,z) having all terms in {0,...,n} and no two consecutive terms equal. - Clark Kimberling, May 31 2012

Let n points in the plane each become the centers of n-1 concentric circles, circles that pass through only one of each of the other points. The maximum number of intersections of these circles is this sequence. [The solution was given by Andrew Weimholt in the Sequence Fans Mailing List] - J. M. Bergot, Mar 10 2014

Both the 'claw graph', a graph with 4 vertices where one vertex is adjacent to the other three, and the path graph on 4 vertices (per Clark Kimberling's comment), have this sequence as their chromatic polynomial, or the number of proper colorings of the graph using at most n colors. This is the standard example of two graphs which are not isomorphic, but which have the same chromatic polynomial.

Examples

			From _Jack W Grahl_, Jul 16 2018: (Start)
Consider the claw graph, which has vertices A, B, C, D, and edges AB, AC, AD. To color this graph with 3 colors, we can choose any of the 3 colors for A. Then each of the other vertices can be colored with any of the remaining two colors, giving 3 * 2 * 2 * 2 = 24 choices in all.
Similarly, consider the path graph with the same vertices and edges AB, BC, CD. We have 3 choices for the color of A, then 2 choices for the color of B (any color except that chosen for A), 2 choices for the color of C (any color except B's) etc. (End)
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 2..1000
• Andrew Weimholt, Re: Intersecting circles, SeqFan post Mar 09 2014.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A019582, A179824.

Programs

GAP

List([2..40], n-> n*(n-1)^3 ); # G. C. Greubel, Aug 10 2019

Haskell

let f n = [ (x,a,b,c) | let t = [1..n], x <- t, a <- t, x /= a, b <- t, x /= b, c <- t, x /= c ] in map (length.f) [2..]

Haskell

let f n = n*(n-1)^3 in map f [2..]

Magma

I:=[2,24,108,320,750]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Mar 12 2014

Mathematica

CoefficientList[Series[2 (1 + 7 x + 4 x^2)/(1 - x)^5, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 12 2014 *)
Table[n^3+n^4,{n,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{2,24, 108,320,750},40] (* Harvey P. Dale, Sep 05 2015 *)

PARI

a(n) = n*(n-1)^3 \\ Charles R Greathouse IV, Mar 11 2014

Sage

[n*(n-1)^3 for n in (2..40)] # G. C. Greubel, Aug 10 2019

Formula

a(n) = n*(n-1)^3. - Jaime Soffer (jaime.soffer(AT)gmail.com), Jul 30 2010

G.f.: 2*x^2*(1 + 7*x + 4*x^2)/(1-x)^5. - Colin Barker, Jan 30 2012

a(n) = 2*A019582(n). - R. J. Mathar, Jun 09 2013

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Mar 12 2014

Sum_{n>=2} 1/a(n) = A249649. - R. J. Mathar, Oct 18 2019

Sum_{n>=2} (-1)^n/a(n) = 3*zeta(3)/4 + 2*log(2) - Pi^2/12 - 1. - Amiram Eldar, Nov 05 2020

Extensions

Name edited by Jack W Grahl, Jul 16 2018