OFFSET

2,1

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.

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).

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

EXAMPLE

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)

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 *)

PROG

(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..]

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

(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

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

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

CROSSREFS

KEYWORD

nonn,easy

AUTHOR

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

EXTENSIONS

Name edited by Jack W Grahl, Jul 16 2018

STATUS

approved