A006528 a(n) = (n^4 + n^2 + 2*n)/4.

0, 1, 6, 24, 70, 165, 336, 616, 1044, 1665, 2530, 3696, 5226, 7189, 9660, 12720, 16456, 20961, 26334, 32680, 40110, 48741, 58696, 70104, 83100, 97825, 114426, 133056, 153874, 177045, 202740, 231136, 262416, 296769, 334390, 375480, 420246, 468901, 521664, 578760

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Number of ways to color vertices of a square using <= n colors, allowing only rotations.

Also product of first and last terms in n-th row of a triangle of form: row(1)= 1; row(2)= 2,3; row(3) = 4, 5, 6, ... . - Dave Durgin, Aug 17 2012

References

• Nick Baxter, The Burnside di-lemma: combinatorics and puzzle symmetry, in Tribute to a Mathemagician, Peters, 2005, pp. 199-210.
• M. Gardner, New Mathematical Diversions from Scientific American. Simon and Schuster, NY, 1966, p. 246.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Row n=2 of A343095.

Cf. A002817 (square colorings).

Programs

Magma

I:=[0, 1, 6, 24, 70]; [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..40]]; // Vincenzo Librandi, Apr 30 2012

Maple

A006528:=-z*(1+z+4*z**2)/(z-1)**5; # Simon Plouffe in his 1992 dissertation
a:=n->add(n+add(binomial(n,2), j=1..n),j=0..n):seq(a(n)/2, n=0..35); # Zerinvary Lajos, Aug 26 2008

Mathematica

Table[CycleIndex[CyclicGroup[4],t]/.Table[t[i]->n,{i,1,4}],{n,0,20}] (* Geoffrey Critzer, Mar 13 2011*)
Table[(n^4+n^2+2*n)/4,{n,0,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,6,24,70},40] (* Harvey P. Dale, Jan 13 2019 *)

PARI

a(n) = n*(n+1)*(n^2-n+2)/4; /* Joerg Arndt, Apr 30 2012 */

Formula

a(n) = n*(n+1)*(n^2-n+2)/4.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Apr 30 2012

From Stefano Spezia, Oct 11 2023: (Start)

O.g.f.: x*(1 + x + 4*x^2)/(1 - x)^5.

E.g.f.: exp(x)*x*(4 + 8*x + 6*x^2 + x^3)/4. (End)