A004652 Expansion of x*(1+x^2+x^4)/((1-x)*(1-x^2)*(1-x^3)).

0, 1, 1, 3, 4, 7, 9, 13, 16, 21, 25, 31, 36, 43, 49, 57, 64, 73, 81, 91, 100, 111, 121, 133, 144, 157, 169, 183, 196, 211, 225, 241, 256, 273, 289, 307, 324, 343, 361, 381, 400, 421, 441, 463, 484, 507, 529, 553, 576, 601, 625, 651, 676, 703, 729, 757, 784, 813

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

As a Molien series this arises as (1+x^12)/((1-x^4)*(1-x^8)^2).

Starting (1, 3, 4, ...) = row sums of an infinite triangle with alternate columns of (1, 2, 3, ...) and (1, 0, 0, 0, ...). - Gary W. Adamson, May 14 2010

a(n) is also the number of inequivalent (modulo C_4 rotations) square n X n grids with squares coming in two colors and one square has one of the colors. See the formula from A054772. - Wolfdieter Lang, Oct 03 2016

Also the genus of the complete bipartite graph K_{n+2,n+2}. - Eric W. Weisstein, Jan 19 2018

Examples

			From _Gary W. Adamson_, May 14 2010: (Start)
First few rows of the generating triangle =
1;
2, 1;
3, 0, 1;
4, 0, 2, 1;
5, 0, 3, 0, 1;
6, 0, 4, 0, 2, 1;
7, 0, 5, 0, 3, 0, 1;
8, 0, 6, 0, 4, 0, 2, 1;
...
Example: a(7) = 13 = (6 + 0 + 4 + 0 + 2 + 1). (End)
x + x^2 + 3*x^3 + 4*x^4 + 7*x^5 + 9*x^6 + 13*x^7 + 16*x^8 + 21*x^9 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• A. R. Calderbank and N. J. A. Sloane, Double circulant codes over Z_4, J. Algeb. Combin., 6 (1997) 119-131 (Abstract, pdf, ps).
• G. Nebe, E. M. Rains and N. J. A. Sloane, Self-Dual Codes and Invariant Theory, Springer, Berlin, 2006.
• J. E. Strapasson, S. I. R. Costa, and M. M. S. Alves, On Genus of Circulant Graphs, arXiv:1004.0244 [math.GN], 2010-2016. - _Jonathan Vos Post_, Apr 05 2010
• Eric Weisstein's World of Mathematics, Complete Bipartite Graph
• Eric Weisstein's World of Mathematics, Graph Genus
• Index entries for Molien series
• Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Crossrefs

First differences give A028242. Cf. A035104, A035106.

A002061(n)=a(2*n-1). A035104(n)=a(n+7)-12. A035106(n)=a(n+3)-1.

Cf. A135840, A000290.

Column 1 of A195040. - Omar E. Pol, Sep 28 2011

Cf. A054772, column 2.

Programs

Haskell

a004652 = ceiling . (/ 4) . fromIntegral . (^ 2)
a004652_list = 0 : 1 : zipWith (+) a004652_list [1..]
-- Reinhard Zumkeller, Dec 18 2013

Magma

[Ceiling(n^2/4): n in [0..60] ]; // Vincenzo Librandi, Aug 19 2011

Maple

with(combstruct):ZL:=[st,{st=Prod(left,right),left=Set(U,card=r),right=Set(U,card=2)}, unlabeled]: subs(r=1,stack): seq(count(subs(r=2,ZL),size=m+3),m=0..57) ; # Zerinvary Lajos, Mar 09 2007

Mathematica

CoefficientList[Series[x (1 - x + x^2)/((1 - x)^2*(1 - x^2)), {x, 0, 57}], x] (* Michael De Vlieger, Oct 03 2016 *)
Table[Ceiling[n^2/4], {n, 0, 20}] (* Eric W. Weisstein, Jan 19 2018 *)
Ceiling[Range[0, 20]^2/4] (* Eric W. Weisstein, Jan 19 2018 *)
LinearRecurrence[{2, 0, -2, 1}, {1, 1, 3, 4}, {0, 20}] (* Eric W. Weisstein, Jan 19 2018 *)

PARI

{a(n) = ceil(n^2 / 4)}

Formula

a(n) = ceiling(n^2/4).

a(-n) = a(n).

G.f.: x * (1 - x + x^2) / ((1 - x)^2 * (1 - x^2)).

a(n) = a(n-1) + a(n-2) - a(n-3) + 1. a(2*n) = n^2, a(2*n-1) = n^2 - n + 1. - Michael Somos, Apr 21 2000

Interleaves square numbers with centered polygonal numbers: a(2*n)=A000290(n), a(2*n+1)=A002061(n+1). - Paul Barry, Mar 13 2003

For n > 1: a(n) is the digit reversal of n in base A008619(n), where a(n) is written in base 10. - Naohiro Nomoto, Mar 15 2004

a(n) = a(n-2) + n - 1. - Paul Barry, Jul 14 2004

Euler transform of length 6 sequence [ 1, 2, 1, 0, 0, -1]. - Michael Somos, Apr 03 2007

Starting (1, 3, 4, 7, 9, 13, ...), row sums of triangle A135840. - Gary W. Adamson, Dec 01 2007

a(n) = (3/8)*(-1)^(n+1) + 5/8 - (3/4)*(n+1) + (1/4)*(n+2)*(n+1). - Richard Choulet, Nov 27 2008

a(n) = n^2/4 - 3*((-1)^n-1)/8. - Omar E. Pol, Sep 28 2011

a(n) = -n + floor( (n+1)(n+3)/4 ). - Wesley Ivan Hurt, Jun 23 2013

a(n) = A054772(n, 1) = A054772(n, n^2-1), n >= 1. - Wolfdieter Lang, Oct 03 2016

E.g.f.: (x*(x + 1)*exp(x) + 3*sinh(x))/4. - Ilya Gutkovskiy, Oct 03 2016

a(n) = binomial(floor((n+3)/2),2) + binomial(floor((n+(-1)^n)/2),2). - Yuchun Ji, Feb 03 2021