A008747 Expansion of (1+x^4)/((1-x)*(1-x^2)*(1-x^3)).

1, 1, 2, 3, 5, 6, 9, 11, 14, 17, 21, 24, 29, 33, 38, 43, 49, 54, 61, 67, 74, 81, 89, 96, 105, 113, 122, 131, 141, 150, 161, 171, 182, 193, 205, 216, 229, 241, 254, 267, 281, 294, 309, 323, 338, 353, 369, 384, 401, 417, 434, 451, 469, 486, 505, 523, 542, 561

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

For n>=1, the set {A008747(6n+-1)} is the set of numbers of the form a^2 + 5*(a+1)^2 for -inf < a < inf. Furthermore the set A008747(6n) is A033581(n). - Kieren MacMillan, Dec 19 2007

For n>1, a(n-1) is the number of aperiodic necklaces (Lyndon words) with k<=3 black beads and n-k white beads. For n=4 we have for example a(3)=3 aperiodic necklaces: BWWW, BBWW and BBBW. BWBW is periodic and is not counted. - Herbert Kociemba, Oct 23 2016

Examples

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 6*x^5 + 9*x^6 + 11*x^7 + 14*x^8 + ...
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000
• Robert Morris, Minimal percolating sets in bootstrap percolation, arXiv:math/0702370 [math.CO], 2007-2008.
• Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1,-1,1).

Crossrefs

Cf. A008747, A033581, A051274, A061347, A071619, A077043.

Programs

GAP

a:=[1,1,2,3,5,6];; for n in [7..60] do a[n]:=a[n-1]+a[n-2]-a[n-4] -a[n-5]+a[n-6]; od; a; # G. C. Greubel, Aug 03 2019

Magma

R:=PowerSeriesRing(Integers(), 60); Coefficients(R!( (1+x^4)/((1-x)*(1-x^2)*(1-x^3)) )); // G. C. Greubel, Aug 03 2019

Maple

A008747:=n->ceil((n+1)^2/6): seq(A008747(n), n=0..100); # Wesley Ivan Hurt, Oct 25 2016

Mathematica

CoefficientList[Series[(1+x^4)/((1-x)(1-x^2)(1-x^3)),{x,0,60}],x] (* or *) LinearRecurrence[{1,1,0,-1,-1,1},{1,1,2,3,5,6},60] (* Harvey P. Dale, Sep 05 2012 *)

PARI

Vec((1+x^4)/((1-x)*(1-x^2)*(1-x^3))+O(x^60)) \\ Charles R Greathouse IV, Sep 25 2012

Sage

((1+x^4)/((1-x)*(1-x^2)*(1-x^3))).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Aug 03 2019

Formula

G.f.: (1+x^4)/((1-x)*(1-x^2)*(1-x^3)).

a(n) = ceiling((n+1)^2/6).

a(n) = (12*n + 23 + 6*n^2 + 9*(-1)^n + 4*A061347(n))/36. - R. J. Mathar, Mar 15 2011

a(0)=1, a(1)=1, a(2)=2, a(3)=3, a(4)=5, a(5)=6, a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6) for n > 5. - Harvey P. Dale, Sep 05 2012

From Michael Somos, Oct 25 2016: (Start)

Euler transform of length 8 sequence [ 1, 1, 1, 1, 0, 0, 0, -1].

a(n) = a(-2-n) for all n in Z.

a(2*n-1) = A071619(n).

a(3*n-1) = 2*A077043(n).

a(n) - a(n-1) = A051274(n). (End)