A003230 Expansion of 1/((1-x)*(1-2*x)*(1-x-2*x^3)).

1, 4, 11, 28, 67, 152, 335, 724, 1539, 3232, 6727, 13900, 28555, 58392, 118959, 241604, 489459, 989520, 1997015, 4024508, 8100699, 16289032, 32726655, 65705268, 131837763, 264399936, 530028199, 1062139180, 2127809963

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

The number of simple squares in the (n+4)-th iteration of the Harter-Heighway dragon (see Wikipedia reference below). - Roland Kneer, Jul 01 2013

The number of double points of the (n+4)-th iteration of the Harter-Heighway dragon. - Manfred Lindemann, Nov 11 2015

References

• D. E. Daykin and S. J. Tucker, Introduction to Dragon Curves. Unpublished, 1976. See links in A003229 for an earlier version.
• 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
• D. E. Daykin, Letter to N. J. A. Sloane, Dec 1973
• D. E. Daykin, Letter to N. J. A. Sloane, Mar 1974
• 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
• Wikipedia, Dragon curve: Harter-Heighway dragon
• Index entries for linear recurrences with constant coefficients, signature (4,-5,4,-6,4).

Crossrefs

Cf. A003229, A077949, A003477.

Programs

Maple

A003230:=-1/(z-1)/(2*z-1)/(-1+z+2*z**3); # Simon Plouffe in his 1992 dissertation
S:=series(1/((1-x)*(1-2*x)*(1-x-2*x^3)),x,101): a:=n->coeff(S,x,n):
seq(a(n),n=0..100); # Manfred Lindemann, Nov 13 2015

Mathematica

CoefficientList[Series[1/((1-x)*(1-2x)*(1-x-2x^3)),{x,0,40}],x] (* Vincenzo Librandi, Jun 11 2012 *)

PARI

Vec(1/((1-x)*(1-2*x)*(1-x-2*x^3))+O(x^66)) \\ Joerg Arndt, Jun 29 2013

Formula

a(n+3) = a(n+2) + 2*a(n) + 2^(n+4) - 1, with a(-3)=a(-2)=a(-1)=0. - Manfred Lindemann, Nov 11 2015

a(n+2) - a(n+1) = A003477(n+2) + A003477(n). - Manfred Lindemann, Dec 08 2015

a(n) = q(n) + q(n-1) + 2*Sum_{i=0..n-2}(q(i)), where q(i)=A003477 and q(-1)=0. - Manfred Lindemann, Dec 08 2015

From Manfred Lindemann, Nov 11 2015: (Start)

With thrt:=(54+6*sqrt(87))^(1/3), ROR:=(thrt/6-1/thrt) and RORext:=(thrt/6+1/thrt) becomes ROC:=(1/2)*(i*sqrt(3)*RORext-ROR), where i^2=-1.

Now ROR, ROC and conjugate(ROC) are the zeros of 1-x-2*x^3.

With AR:=(2*ROR^2+ROR+2)/(2*ROR-3), AC:=(2*ROC^2+ROC+2)/(2*ROC-3) and the zeros of (1-2*x) and (1-x)

a(n) = (1/2)*(AR*ROR^-(n+4)+AC*ROC^-(n+4)+conjugate(AC*ROC^-(n+4))+1*(1/2)^-(n+4)+1*1^-(n+4)).

Simplified: a(n) = (1/2)*(AR*ROR^-(n+4)+2*Re(AC*ROC^-(n+4))+2^(n+4)+1).

(End)

Extensions

More terms from James Sellers, Aug 21 2000

Maple program corrected by Robert Israel, Nov 11 2015