A003477 Expansion of 1/((1-2x)(1+x^2)(1-x-2x^3)).

1, 3, 6, 14, 33, 71, 150, 318, 665, 1375, 2830, 5798, 11825, 24039, 48742, 98606, 199113, 401455, 808382, 1626038, 3267809, 6562295, 13169814, 26416318, 52962681, 106145855, 212665582, 425965126, 853005201, 1707833095, 3418756806

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

The number of simple squares in the biggest 'cloud' of the Harter-Heighway dragon of degree (n+4). Equals the number of double points in the biggest 'cloud' of the very same. - Manfred Lindemann, Dec 06 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 Unpublished, 1976.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, 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
• Kevin Ryde, Iterations of the Dragon Curve, see index "BlobA".
• Index entries for linear recurrences with constant coefficients, signature (3,-3,5,-6,2,-4).

Crossrefs

Cf. A003230, A077949. - Manfred Lindemann, Dec 06 2015

Cf. A077854.

Programs

Maple

A003477:=1/(2*z-1)/(-1+z+2*z**3)/(1+z**2); # Simon Plouffe in his 1992 dissertation
S:=series(1/((1-x-2*x^3)*(1-2*x)*(1+x^2)), x, 101): a:=n->coeff(S, x, n):
seq(a(n), n=0..100); # Manfred Lindemann, Dec 06 2015
a:= gfun:-rectoproc({a(n) = 3*a(n-1)-3*a(n-2)+5*a(n-3)-6*a(n-4)+2*a(n-5)-4*a(n-6),seq(a(i)=[1,3,6,14,33,71][i+1],i=0..5)},a(n),remember):
seq(a(n),n=0..100); # Robert Israel, Dec 14 2015

Mathematica

CoefficientList[Series[1/((1-2x)(1+x^2)(1-x-2x^3)),{x,0,40}],x] (* Vincenzo Librandi, Jun 11 2012 *)
LinearRecurrence[{3, -3, 5, -6, 2, -4}, {1, 3, 6, 14, 33, 71}, 31] (* Arie Bos, Dec 03 2019 *)

PARI

Vec(1/((1-2*x)*(1+x^2)*(1-x-2*x^3))+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

Formula

a(0) = 1; for n > 0, a(n) = 3*a(n-1) - 3*a(n-2) + 5*a(n-3) - 6*a(n-4) + 2*a(n-5) - 4*a(n-6) (where a(n)=0 for -5 <= n <= -1). - Jon E. Schoenfield, Apr 23 2010

From Manfred Lindemann, Dec 06 2015: (Start)

a(n) = 3*a(n-1) - 2*a(n-2) + 2*a(n-3) - 4*a(n-4) + Re(i^(n-4)), a(-5)=a(-4)=a(-3)=a(-2)=0 for all integers n element Z.

a(n+2)+a(n) = A003230(n+2)-A003230(n+1). [Daykin and Tucker equation (5)]

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 BR:=1/(2*ROR-3), BC:=1/(2*ROC-3) and the zeros of (1-2*x) and (1+x^2) becomes

a(n) = (1/2)*(BR*ROR^-(n+4) + BC*ROC^-(n+4) + conjugate(BC*ROC^-(n+4)) + (2/5)*(1/2)^-(n+4) + (3/10 + i*(1/10))*i^-(n+4) + conjugate((3/10 + i*(1/10))*i^-(n+4))).

Simplified: a(n) = (BR/2)*ROR^-(n+4) + Re(BC*ROC^-(n+4)) + (1/5)*(1/2)^-(n+4) + Re((3/10 + i*(1/10))*i^-(n+4)).

(End)

Conjecture: a(n) = A077854(n) + 2*(a(n-3) + a(n-4) + ... + a(1)). - Arie Bos, Nov 29 2019

Extensions

More terms from Jon E. Schoenfield, Apr 23 2010