A133044 Area of the spiral of equilateral triangles with side lengths which follow the Padovan sequence, divided by the area of the initial triangle.

1, 2, 3, 7, 11, 20, 36, 61, 110, 191, 335, 591, 1032, 1816, 3185, 5586, 9811, 17207, 30203, 53004, 93004, 163229, 286430, 502655, 882111, 1547967, 2716528, 4767152, 8365761, 14680930, 25763171, 45211271, 79340235, 139232356, 244335860, 428779421, 752455502, 1320467391

Offset: 1

Author

Omar E. Pol, Nov 04 2007

Keywords

• nonn

Comments

First differs from A014529 at a(8).

Links

• G. C. Greubel, Table of n, a(n) for n = 1..2000
• Index entries for linear recurrences with constant coefficients, signature (1,1,1,-1,1,-1).

Crossrefs

Cf. A000931, A014529, A133043.

Programs

Mathematica

RecurrenceTable[{a[n + 6] == a[n + 5] + a[n + 4] + a[n + 3] - a[n + 2] + a[n + 1] - a[n], a[1] == 1, a[2] == 2, a[3] == 3, a[4] == 7, a[5] == 11, a[6] == 20}, a, {n, 1, 2000}] (* G. C. Greubel, Dec 17 2015 *)
Rest@ CoefficientList[Series[x (x^3 + x + 1)/((x^3 - x^2 + 2 x - 1) (x^3 - x - 1)), {x, 0, 38}], x] (* Michael De Vlieger, Feb 21 2018 *)

PARI

Vec((x^3+x+1)/((x^3-x^2+2*x-1)*(x^3-x-1)) + O(x^40)) \\ Andrew Howroyd, Feb 21 2018

Formula

From Colin Barker, Sep 18 2013: (Start)

Conjecture: a(n) = a(n-1) + a(n-2) + a(n-3) - a(n-4) + a(n-5) - a(n-6).

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

(End)

From Félix Breton, Dec 17 2015: (Start)

a(n) = 2*p(n+4)*p(n+5) - p(n+2)^2 where p is the Padovan sequence (A000931). This establishes Colin Barker's conjecture, because

a(n) = a(n-1) + p(n+4)^2

= a(n-1) + (p(n+1) + p(n+2))^2

= a(n-1) + p(n+1)^2 + p(n+2)^2 + 2*p(n+1)*p(n+2) - p(n-1)^2 + p(n-1)^2

= a(n-1) + (a(n-3)-a(n-4)) + (a(n-2)-a(n-3)) + a(n-3) + (a(n-5)-a(n-6))

= a(n-1) + a(n-2) + a(n-3) - a(n-4) + a(n-5) - a(n-6). (End)

Extensions

a(27) and beyond taken from G. C. Greubel's table. - Omar E. Pol, Dec 18 2015

a(589) in b-file corrected by Andrew Howroyd, Feb 21 2018