A321257 Start with an equilateral triangle, and repeatedly append along the triangles of the previous step equilateral triangles with half their side length that do not overlap with any prior triangle; a(n) gives the number of triangles appended at n-th step.

1, 6, 21, 60, 147, 330, 705, 1464, 2991, 6054, 12189, 24468, 49035, 98178, 196473, 393072, 786279, 1572702, 3145557, 6291276, 12582723, 25165626, 50331441, 100663080, 201326367, 402652950, 805306125, 1610612484, 3221225211, 6442450674, 12884901609, 25769803488

Offset

1,2

Comments

The following diagram depicts the first three steps of the construction:

- - - -

/ \ / \ / \ / \

/ 3 \ / 3 \ / 3 \ / 3 \

-------------------------

/ \ / \ / \

/ 3 \ 2 / \ 2 / 3 \

------- / \ -------

/ \ / \ / \ / \ / \

/ 3 \ / 3 \ / \ / 3 \ / 3 \

------------- -------------

/ \ / \ / \

/ 3 \ 2 / 1 \ 2 / 3 \

------- / \ -------

/ \ / \ / \

/ 3 \ / \ / 3 \

-------------------------------------

/ \ / \ / \

/ 3 \ 2 / 3 \ 2 / 3 \

------- ------- -------

/ \ / \ / \ / \

/ 3 \ / 3 \ / 3 \ / 3 \

-------------------------

A triangle of step n+1 touches one or two triangles of step n.

The construction presents holes from the 3rd step onwards; these will be gradually filled in the subsequent steps.

The limiting construction is a hexagon; its area is 6 times the area of the initial triangle.

See A321237 for a similar sequence.

Links

Table of n, a(n) for n=1..32.

Rémy Sigrist, Illustration of the construction after 7 steps

Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Formula

a(n) = 3*(2^(n-1) + 3*(2^(n-1)-n)) for any n > 1.

Sum_{n > 0} a(n) / 4^(n-1) = 6.

G.f.: x*(1 + 2*x + 2*x^2 + 4*x^3)/((1-2*x)*(1-x)^2). - Vincenzo Librandi, Nov 02 2018

a(n) - 4*a(n-1) + 5*a(n-2) - 2*a(n-3) = 0, with n>1. - Vincenzo Librandi, Nov 02 2018

Maple

1, seq(3*(2^(n-1)+3*(2^(n-1)-n)), n=2..35); # Muniru A Asiru, Nov 02 2018

Mathematica

CoefficientList[Series[(1 + 2 x + 2 x^2 + 4 x^3) / ((1 - 2 x) (1 - x)^2), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 02 2018 *)

Prog

(PARI) a(n) = if (n==1, 1, 3*(2^(n-1) + 3*(2^(n-1)-n)))
(Magma) [1] cat [3*(2^(n-1) + 3*(2^(n-1)-n)): n in [2..35]]; // Vincenzo Librandi, Nov 02 2018
(GAP) Concatenation([1], List([2..35], n->3*(2^(n-1)+3*(2^(n-1)-n)))); # Muniru A Asiru, Nov 02 2018

Crossrefs

Cf. A321237.

Sequence in context: A341221 A143115 A258142 * A305120 A066524 A113070

Adjacent sequences: A321254 A321255 A321256 * A321258 A321259 A321260

Keyword

nonn

Author

Rémy Sigrist, Nov 01 2018

Status

approved