A064412 At stage 1, start with a unit equilateral equiangular triangle. At each successive stage add 3*(n-1) new triangles around outside with edge-to-edge contacts. Sequence gives number of triangles (regardless of size) at n-th stage.

1, 5, 14, 32, 60, 103, 160, 238, 335, 459, 606, 786, 994, 1241, 1520, 1844, 2205, 2617, 3070, 3580, 4136, 4755, 5424, 6162, 6955, 7823, 8750, 9758, 10830, 11989, 13216, 14536, 15929, 17421, 18990, 20664, 22420, 24287, 26240, 28310, 30471, 32755, 35134, 37642

Offset

1,2

Comments

Number of unit triangles at n-th stage = 3n(n-1)/2 + 1, A005448.

References

Anthony Gardiner, "Mathematical Puzzling," Dover Publications, Inc., Mineola, NY., 1987, page 88.

Links

Table of n, a(n) for n=1..44.

N. J. A. Sloane, Illustration of initial terms

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).

Formula

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

a(2n+1) = (7n^3+12n^2+7n+2)/2; a(2n) = (28n^3+6n^2+4n+1+(-1)^(n+1))/8. - Len Smiley, Oct 07 2001

a(n) = (14*n^3+6*n^2+5*n+7+3*(n-1)*(-1)^n-2*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((6*n-1+(-1)^n)/4)))/32. - Luce ETIENNE, Jun 27 2014

Example

a(4) = 32: 19 triangles of side 1, 10 of side 2 and 3 of side 3.

Maple

A064412:=n->(14*n^3+6*n^2+5*n+7+3*(n-1)*(-1)^n-2*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((6*n-1+(-1)^n)/4)))/32; seq(A064412(n), n=1..30); # Wesley Ivan Hurt, Jun 27 2014

Mathematica

CoefficientList[Series[(1 + x + x^2) (1 + 2 x + x^2 + 3 x^3)/((1 - x)^2 (1 - x^2) (1 - x^4)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Jun 27 2014 *)
LinearRecurrence[{2, 0, -2, 2, -2, 0, 2, -1}, {1, 5, 14, 32, 60, 103, 160, 238}, 50] (* Harvey P. Dale, Apr 12 2016 *)

Prog

(PARI) a(n)=polcoeff(x*(1+x+x^2)*(1+2*x+x^2+3*x^3)/((1-x)^2*(1-x^2)*(1-x^4))+x*O(x^n), n)

Crossrefs

Cf. A056640.

Sequence in context: A070134 A295344 A219902 * A211803 A299275 A266759

Adjacent sequences: A064409 A064410 A064411 * A064413 A064414 A064415

Keyword

nonn,easy

Author

Robert G. Wilson v, Sep 29 2001

Status

approved