A051500 a(n) = (3^n + 1)^2/4.

1, 4, 25, 196, 1681, 14884, 133225, 1196836, 10764961, 96864964, 871725625, 7845353476, 70607649841, 635467254244, 5719200505225, 51472790198116, 463255068736321, 4169295489486724, 37523659017960025, 337712929999378756, 3039416366507624401

Offset

0,2

Comments

For n>0, 4*a(n) has the form 10...020...01 when it is converted to base 3. The number of zeros between 2 consecutive nonzero digits is n-1. - Raul Prisacariu, Aug 27 2024

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (13,-39,27).

Formula

a(n) = A007051(n)^2. - Michel Marcus, Jun 08 2013

From Colin Barker, Feb 10 2016: (Start)

a(n) = 13*a(n-1) - 39*a(n-2) + 27*a(n-3) for n > 2.

G.f.: (1-9*x+12*x^2) / ((1-x)*(1-3*x)*(1-9*x)). (End)

E.g.f.: (1/4)*(exp(x) + 2*exp(3*x) + exp(9*x)). - G. C. Greubel, May 22 2023

Mathematica

(3^Range[0, 20]+1)^2/4 (* or *) LinearRecurrence[{13, -39, 27}, {1, 4, 25}, 30] (* Harvey P. Dale, Nov 05 2016 *)

Prog

(PARI) Vec((1-9*x+12*x^2)/((1-x)*(1-3*x)*(1-9*x)) + O(x^30)) \\ Colin Barker, Feb 10 2016
(Python)
def A051500(n): return (3**n+1)**2>>2 # Chai Wah Wu, Nov 14 2022
(Magma) [(3^n+1)^2/4: n in [0..40]]; // G. C. Greubel, May 22 2023
(SageMath) [((3^n+1)//2)^2 for n in range(41)] # G. C. Greubel, May 22 2023

Crossrefs

Cf. A007051.

Sequence in context: A380134 A322442 A036449 * A206179 A151342 A001246

Adjacent sequences: A051497 A051498 A051499 * A051501 A051502 A051503

Keyword

nonn,easy

Author

Colin Mallows

Status

approved