A202804 a(n) = n*(6*n+4).

0, 10, 32, 66, 112, 170, 240, 322, 416, 522, 640, 770, 912, 1066, 1232, 1410, 1600, 1802, 2016, 2242, 2480, 2730, 2992, 3266, 3552, 3850, 4160, 4482, 4816, 5162, 5520, 5890, 6272, 6666, 7072, 7490, 7920, 8362, 8816, 9282, 9760, 10250, 10752, 11266, 11792

Offset

0,2

Comments

Sequence found by reading the line from 0, in the direction 0, 10, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A033579 in the same spiral. - Omar E. Pol, Jul 17 2012

Partial sums give A163815. - Leo Tavares, Feb 25 2022

Links

Jeremy Gardiner, Table of n, a(n) for n = 0..3000

Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.

Leo Tavares, Illustration: Star Halves.

Pavlos Vavolas, Cellular automaton model of cardiac arrhythmias, University of Sheffield Department of Computer Science (2005), (see page 43). [broken link, abstract]

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Formula

a(n) = 2*n(3n+2) = 6n^2 + 4n = 2*A045944(n).

a(n) = A080859(n) - 1. - Omar E. Pol, Jul 18 2012

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Dec 28 2015

G.f.: -2x*(5 + x)/(x - 1)^3. - Indranil Ghosh, Apr 10 2017

a(n) = A003154(n+1) - A005408(n). - Leo Tavares, Feb 25 2022

From Amiram Eldar, Mar 01 2022: (Start)

Sum_{n>=1} 1/a(n) = (Pi/sqrt(3) - 3*log(3) + 3)/8.

Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)) - 3/8. (End)

Maple

A202804:=n->n*(6*n+4): seq(A202804(n), n=0..100); # Wesley Ivan Hurt, Apr 09 2017

Mathematica

Table[n(6n+4), {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 10, 32}, 50] (* Harvey P. Dale, Dec 28 2015 *)

Prog

(PARI) x='x + O('x^50); concat([0], Vec(-2*x*(5 + x)/(x - 1)^3)) \\ Indranil Ghosh, Apr 10 2017

Crossrefs

Cf. A033581, A049453, A033580, A195319.

Cf. A003154, A005408, A163815.

Sequence in context: A063926 A239834 A337148 * A155192 A229720 A350108

Adjacent sequences: A202801 A202802 A202803 * A202805 A202806 A202807

Keyword

nonn,easy

Author

Jeremy Gardiner, Dec 24 2011

Status

approved