A181890 a(n) = 8*n^2 + 14*n + 5.

5, 27, 65, 119, 189, 275, 377, 495, 629, 779, 945, 1127, 1325, 1539, 1769, 2015, 2277, 2555, 2849, 3159, 3485, 3827, 4185, 4559, 4949, 5355, 5777, 6215, 6669, 7139, 7625, 8127, 8645, 9179, 9729, 10295, 10877

Offset

0,1

Comments

A160050(4*n+1) = A033954(n); A160050(4*n+2) = A001107(n); the third quadrisection is a(n).

First 16 terms of clockwise spiral for odd numbers are as follows:

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13--15--17--19

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11 1---3 21

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9---7---5 23

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31--29--27--25

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a(n) comes from the third vertical.

Sequence found by reading the line from 5, in the direction 5, 27, in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Dec 25 2011

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

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

Formula

a(n) = A160050(4*n+3).

a(n) = (2*n+1)*(4*n+5).

a(n) = a(n-1) + 16*n + 6;

a(n) = 2*a(n-1) - a(n-2) + 16.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).

G.f.: (5 + 12*x - x^2)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 25 2011

a(n) = A014635(n+1) - 1. - Omar E. Pol, Dec 25 2011

From Vaclav Kotesovec, Aug 18 2018: (Start)

Sum_{n>=0} 1/a(n) = 2/3 - Pi/12 - log(2)/6 = 0.289342748774193011891907697817...

Sum_{n>=0} (-1)^n / a(n) = (1 + sqrt(2))*Pi/12 - 2/3 - sqrt(2)*log(tan(Pi/8))/6 = 0.173114712692423461587883724528539... (End)

a(n) = A014106(2*n+1). - Rick L. Shepherd, Aug 06 2019

Mathematica

Table[(2n+1)(4n+5), {n, 0, 40}]  (* Harvey P. Dale, Feb 06 2011 *)
CoefficientList[Series[(5 + 12 x - x^2)/(1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, May 23 2014 *)

Prog

(Magma) [8*n^2+14*n+5: n in [0..700]]; // Vincenzo Librandi, Feb 01 2011
(PARI) a(n)=8*n^2+14*n+5 \\ Charles R Greathouse IV, Dec 21 2011

Crossrefs

Cf. A000217, A014106, A014635, A064038, A160050.

Sequence in context: A299578 A308829 A136917 * A048712 A121876 A137116

Adjacent sequences: A181887 A181888 A181889 * A181891 A181892 A181893

Keyword

nonn,easy

Author

Paul Curtz, Feb 01 2011

Status

approved