A059993 Pinwheel numbers: a(n) = 2*n^2 + 6*n + 1.

1, 9, 21, 37, 57, 81, 109, 141, 177, 217, 261, 309, 361, 417, 477, 541, 609, 681, 757, 837, 921, 1009, 1101, 1197, 1297, 1401, 1509, 1621, 1737, 1857, 1981, 2109, 2241, 2377, 2517, 2661, 2809, 2961, 3117, 3277, 3441, 3609, 3781, 3957, 4137, 4321, 4509, 4701, 4897

Offset: 0

Author

Naohiro Nomoto, Mar 14 2001

Keywords

• nonn
• easy

Comments

Nonnegative integers m such that 2*m + 7 is a square. - Vincenzo Librandi, Mar 01 2013

Numbers of the form 4*(h+1)*(2*h-1) + 1, where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ... . - Bruno Berselli, Feb 03 2017

a(n) is also the number of vertices of the Aztec diamond AZ(n) (see Lemma 2.1 of the Imran et al. paper). - Emeric Deutsch, Sep 23 2017

References

• M. Imran and S. Hayat, On computation of topological indices of Aztec diamonds, Sci. Int. (Lahore), Vol. 26(4), 2014, pp. 1407-1412. - Emeric Deutsch, Sep 23 2017

Links

• Harry J. Smith, Table of n, a(n) for n = 0..1000
• Author?, figure. [Wayback Machine link]
• Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. numbers n such that 2*n + 2*k + 1 is a square: A046092 (k=0), A142463 (k=1), A090288 (k=2), this sequence (k=3), A139570 (k=4), A222182 (k=5), A181510 (k=6).

Programs

Magma

[2*n^2+6*n+1: n in [0..50]]; // Vincenzo Librandi, Mar 01 2013

Magma

I:=[1,9]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2)+4: n in [1..50]]; // Vincenzo Librandi, Mar 01 2013

Mathematica

Table[2 n^2 + 6 n + 1, {n, 0, 46}] (* Zerinvary Lajos, Jul 10 2009 *)
LinearRecurrence[{3,-3,1},{1,9,21},50] (* Harvey P. Dale, Oct 01 2018 *)

PARI

a(n) = { 2*n^2 + 6*n + 1 } \\ Harry J. Smith, Jul 01 2009

Formula

a(n) = 4*n + a(n-1) + 4 for n > 0, a(0)=1. - Vincenzo Librandi, Aug 07 2010

G.f.: (1 + 6*x - 3*x^2)/(1-x)^3. - Arkadiusz Wesolowski, Dec 24 2011

a(n) = 2*a(n-1) - a(n-2) + 4. - Vincenzo Librandi, Mar 01 2013

a(n) = Hyper2F1([-2, n], [1], -2). - Peter Luschny, Aug 02 2014

Sum_{n>=0} 1/a(n) = 1/3 + Pi*tan(sqrt(7)*Pi/2)/(2*sqrt(7)). - Amiram Eldar, Dec 13 2022

From Elmo R. Oliveira, Nov 16 2024: (Start)

E.g.f.: exp(x)*(1 + 8*x + 2*x^2).

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