A056220 a(n) = 2*n^2 - 1.

-1, 1, 7, 17, 31, 49, 71, 97, 127, 161, 199, 241, 287, 337, 391, 449, 511, 577, 647, 721, 799, 881, 967, 1057, 1151, 1249, 1351, 1457, 1567, 1681, 1799, 1921, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 3041, 3199, 3361, 3527, 3697, 3871, 4049, 4231, 4417, 4607, 4801

Offset: 0

Author

N. J. A. Sloane, Aug 06 2000

Keywords

• sign
• easy

Comments

Image of squares (A000290) under "little Hankel" transform that sends [c_0, c_1, ...] to [d_0, d_1, ...] where d_n = c_n^2 - c_{n+1}*c_{n-1}. - Henry Bottomley, Dec 12 2000

Surround numbers of an n X n square. - Jason Earls, Apr 16 2001

Numbers n such that 2*n + 2 is a perfect square. - Cino Hilliard, Dec 18 2003, Juri-Stepan Gerasimov, Apr 09 2016

The sums of the consecutive integer sequences 2n^2 to 2(n+1)^2-1 are cubes, as 2n^2 + ... + 2(n+1)^2-1 = (1/2)(2(n+1)^2 - 1 - 2n^2 + 1)(2(n+1)^2 - 1 + 2n^2) = (2n+1)^3. E.g., 2+3+4+5+6+7 = 27 = 3^3, then 8+9+10+...+17 = 125 = 5^3. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 29 2005

X values (other than 0) of solutions to the equation 2*X^3 + 2*X^2 = Y^2. To find Y values: b(n) = 2n*(2*n^2 - 1). - Mohamed Bouhamida, Nov 06 2007

Average of the squares of two consecutive terms is also a square. In fact: (2*n^2 - 1)^2 + (2*(n+1)^2 - 1)^2 = 2*(2*n^2 + 2*n + 1)^2. - Matias Saucedo (solomatias(AT)yahoo.com.ar), Aug 18 2008

Equals row sums of triangle A143593 and binomial transform of [1, 6, 4, 0, 0, 0, ...] with n > 1. - Gary W. Adamson, Aug 26 2008

Start a spiral of square tiles. Trivially the first tile fits in a 1 X 1 square. 7 tiles fit in a 3 X 3 square, 17 tiles fit in a 5 X 5 square and so on. - Juhani Heino, Dec 13 2009

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = coeff(charpoly(A,x),x^(n-2)). - Milan Janjic, Jan 26 2010

For each n > 0, the recursive series, formula S(b) = 6*S(b-1) - S(b-2) - 2*a(n) with S(0) = 4n^2-4n+1 and S(1) = 2n^2, has the property that every even term is a perfect square and every odd term is twice a perfect square. - Kenneth J Ramsey, Jul 18 2010

Fourth diagonal of A154685 for n > 2. - Vincenzo Librandi, Aug 07 2010

First integer of (2*n)^2 consecutive integers, where the last integer is 3 times the first + 1. As example, n = 2: term = 7; (2*n)^2 = 16; 7, 8, 9, ..., 20, 21, 22: 7*3 + 1 = 22. - Denis Borris, Nov 18 2012

Chebyshev polynomial of the first kind T(2,n). - Vincenzo Librandi, May 30 2014

For n > 0, number of possible positions of a 1 X 2 rectangle in a (n+1) X (n+2) rectangular integer lattice. - Andres Cicuttin, Apr 07 2016

This sequence also represents the best solution for Ripà's n_1 X n_2 X n_3 dots problem, for any 0 < n_1 = n_2 < n_3 = floor((3/2)*(n_1 - 1)) + 1. - Marco Ripà, Jul 23 2018

Examples

			a(0) = 0^2-1*1 = -1, a(1) = 1^2 - 4*0 = 1, a(2) = 2^2 - 9*1 = 7, etc.
a(4) = 31 = (1, 3, 3, 1) dot (1, 6, 4, 0) = (1 + 18 + 12 + 0). - _Gary W. Adamson_, Aug 29 2008
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
• Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq., Vol. 13 (2010), Article # 10.7.8.
• Mitch Phillipson, Manda Riehl, and Tristan Williams, Enumeration of Wilf classes in Sn ~ Cr for two patterns of length 3, PU. M. A., Vol. 21, No. 2 (2010), pp. 321-338.
• Marco Ripà, The rectangular spiral or the n1 X n2 X ... X nk Points Problem , Notes on Number Theory and Discrete Mathematics, Vol. 20, No. 1 (2014), pp. 59-71.
• Leo Tavares, Illustration: Twin Squares
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000105, A000217, A000290, A000384, A001082, A001653, A001844, A002378, A002593, A003215, A005563, A028347, A036666, A046092, A047875, A062717, A069074, A077585, A087475, A119258, A143593, A154685, A162610, A188653, A225227.

Cf. A066049 (indices of prime terms)

Column 2 of array A188644 (starting at offset 1).

Programs

GAP

List([0..50], n-> 2*n^2-1); # Muniru A Asiru, Jul 24 2018

Magma

[2*n^2-1 : n in [0..50]]; // Vincenzo Librandi, May 30 2014

Maple

A056220:=n->2*n^2-1; seq(A056220(n), n=0..50); # Wesley Ivan Hurt, Jun 16 2014

Mathematica

Array[2 #^2 - 1 &, 50, 0] (* Robert G. Wilson v, Jul 23 2018 *)
CoefficientList[Series[(x^2 +4x -1)/(1-x)^3, {x, 0, 50}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {-1, 1, 7}, 51] (* Robert G. Wilson v, Jul 24 2018 *)

PARI

a(n)=2*n^2-1;

Sage

[2*n^2-1 for n in (0..50)] # G. C. Greubel, Jul 07 2019

Formula

G.f.: (-1 + 4*x + x^2)/(1-x)^3. - Henry Bottomley, Dec 12 2000

a(n) = A119258(n+1,2) for n > 0. - Reinhard Zumkeller, May 11 2006

From Doug Bell, Mar 08 2009: (Start)

a(0) = -1,

a(n) = sqrt(A001844(n)^2 - A069074(n-1)),

a(n+1) = sqrt(A001844(n)^2 + A069074(n-1)) = sqrt(a(n)^2 + A069074(n-1)*2). (End)

a(n) + a(n+1) + 1 = (2n+1)^2. - Doug Bell, Mar 09 2009

a(n) = a(n-1) + 4*n - 2 (with a(0)=-1). - Vincenzo Librandi, Dec 25 2010

a(n) = A188653(2*n) for n > 0. - Reinhard Zumkeller, Apr 13 2011

a(n) = A162610(2*n-1,n) for n > 0. - Reinhard Zumkeller, Jan 19 2013

a(n) = ( Sum_{k=0..2} (C(n+k,3)-C(n+k-1,3))*(C(n+k,3)+C(n+k+1,3)) ) - (C(n+2,3)-C(n-1,3))*(C(n,3)+C(n+3,3)), for n > 3. - J. M. Bergot, Jun 16 2014

a(n) = j^2 + k^2 - 2 or 2*j*k if n >= 2 and j = n + sqrt(2)/2 and k = n - sqrt(2)/2. - Avi Friedlich, Mar 30 2015

a(n) = A002593(n)/n^2. - Bruce J. Nicholson, Apr 03 2017

a(n) = A000384(n) + n - 1. - Bruce J. Nicholson, Nov 12 2017

a(n)*a(n+k) + 2k^2 = m^2 (a perfect square), m = a(n) + (2n*k), for n>=1. - Ezhilarasu Velayutham, May 13 2019

From Amiram Eldar, Aug 10 2020: (Start)

Sum_{n>=1} 1/a(n) = 1/2 - sqrt(2)*Pi*cot(Pi/sqrt(2))/4.

Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(2)*Pi*csc(Pi/sqrt(2))/4 - 1/2. (End)

From Amiram Eldar, Feb 04 2021: (Start)

Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(2))*csc(Pi/sqrt(2)).

Product_{n>=2} (1 - 1/a(n)) = (Pi/(4*sqrt(2)))*csc(Pi/sqrt(2)). (End)

a(n) = A003215(n) - A000217(n-2)*2. - Leo Tavares, Jun 29 2021

Let T(n) = n*(n+1)/2. Then a(n)^2 = T(2n-2)*T(2n+1) + n^2. - Charlie Marion, Feb 12 2023

E.g.f.: exp(x)*(2*x^2 + 2*x - 1). - Stefano Spezia, Jul 08 2023