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A056220 2*n^2-1. 51
-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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

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}.

Also surround numbers of an n X n square. - Jason Earls (zevi_35711(AT)yahoo.com), Apr 16 2001

Also 8n + 8 is a square. - Cino Hilliard, Dec 18 2003

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

Sequence allows us to find X values of the equation: 2*X^3 + 2*X^2 = Y^2. To find Y values: b(n)=2n(2*n^2 - 1). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), 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

Sqrt(a(n) + a(n+1) + 1) = 2n+1. - Doug Bell (bell.doug(AT)gmail.com), Mar 09 2009

Apart the first term which is -1 the number of units of a(n) belongs to a periodic sequence: 1, 7, 7, 1, 9. We conclude that a(n) and a(n+5) have the same number of units. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 05 2009

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

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

Also 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

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

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

For n>3 a(n)=sum[(C(n+k,3)-(C(n+k-1,3))*(C(n+k,3)+C(n+k+1,3)) {0<=k<=2}]-(C(n+2,3)-C(n-1,3))*(C(n,3)+C(n+3,3)). - J. M. Bergot, Jun 16 2014

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

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 (2010), No. 2, pp. 321-338.

Index to sequences with linear recurrences with constant coefficients, signature (3,-3,1).

FORMULA

G.f.: (-1+4x+x^2)/(1-x)^3.

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

From Doug Bell (bell.doug(AT)gmail.com), 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)+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

EXAMPLE

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

MAPLE

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

MATHEMATICA

Table[2*n^2 + 4*n + 1, {n, -1, 46}] (* Zerinvary Lajos, Jul 10 2009 *)

Table[ChebyshevT[2, n], {n, 0, 60}] (* Vincenzo Librandi, May 30 2014 *)

PROG

(PARI) a(n)=if(n<0, 0, 2*n^2-1)

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

CROSSREFS

Cf. A047875, A000105, A077585, A005563, A046092, A001082, A002378, A036666, A062717, A028347, A087475, A000217.

Cf. A143593. - Gary W. Adamson, Aug 26 2008

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

Cf. A001653.

Sequence in context: A046118 A120092 A130284 * A024840 A024835 A225251

Adjacent sequences:  A056217 A056218 A056219 * A056221 A056222 A056223

KEYWORD

sign,easy

AUTHOR

N. J. A. Sloane, Aug 06 2000

EXTENSIONS

Formula and additional comments from Henry Bottomley, Dec 12 2000

STATUS

approved

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Last modified December 19 08:59 EST 2014. Contains 252186 sequences.