

A056220


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


71



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_{n1}.  Henry Bottomley, Dec 12 2000
Also surround numbers of an n X n square.  Jason Earls, Apr 16 2001
Also numbers n such that 2 * n + 2 is a perfect square.  Cino Hilliard, Dec 18 2003, JuriStepan Gerasimov, Apr 09 2016
The sums of the consecutive integer sequences 2n^2 to 2(n+1)^21 are cubes, as 2n^2 + ... + 2(n+1)^21 = (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
Sqrt(a(n) + a(n+1) + 1) = 2n+1.  Doug Bell, 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, 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,i1]=1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = coeff(charpoly(A,x),x^(n2)).  Milan Janjic, Jan 26 2010
For each n > 0, the recursive series, formula S(b) = 6*S(b1)  S(b2)  2*a(n) with S(0) = 4n^24n+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*n1,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_{k=0..2} ( (C(n+k,3)(C(n+k1,3))*(C(n+k,3)+ C(n+k+1,3)) )  (C(n+2,3)C(n1,3))*(C(n,3)+C(n+3,3)).  J. M. Bergot, Jun 16 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


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.
M. Janjic, Hessenberg Matrices and Integer Sequences, J. Int. Seq. 13 (2010) # 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 (2010), No. 2, pp. 321338.
M. Ripà, The rectangular spiral or the n1 X n2 X ... X nk Points Problem , Notes on Number Theory and Discrete Mathematics, 2014, 20(1), 5971.
Index entries for linear recurrences with constant coefficients, signature (3,3,1).


FORMULA

G.f.: (1 + 4*x + x^2)/(1x)^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(n1)),
a(n+1) = sqrt(A001844(n)^2 + A069074(n1)) = sqrt(a(n)^2 + A069074(n1)*2). (End)
a(n) = a(n1) + 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) = 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*cosec(Pi/sqrt(2))/4  1/2. (End)


EXAMPLE

a(0) = 0^21*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^21; 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)/(1x)^3, {x, 0, 50}], x] (* or *)
LinearRecurrence[{3, 3, 1}, {1, 1, 7}, 51] (* Robert G. Wilson v, Jul 24 2018 *)


PROG

(PARI) a(n)=2*n^21;
(MAGMA) [2*n^21 : n in [0..50]]; // Vincenzo Librandi, May 30 2014
(GAP) List([0..50], n> 2*n^21); # Muniru A Asiru, Jul 24 2018
(Sage) [2*n^21 for n in (0..50)] # G. C. Greubel, Jul 07 2019


CROSSREFS

Cf. A047875, A000105, A077585, A005563, A046092, A001082, A002378, A036666, A062717, A028347, A087475, A000217, A143593, A001653, A000384, A225227.
Cf. A066049 (indices of prime terms)
Column 2 of array A188644 (starting at offset 1).
Sequence in context: A285738 A120092 A130284 * A024840 A024835 A225251
Adjacent sequences: A056217 A056218 A056219 * A056221 A056222 A056223


KEYWORD

sign,easy


AUTHOR

N. J. A. Sloane, Aug 06 2000


STATUS

approved



