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A056220 a(n) = 2*n^2 - 1. 88
-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 (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}. - 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
LINKS
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.
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
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
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 *)
PROG
(PARI) a(n)=2*n^2-1;
(Magma) [2*n^2-1 : n in [0..50]]; // Vincenzo Librandi, May 30 2014
(GAP) List([0..50], n-> 2*n^2-1); # Muniru A Asiru, Jul 24 2018
(Sage) [2*n^2-1 for n in (0..50)] # G. C. Greubel, Jul 07 2019
CROSSREFS
Cf. A066049 (indices of prime terms)
Column 2 of array A188644 (starting at offset 1).
Sequence in context: A285738 A120092 A130284 * A024840 A024835 A225251
KEYWORD
sign,easy
AUTHOR
N. J. A. Sloane, Aug 06 2000
STATUS
approved

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Last modified April 24 11:49 EDT 2024. Contains 371936 sequences. (Running on oeis4.)