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 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_{n-1}. 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, 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 (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, 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_{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)). - 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, 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. 321-338. 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), 59-71. Index entries for linear recurrences with constant coefficients, signature (3,-3,1). FORMULA G.f.: (-1 + 4*x + x^2)/(1-x)^3. 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) + 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 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. 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 EXTENSIONS Formula and additional comments from Henry Bottomley, Dec 12 2000 STATUS approved

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Last modified October 22 22:34 EDT 2019. Contains 328335 sequences. (Running on oeis4.)