The OEIS is supported by the many generous donors to the OEIS Foundation.

 Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 59th year, we have over 358,000 sequences, and we’ve crossed 10,300 citations (which often say “discovered thanks to the OEIS”). Other ways to Give
 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A056220 a(n) = 2*n^2 - 1. 83
 -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}. - 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 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,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 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). 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*cosec(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 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 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified December 8 08:22 EST 2022. Contains 358693 sequences. (Running on oeis4.)