A035608 Expansion of g.f. x*(1 + 3*x)/((1 + x)*(1 - x)^3).
0, 1, 5, 10, 18, 27, 39, 52, 68, 85, 105, 126, 150, 175, 203, 232, 264, 297, 333, 370, 410, 451, 495, 540, 588, 637, 689, 742, 798, 855, 915, 976, 1040, 1105, 1173, 1242, 1314, 1387, 1463, 1540, 1620, 1701, 1785, 1870, 1958, 2047, 2139, 2232, 2328, 2425, 2525, 2626
Offset: 0
References
- J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 115.
Links
- William A. Tedeschi, Table of n, a(n) for n = 0..10000
- Emilio Apricena, A version of the Ulam spiral (This is called the square spiral. - _N. J. A. Sloane_, Jul 27 2018)
- Emanuele Munarini, Topological indices for the antiregular graphs, Le Mathematiche, Vol. 76, No. 1 (2021), see p. 283.
- Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Crossrefs
Partial sums of A042948.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Programs
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Magma
[n^2 + n - 1 - Floor((n-1)/2): n in [0..25]]; // G. C. Greubel, Oct 29 2017
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Maple
A035608:=n->floor((n + 1/4)^2): seq(A035608(n), n=0..100); # Wesley Ivan Hurt, Oct 29 2017
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Mathematica
Table[n^2 + Floor[n/2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Apr 12 2011 *) CoefficientList[Series[x (1 + 3 x)/((1 + x) (1 - x)^3), {x, 0, 60}], x] (* or *) LinearRecurrence[{2, 0, -2, 1}, {0, 1, 5, 10}, 60] (* Harvey P. Dale, Feb 21 2013 *)
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PARI
a(n)=n^2+n-1-(n-1)\2
Formula
a(n) = n^2 + n - 1 - floor((n-1)/2).
a(n) = A011848(2*n+1).
a(n) = floor((n + 1/4)^2). - Reinhard Zumkeller, Jan 27 2010
a(n) = (-1)^n*Sum_{i=0..n} (-1)^i*(2*i^2 + 3*i + 1). Omits the leading 0. - William A. Tedeschi, Aug 25 2010
a(n) = n^2 + floor(n/2), from Mathematica section. - Vladimir Joseph Stephan Orlovsky, Apr 12 2011
a(0)=0, a(1)=1, a(2)=5, a(3)=10; for n > 3, a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Harvey P. Dale, Feb 21 2013
For n > 1: a(n) = a(n-2) + 4*n - 3; see also row sums of triangle A253146. - Reinhard Zumkeller, Dec 27 2014
From Amiram Eldar, Mar 20 2022: (Start)
Sum_{n>=1} 1/a(n) = 4 - 2*log(2) - Pi/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/3 - 4*(1-log(2)). (End)
E.g.f.: (x*(2*x + 3)*cosh(x) + (2*x^2 + 3*x - 1)*sinh(x))/2. - Stefano Spezia, Apr 24 2024
Comments