cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A035608 Expansion of g.f. x*(1 + 3*x)/((1 + x)*(1 - x)^3).

Original entry on oeis.org

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

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Author

Keywords

Comments

Maximum value of Voronoi's principal quadratic form of the first type when variables restricted to {-1,0,1}. - Michael Somos, Mar 10 2004
This is the main row of a version of the "square spiral" when read alternatively from left to right (see link). See also A001107, A007742, A033954, A033991. It is easy to see that the only prime in the sequence is 5. - Emilio Apricena (emilioapricena(AT)yahoo.it), Feb 08 2009
From Mitch Phillipson, Manda Riehl, Tristan Williams, Mar 06 2009: (Start)
a(n) gives the number of elements of S_2 \wr C_k that avoid the pattern 12, using the following ordering:
In S_j, a permutation p avoids a pattern q if it has no subsequence that is order-isomorphic to q. For example, p avoids the pattern 132 if it has no subsequence abc with a < c < b. We extend this notion to S_j \wr C_n as follows. Element psi =[ alpha_1^beta_1, ... alpha_j^beta_j ] avoids tau = [ a_1 ... a_m ] (tau in S_m) if psi' = [ alpha_1*beta_1 ... alpha_j*beta_j ] avoids tau in the usual sense. For n=2, there are 5 elements of S_2 \wr C_2 that avoid the pattern 12. They are: [ 2^1,1^1 ], [ 2^2,1^1 ], [ 2^2,1^2 ], [ 2^1,1^2 ], [ 1^2,2^1 ].
For example, if psi = [2^1,1^2], then psi'=[2,2] which avoids tau=[1,2] because no subsequence ab of psi' has a < b. (End)

References

  • J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 115.

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.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

  • Magma
    [n^2 + n - 1 - Floor((n-1)/2): n in [0..25]]; // G. C. Greubel, Oct 29 2017
  • Maple
    A035608:=n->floor((n + 1/4)^2): seq(A035608(n), n=0..100); # Wesley Ivan Hurt, Oct 29 2017
  • 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 *)
  • 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) = A002378(n) - A004526(n+1). - Reinhard Zumkeller, Jan 27 2010
a(n) = 2*A006578(n) - A002378(n)/2 = A139592(n)/2. - Reinhard Zumkeller, Feb 07 2010
a(n) = A002265(n+2) + A173562(n). - Reinhard Zumkeller, Feb 21 2010
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
a(n) = 3*A002620(n) + A002620(n+1). - R. J. Mathar, Jul 18 2015
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