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.

A004396 One even number followed by two odd numbers.

Original entry on oeis.org

0, 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 13, 13, 14, 15, 15, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 25, 25, 26, 27, 27, 28, 29, 29, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 43, 43, 44, 45, 45, 46, 47, 47
Offset: 0

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Keywords

Comments

Maximal number of points on a triangular grid of edge length n-1 with no 2 points on same row, column, or diagonal. See Problem 252 in The Inquisitive Problem Solver. - R. K. Guy [Comment revised by N. J. A. Sloane, Jul 01 2016]
See also Problem C2 of 2009 International Mathematical Olympiad. - Ruediger Jehn, Oct 19 2021
Dimension of the space of weight 2n+4 cusp forms for Gamma_0(3).
Starting at 3, 3, ..., gives maximal number of acute angles in an n-gon. - Takenov Nurdin (takenov_vert(AT)e-mail.ru), Mar 04 2003
Let b(1) = b(2) = 1, b(k) = b(k-1)+( b(k-2) reduced (mod 2)); then a(n) = b(n-1). - Benoit Cloitre, Aug 14 2002
(1+x+x^2+x^3 ) / ( (1-x^2)*(1-x^3)) is the Poincaré series [or Poincare series] (or Molien series) for Sigma_4.
For n > 6, maximum number of knight moves to reach any square from the corner of an (n-2) X (n-2) chessboard. Likewise for n > 6, the maximum number of knight moves to reach any square from the middle of an (2n-5) X (2n-5) chessboard. - Ralf Stephan, Sep 15 2004
A transform of the Jacobsthal numbers A001045 under the mapping of g.f.s g(x)->g(x/(1+x^2)). - Paul Barry, Jan 16 2005
For n >= 1; a(n) = number of successive terms of A040001 that add to n; or length of n-th term of A028359. - Jaroslav Krizek, Mar 28 2010
For n > 0: a(n) = length of n-th row in A082870. - Reinhard Zumkeller, Apr 13 2014
Also the independence number of the n-triangular honeycomb queen graph. - Eric W. Weisstein, Jul 14 2017
In a game of basketball points can be accumulated by making field goals (two or three points) or free throws (one point). a(n) is the number of different ways to score n-1 points. For example, a score of 4 can be achieved in 3 different ways, with 2 shots (3+1 or 2+2), 3 shots (2+1+1) or 4 shots (1+1+1+1), so a(5) = 3. - Ivan N. Ianakiev, Mar 31 2025

Examples

			G.f. = x + x^2 + 2*x^3 + 3*x^4 + 3*x^5 + 4*x^6 + 5*x^7 + 5*x^8 + 6*x^9 + 7*x^10 + ...
		

References

  • J. Kurschak, Hungarian Mathematical Olympiads, 1976, Mir, Moscow.
  • Paul Vanderlind, Richard K. Guy, and Loren C. Larson, The Inquisitive Problem Solver, MAA, 2002. See Problem 252.

Crossrefs

Programs

  • Haskell
    a004396 n = a004396_list !! n
    a004396_list = 0 : 1 : 1 : map (+ 2) a004396_list
    -- Reinhard Zumkeller, Nov 06 2012
    
  • Magma
    [(Floor(n/3) + Ceiling(n/3)): n in [0..70]]; // Vincenzo Librandi, Aug 07 2011
    
  • Maple
    A004396:=n->floor((2*n + 1)/3); seq(A004396(n), n=0..100); # Wesley Ivan Hurt, Nov 30 2013
  • Mathematica
    Table[Floor[(2 n + 1)/3], {n, 0, 75}]
    With[{n = 50}, Riffle[Range[0, n], Range[1, n, 2], {3, -1, 3}]] (* Harvey P. Dale, May 14 2015 *)
    CoefficientList[Series[(x + x^3)/((1 - x) (1 - x^3)), {x, 0, 71}], x] (* Michael De Vlieger, Oct 27 2016 *)
    a[ n_] := Quotient[2 n + 1, 3]; (* Michael Somos, Oct 23 2017 *)
    a[ n_] := Sign[n] SeriesCoefficient[ (x + x^3) / ((1 - x) (1 - x^3)), {x, 0, Abs@n}]; (* Michael Somos, Oct 23 2017 *)
    LinearRecurrence[{1, 0, 1, -1}, {1, 1, 2, 3}, {0, 20}] (* Eric W. Weisstein, Jul 14 2017 *)
    f[-1]=0; f[n_]:=Length[Union[Plus@@@FrobeniusSolve[{1,2,3},n]]]; f/@Range[-1,100] (* Ivan N. Ianakiev, Mar 31 2025 *)
  • PARI
    a(n)=2*n\/3 \\ Charles R Greathouse IV, Apr 17 2012
    
  • Sage
    def a(n) : return( dimension_cusp_forms( Gamma0(3), 2*n+4) ); # Michael Somos, Jul 03 2014

Formula

G.f.: (x+x^3)/((1-x)*(1-x^3)).
a(n) = floor( (2*n + 1)/3 ).
a(n) = a(n-1) + (1/2)*((-1)^floor((4*n+2)/3) + 1), a(0) = 0. - Mario Catalani (mario.catalani(AT)unito.it), Oct 20 2003
a(n) = 2n/3 - cos(2*Pi*n/3 + Pi/3)/3 + sqrt(3)*sin(2*Pi*n/3 + Pi/3)/9. - Paul Barry, Mar 18 2004
a(n) = A096777(n+1) - A096777(n) for n > 0. - Reinhard Zumkeller, Jul 09 2004
From Paul Barry, Jan 16 2005: (Start)
G.f.: x*(1+x^2)/(1-x-x^3+x^4).
a(n) = a(n-1) + a(n-3) - a(n-4) for n>3.
a(n) = Sum_{k = 0..n} binomial(n-k-1, k)*(-1)^k*A001045(n-2k). (End)
a(n) = (A006369(n) - (A006369(n) mod 2) * (-1)^(n mod 3)) / (1 + A006369(n) mod 2). - Reinhard Zumkeller, Jan 23 2005
a(n) = A004773(n) - A004523(n). - Reinhard Zumkeller, Aug 29 2005
a(n) = floor(n/3) + ceiling(n/3). - Jonathan Vos Post, Mar 19 2006
a(n+1) = A008620(2n). - Philippe Deléham, Dec 14 2006
a(A032766(n)) = n. - Reinhard Zumkeller, Oct 30 2009
a(n) = floor((2*n^2+4*n+2)/(3*n+4)). - Gary Detlefs, Jul 13 2010
Euler transform of length 4 sequence [1, 1, 1, -1]. - Michael Somos, Jul 03 2014
a(n) = n - floor((n+1)/3). - Wesley Ivan Hurt, Sep 17 2015
a(n) = A092200(n) - floor((n+5)/3). - Filip Zaludek, Oct 27 2016
a(n) = -a(-n) for all n in Z. - Michael Somos, Oct 30 2016
E.g.f.: (2/9)*(3*exp(x)*x + sqrt(3)*exp(-x/2)*sin(sqrt(3)*x/2)). - Stefano Spezia, Sep 20 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/2. - Amiram Eldar, Sep 29 2022