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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.

A006566 Dodecahedral numbers: a(n) = n*(3*n - 1)*(3*n - 2)/2.

Original entry on oeis.org

0, 1, 20, 84, 220, 455, 816, 1330, 2024, 2925, 4060, 5456, 7140, 9139, 11480, 14190, 17296, 20825, 24804, 29260, 34220, 39711, 45760, 52394, 59640, 67525, 76076, 85320, 95284, 105995, 117480, 129766, 142880, 156849, 171700, 187460, 204156
Offset: 0

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Comments

Schlaefli symbol for this polyhedron: {5,3}.
A093485 = first differences; A124388 = second differences; third differences = 27. - Reinhard Zumkeller, Oct 30 2006
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
From Peter Bala, Sep 09 2013: (Start)
a(n) = binomial(3*n,3). Two related sequences are binomial(3*n+1,3) (A228887) and binomial(3*n+2,3) (A228888). The o.g.f.'s for these three sequences are rational functions whose numerator polynomials are obtained from the fourth row [1, 4, 10, 16, 19, 16, 10, 4, 1] of the triangle of trinomial coefficients A027907 by taking every third term:
Sum_{n >= 1} binomial(3*n,3)*x^n = (x + 16*x^2 + 10*x^3)/(1-x)^4;
Sum_{n >= 1} binomial(3*n+1,3)*x^n = (4*x + 19*x^2 + 4*x^3)/(1-x)^4;
Sum_{n >= 1} binomial(3*n+2,3)*x^n = (10*x + 16*x^2 + x^3)/(1-x)^4. (End)

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A000292 (tetrahedral numbers), A000578 (cubes), A005900 (octahedral numbers), A006564 (icosahedral numbers).

Programs

  • Haskell
    a006566 n = n * (3 * n - 1) * (3 * n - 2) `div` 2
    a006566_list = scanl (+) 0 a093485_list  -- Reinhard Zumkeller, Jun 16 2013
    
  • Magma
    [n*(3*n-1)*(3*n-2)/2: n in [0..40]]; // Vincenzo Librandi, Dec 11 2015
  • Maple
    A006566:=(1+16*z+10*z**2)/(z-1)**4; # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    Table[n(3n-1)(3n-2)/2,{n,0,100}] (* Vladimir Joseph Stephan Orlovsky, Apr 13 2011 *)
    LinearRecurrence[{4,-6,4,-1},{0,1,20,84},40] (* Harvey P. Dale, Jul 24 2013 *)
    CoefficientList[Series[x (1 + 16 x + 10 x^2)/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 11 2015 *)
  • PARI
    a(n)=n*(3*n-1)*(3*n-2)/2
    

Formula

G.f.: x(1 + 16x + 10x^2)/(1 - x)^4.
a(n) = A000292(3n-3) = A054776(n)/6 = n*A060544(n).
a(n) = C(n+2,3) + 16 C(n+1,3) + 10 C(n,3).
a(0)=0, a(1)=1, a(2)=20, a(3)=84, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Jul 24 2013
a(n) = binomial(3*n,3). a(-n) = - A228888(n). Sum_{n>=1} 1/a(n) = 1/2*( sqrt(3)*Pi - 3*log(3) ). Sum_{n>=1} (-1)^n/a(n) = 1/3*sqrt(3)*Pi - 4*log(2). - Peter Bala, Sep 09 2013
a(n) = A006564(n) + A035006(n). - Peter M. Chema, May 04 2016
E.g.f.: x*(2 + 18*x + 9*x^2)*exp(x)/2. - Ilya Gutkovskiy, May 04 2016
From Amiram Eldar, Jan 09 2024: (Start)
Sum_{n>=1} 1/a(n) = (sqrt(3)*Pi - 3*log(3))/2 (A295421).
Sum_{n>=1} (-1)^(n+1)/a(n) = (12*log(2) - sqrt(3)*Pi)/3. (End)

Extensions

More terms from Henry Bottomley, Nov 23 2001