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A006566
Dodecahedral numbers: a(n) = n*(3*n - 1)*(3*n - 2)/2.
(Formerly M5089)
36
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, 221815, 240464, 260130, 280840
OFFSET
0,3
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
Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing (2012), page 114.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Tanya Khovanova, Recursive Sequences.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75.
Victor Meally, Letter to N. J. A. Sloane, no date.
Ed Pegg Jr, Dodecahedral 2024.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
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(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(0)=0, a(1)=1, a(2)=20, a(3)=84. - Harvey P. Dale, Jul 24 2013
From Peter Bala, Sep 09 2013: (Start)
a(n) = binomial(3*n,3).
a(-n) = - A228888(n).
Sum_{n>=1} 1/a(n) = (1/2)*( sqrt(3)*Pi - 3*log(3) ) (A295421).
Sum_{n>=1} (-1)^n/a(n) = 1/3*sqrt(3)*Pi - 4*log(2). (End)
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
Product_{n>=2} (1 - 1/a(n)) = (2*sqrt(6)/11)*sinh(sqrt(2)*Pi/3). - Amiram Eldar, Nov 02 2025
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 *)
PROG
(PARI) a(n)=n*(3*n-1)*(3*n-2)/2
(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
CROSSREFS
Subsequence of A000292.
Cf. A000292 (tetrahedral numbers), A000578 (cubes), A005900 (octahedral numbers), A006564 (icosahedral numbers).
Sequence in context: A044207 A044588 A172221 * A205312 A268888 A211158
KEYWORD
nonn,easy,nice,changed
EXTENSIONS
More terms from Henry Bottomley, Nov 23 2001
STATUS
approved