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a(n) = floor(n^2/12).
13

%I #63 Dec 07 2023 01:11:49

%S 0,0,0,0,1,2,3,4,5,6,8,10,12,14,16,18,21,24,27,30,33,36,40,44,48,52,

%T 56,60,65,70,75,80,85,90,96,102,108,114,120,126,133,140,147,154,161,

%U 168,176,184,192,200,208,216,225,234,243,252,261,270,280,290,300,310,320,330,341,352

%N a(n) = floor(n^2/12).

%C With a different offset, Molien series for 3-dimensional group [2,n] = *22n.

%H Vincenzo Librandi, <a href="/A008724/b008724.txt">Table of n, a(n) for n = 0..10000</a>

%H P. T. Ho, <a href="http://dx.doi.org/10.1016/j.disc.2005.09.010">The crossing number of K_{4,n} on the real projective plane</a>, Discr. Math., 304 (2005), pp. 23-33.

%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=189">Encyclopedia of Combinatorial Structures 189</a>.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/ToroidalCrossingNumber.html">Toroidal Crossing Number</a>.

%H <a href="/index/Mo#Molien">Index entries for Molien series</a>.

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,0,0,1,-2,1).

%F a(n) = a(n-6) + n - 3. - _Paul Barry_, Jul 14 2004

%F a(n) = Sum_{j=0..n+2} floor(j/6), a(n-2) = (1/2)*floor(n/6)*(2*n - 4 - 6*floor(n/6)). - _Mitch Harris_, Sep 08 2008

%F G.f.: x^4/((1-x)^2*(1-x^6)).

%F Sum_{n>=4} 1/a(n) = Pi^2/18 - Pi/(2*sqrt(3)) + 49/12. - _Amiram Eldar_, Aug 14 2022

%F a(n) = a(-n) = A174709(n+2). - _Michael Somos_, Dec 05 2023

%p A008724 := proc(n)

%p floor(n^2/12) ;

%p end proc:

%p seq(A008724(n),n=0..30) ; # _R. J. Mathar_, Mar 28 2017

%t Floor[Range[0, 70]^2/12] (* _G. C. Greubel_, Sep 09 2019 *)

%o (Magma) a008724:=func< n | Floor(n^2/12) >; [ a008724(n): n in [0..70] ];

%o (PARI) a(n)=n^2\12 \\ _Charles R Greathouse IV_, Jul 02 2013

%o (Sage) [floor(n^2/12) for n in (0..70)] # _G. C. Greubel_, Sep 09 2019

%o (GAP) List([0..70], n-> Int(n^2/12) ); # _G. C. Greubel_, Sep 09 2019

%Y Cf. A001399, A174709.

%K nonn,easy

%O 0,6

%A _N. J. A. Sloane_

%E Minor edits by _Klaus Brockhaus_, Nov 24 2010