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a(n) = n*(11*n^2 - 5)/6.
17

%I #35 Sep 08 2022 08:44:33

%S 0,1,13,47,114,225,391,623,932,1329,1825,2431,3158,4017,5019,6175,

%T 7496,8993,10677,12559,14650,16961,19503,22287,25324,28625,32201,

%U 36063,40222,44689,49475,54591,60048

%N a(n) = n*(11*n^2 - 5)/6.

%C 3-dimensional analog of centered polygonal numbers, that is: centered hendecagonal pyramidal numbers (see Deza paper in References).

%D E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

%H Vincenzo Librandi, <a href="/A004467/b004467.txt">Table of n, a(n) for n = 0..5000</a>

%H T. P. Martin, <a href="http://dx.doi.org/10.1016/0370-1573(95)00083-6">Shells of atoms</a>, Phys. Reports, 273 (1996), 199-241, eq. (11).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, -6, 4, -1).

%F G.f.: x*(1+9*x+x^2)/(1-x)^4. - _Colin Barker_, Jan 08 2012

%F a(0)=0, a(1)=1, a(2)=13, a(3)=47; for n>3, a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - _Harvey P. Dale_, Sep 22 2013

%F E.g.f.: (x/6)*(6 + 33*x + 11*x^2)*exp(x). - _G. C. Greubel_, Sep 01 2017

%t Table[n(11n^2-5)/6,{n,0,80}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 18 2011 *)

%t LinearRecurrence[{4,-6,4,-1},{0,1,13,47},80] (* _Harvey P. Dale_, Sep 22 2013 *)

%o (Magma) [n*(11*n^2-5)/6: n in [0..50]]; // _Vincenzo Librandi_, May 15 2011

%o (PARI) a(n)=n*(11*n^2-5)/6 \\ _Charles R Greathouse IV_, Sep 28 2011

%Y 1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.

%K nonn,easy

%O 0,3

%A Albert D. Rich (Albert_Rich(AT)msn.com).